Brilliant Preparatory Section, Sitamarhi XII Maths Advanced Study Material 1. Applications of Matrices and Determinants 2. Vector Algebra 3. Complex Numbers 4. Analytical Geometry 5. Differential Calculus Applications − I 6. Differential Calculus Applications – II 7. Integral Calculus and its applications 8. Differential Equations 9. Discrete Mathematics 10.Probability Distributions
Session : 2015-17 Office: Rajopatti, Dumra Road, Sitamarhi(Bihar), Pin-843301 Ph.06226-252314, Mobile: 9431636758, 9931610902
1. APPLICATIONS OF MATRICES AND DETERMINANTS 1.1. Introduction : The students are already familiar with the basic definitions, the elementary operations and some basic properties of matrices. The concept of division is not defined for matrices. In its place and to serve similar purposes, the notion of the inverse of a matrix is introduced. In this section, we are going to study about the inverse of a matrix. To define the inverse of a matrix, we need the concept of adjoint of a matrix.
1.2 Adjoint : Let A = [aij] be a square matrix of order n. Let Aij be the cofactor of aij. Then the nth order matrix [Aij]T is called the adjoint of A. It is denoted by adjA. Thus the adjA is nothing but the transpose of the cofactor matrix [Aij] of A. Result : If A is a square matrix of order n, then A (adjA) = | A | In = (adj A) A, where In is the identity matrix of order n. Proof : Let us prove this result for a square matrix A of order 3. a11 a12 a13
a31 A 11 Then adj A = A12 A13
a33 A31 A32 A33
Let A = a21 a22 a23 a32 A21 A22 A23
The (i, j)th = ai1 Aj1 + ai2 Aj2 + ai3 Aj3 = ∆ = | A | if i = j element of A (adj A) = 0 if i ≠ j 0 | A | 0 1 0 0
∴ A (adj A) = 0 | A | 0 = | A | 0 1 0 = | A | I3 0 0 0 1 0 | A | Similarly we can prove that (adj A)A = | A | I3
1
∴ A (adj A) = | A | I3 = (adj A) A In general we can prove that A (adj A) = | A | In = (adj A) A.
a b Example 1.1 : Find the adjoint of the matrix A = c d Solution: The cofactor of a is d, the cofactor of b is − c, the cofactor of c is − b and the cofactor of d is a. The matrix formed by the cofactors taken in order is the cofactor matrix of A. d − c ∴ The cofactor matrix of A is = . − b a Taking transpose of the cofactor matrix, we get the adjoint of A. d − b ∴ The adjoint of A = − c a 1 1 1
2
Example 1.2 : Find the adjoint of the matrix A = 1
2 −1
Solution: The cofactors are given by
2 − 3 =3 − 1 3 1 − 3 − = −9 2 3 1 2 =−5 2 − 1 1 1 − =−4 − 1 3 1 1 =1 2 3 1 1 − =3 2 − 1 1 1 =−5 2 − 3
Cofactor of 1 = A11 = Cofactor of 1 = A12 = Cofactor of 1 = A13 = Cofactor of 1 = A21 = Cofactor of 2 = A22 = Cofactor of − 3 = A23 = Cofactor of 2 = A31 =
2
3
−3
1 1 =4 1 − 3 1 1 = =1 1 2
Cofactor of − 1 = A32 = − Cofactor of 3 = A33
3 The Cofactor matrix of A is [Aij] = − 4 − 5 3 T ∴ adj A = (Aij) = − 9 − 5
−9 −5 1 4 −4 1
1 −5 4 1 3
3 − 1 2 Example 1.3 : If A = , verify the result A (adj A) = (adj A) A = | A | I2 1 − 4 − 1 2 − 1 2 Solution: A= , | A | = =2 1 − 4 1 − 4 − 4 − 2 adj A = − 1 − 1 − 1 2 − 4 − 2 2 0 1 0 A (adj A) = … (1) = =2 = 2I2 0 1 1 − 4 − 1 − 1 0 2 − 4 − 2 − 1 2 2 0 1 0 … (2) (adj A) A = = = 2 = 2I2 0 1 − 1 − 1 1 − 4 0 2 From (1) and (2) we get ∴ A (adj A) = (adj A) A = | A | I2.
Example 1.4 : If A = 1 2 1
1
3 1
− 3 , verify A (adj A) = (adj A) A = | A | I3
2 −1
Solution: In example 1.2, we have found 3 −4 −5
− 5
adj A = − 9
1 3
1 4
3
| A | = 1 2 1 A (adj A) = 1 2 1
1
= 1(6 − 3) − 1 (3 + 6) + 1(− 1 − 4) = − 11 3 1 − 11 0 0 3 − 4 − 5 − 3 − 9 1 4 = 0 − 11 0 0 0 − 11 3 − 5 3 1 1
−3
2 −1 1 2
−1 1 0 0
= − 11 0 1 0 = − 11 I3 = | A | I3 0 0 1 1 1 1 3 −4 −5
(adj A) A = − 9 − 5
1
4 1 1 2
3 1 0 0
2 −1
…(1)
− 11 − 3 = 0 3 0
0 − 11 0
= − 11 0 1 0 = − 11 I3 = | A | I3 0 0 1
− 11 0
0
…(2)
From (1) and (2) we get A(adj A) = (adj A) A = | A | I3
1.3 Inverse : Let A be a square matrix of order n. Then a matrix B, if it exists, such that AB = BA = In is called inverse of the matrix A. In this case, we say that A is an invertible matrix. If a matrix A possesses an inverse, then it must be unique. To see this, assume that B and C are two inverses of A, then AB = BA = In … (1) … (2)
AC = CA = In Now AB = In ⇒ ⇒
C(AB) = CIn ⇒ (CA)B = C
(Q associative property)
InB = C ⇒ B = C
i.e., The inverse of a matrix is unique. Next, let us find a formula for computing the inverse of a matrix. We have already seen that, if A is a square matrix of order n, then A(adj A) = (adj A)A = | A | In
4
If we assume that A is non-singular, then | A | ≠ 0. Dividing the above equation by | A |, we get 1 1 A | A | (adj A) = | A | (adj A) A = In. From this equation it is clear that the inverse of A is nothing but 1 −1 | A | (adj A). We denote this by A . Thus we have the following formula for computing the inverse of a matrix through its adjoint. If A is a non-singular matrix, there exists an inverse which is given by 1 A−1 = | A | (adj A).
1.3.1 Properties : 1. Reversal Law for Inverses : If A, B are any two non-singular matrices of the same order, then AB is also non-singular and (AB)−1 = B−1 A−1 i.e., the inverse of a product is the product of the inverses taken in the reverse order. Proof : Since A and B are non-singular, | A | ≠ 0 and | B | ≠ 0. We know that | AB | = | A | | B | | A | ≠ 0, | B | ≠ 0 ⇒ | A | | B | ≠ 0 ⇒ | AB | ≠ 0 Hence AB is also non-singular. So AB is invertible. (AB) (B−1A−1) = A (BB−1)A−1 = AIA−1 = AA−1 = I Similarly we can show that (B−1A−1) (AB) = I ∴ (AB) (B−1A−1) = (B−1A−1) (AB) = I ∴ B−1 A−1is the inverse of AB. ∴ (AB)−1 = B−1 A−1 2. Reversal Law for Transposes (without proof) : If A and B are matrices conformable to multiplication, then (AB)T = BTAT.
5
i.e., the transpose of the product is the product of the transposes taken in the reverse order. −1
3. For any non-singular matrix A, (AT)
= (A− 1)
T
Proof : We know that AA−1 = I = A−1 A T
Taking transpose on both sides of AA−1 = I, we have (AA−1) = IT By reversal law for transposes we get T
(A−1) AT = I
… (1) −1
Similarly, by taking transposes on both sides of A A = I, we have T
AT(A−1) = I
… (2)
From (1) & (2) T T (A−1) AT = AT (A−1) = I T
∴ (A−1) is the inverse of AT i.e.,
(AT)
−1
T
= (A− 1)
1.3.2 Computation of Inverses The following examples illustrate the method of computing the inverses of the given matrices. Example 1.5 : Find the inverses of the following matrices : 3 1 −1 − 1 2 2 − 1 cos α sin α (i) (ii) (iii) (iv) 2 − 2 0 1 − 4 − 4 2 − sin α cos α 1 2 −1
Solution:
− 1 2 − 1 2 (i) Let A = , Then | A | = =2≠0 1 − 4 1 − 4 A is a non-singular matrix. Hence it is invertible. The matrix formed by the cofactors is − 4 − 1 [Aij] = − 2 − 1 − 4 − 2 adj A = [Aij]T = − 1 − 1
6
−1
A
− 2 − 1 1 1 − 4 − 2 = | A | (adj A) = 2 = 1 1 − 1 − 1 − 2 − 2
2 − 1 2 − 1 . then | A | = =0 − 4 2 − 4 2
(ii) Let A =
A is singular. Hence A−1 does not exist. cos α sin α cos α sin α (iii) Let A = . Then | A | = − sin α cos α − sin α cos α = cos2α + sin2α = 1 ≠ 0 ∴ A is non singular and hence it is invertible cos α − sin α Adj A = sin α cos α 1 1 cos α − sin α cos α − sin α A−1 = | A | (Adj A) = 1 = sin α cos α sin α cos α 3 1 −1 3 1 −1
1
(iv) Let A = 2 − 2 2
. − 1 0
Then | A | =
2 1
−2 2
=2≠0 − 1 0
A is non-singular and hence A− 1 exists − 2 0 Cofactor of 3 = A11 = =2 2 − 1 2 0 Cofactor of 1 = A12 = − = 2 1 − 1 2 − 2 Cofactor of − 1 = A13 = =6 1 2 1 − 1 Cofactor of 2 = A21 = − =−1 2 − 1 3 − 1 Cofactor of − 2 = A22 = = −2 1 − 1 3 1 Cofactor of 0 = A23 = − = −5 1 2
7
1 − 1 =−2 − 2 0 3 − 1 =− =−2 2 0 3 1 = =−8 2 − 2
Cofactor of 1 = A31 = Cofactor of 2 = A32 Cofactor of − 1 = A33
[Aij] = − 1 − 2 2
2 − 1 − 2 − 2 − 5 ; adj A = 2 − 2 − 2 6 − 5 − 8 − 2 − 8 2 −1 −2 1 1 A−1 = | A | (adj A) = 2 2 − 2 − 2 6 − 5 − 8 2
6
= 1 3
−1 − 4
1 1 −2 −1 −1 5 −2
0 − 1 1 2 Example 1.6 : If A = verify that (AB)−1 = B−1 A−1. and B = 1 1 1 2 Solution: | A | = − 1 ≠ 0 and | B | = 1 ≠ 0 So A and B are invertible.
1 2 0 − 1 2 3 = 1 1 1 2 1 1 2 3 | AB | = = − 1 ≠ 0. So AB is invertible. 1 1 1 − 2 adj A = − 1 1 − 1 2 1 A−1 = | A | (adj A) = 1 − 1 AB =
8
2 1 − 1 0
adj B =
2 1 1 B−1 = | B | (adj B) = − 1 0 1 − 3 adj AB = − 1 2 − 1 3 1 (AB)−1 = | AB | (adj AB) = 1 − 2
… (1)
2 1 − 1 2 − 1 3 = − 1 0 1 − 1 1 − 2
B−1 A−1 =
… (2)
From (1) and (2) we have (AB)−1 = B−1 A−1. EXERCISE 1.1 (1) Find the adjoint of the following matrices : 1 2 3 2 5 3 3 − 1 (ii) 0 5 0 (i) (iii) 3 1 2 2 − 4 2 4 3 1 2 1 1 2 (2) Find the adjoint of the matrix A = and verify the result 3 − 5 A (adj A) = (adj A)A = | A | . I 3 −3 4
0
1
(3) Find the adjoint of the matrix A = 2 − 3 4 and verify the result −1 A (adj A) = (adj A)A = | A | . I (4) Find the inverse of each of the following matrices : 1 0 3 1 3 7
(i) 2 1 − 1 1 − 1 1 8 − 1 − 3 2 (iv) − 5 1 10 − 1 − 4
(ii) 4 2 3 1 2 1
2 2 1 (v) 1 3 1 1 2 2
9
1 (iii) − 1 0
2
−2
3
0
−2
1
2 − 1 5 2 verify that and B = 7 3 − 1 1
(5) If A =
(i) (AB)−1 = B−1 A−1 (6)
(7)
(ii) (AB)T = BTAT
3 − 3 4 Find the inverse of the matrix A = 2 − 3 4 and verify that A3 = A− 1 0 − 1 1 − 1 − 2 − 2 T 1 − 2 is 3A . Show that the adjoint of A = 2 2 −2 1
− 4 − 3 − 3 0 1 is A itself. (8) Show that the adjoint of A = 1 4 4 3
− 2 1 2, prove that A−1 = AT. 1 −2 2 − 1 2 − 2 4 − 3 4 , show that A = A−1 4 −4 5 2
1 (9) If A = 3
(10) For A =
2
1
1.3.3 Solution of a system of linear equations by Matrix Inversion method : Consider a system of n linear non-homogeneous equations in n unknowns x1, x2, x3 ………xn. a11 x1 + a12 x2 + ………… + a1n xn = b1 a21 x1 + a22 x2 + ………… + a2n xn = b2 …………………………………………… …………………………………………… an1x1 + an2 x2 + …………… + ann xn = bn
10
a11 a12 … a1n
a … This is of the form … an
21
1
… … … … … a
a22 … … … an2
a2n
nn
Thus we get the matrix equation AX = B
x b … = … … … x b x1
b1
2
2
n
n
… (1) where
a11 a12 … a1n
x1
b1
a22 …
a2n
2
2
nn
n
n
a … A= … a
21
n1
… … an2
x b … … ; X= … ;B= … … … … … x b … a
If the coefficients matrix A is non-singular, then A−1 exists. Pre-multiply both sides of (1) by A−1 we get A−1 (AX) = A−1B (A−1A)X = A−1B IX = A−1B X = A−1B is the solution of (1) Thus to determine the solution vector X we must compute A−1. Note that this solution is unique. Example 1.7 : Solve by matrix inversion method x + y = 3, 2x + 3y = 8 Solution: The given system of equations can be written in the form of 1 1 x 3 = 2 3 y 8
Here
AX = B 1 1 |A| = =1≠0 2 3
11
Since A is non-singular, A−1 exists. 3 − 1 A−1 = − 2 1 The solution is X = A−1B
x 3 − 1 3 = y − 2 1 8 x 1 = y 2 x = 1, y = 2 Example 1.8 : Solve by matrix inversion method 2x − y + 3z = 9, x + y + z = 6, x−y+z=2 Solution : The matrix equation is 2 − 1 3 x
1 1 − 1 1 2 A X = B, where A = 1 1 1 1
9 y = 6 z 2 −1 3
9 x 1 1, X = y and B = 6 z 2 − 1 1 2 − 1 3 | A | = 1 1 1 = − 2 ≠ 0 1 − 1 1
A is a non-singular matrix and hence A−1 exists. The cofactors are A11 = 2, A12 = 0, A13 = − 2 A21 = − 2, A22 = − 1, A23 = 1, A31 = − 4, A32 = + 1, A33 = 3 The matrix formed by the cofactors is 2 0
[Aij] = − 2 − 4
12
−1 1
−2
3 1
2 The adjoint of A = 0 − 2
−2 −4
= adj A 3
−1
1
1 1 Inverse of A = | A | (adj A) 2 −2 −4 1 −1 0 −1 1 A = −2 −2 1 3
The solution is given by X = A−1B 2 − 2 − 4 9 x y = − 1 0 − 1 1 6 2 z −2 1 3 2 −2 1 1 − 4 = 2 = −2 3 −6 ∴ x = 1, y = 2, z = 3 EXERCISE 1.2 Solve by matrix inversion method each of the following system of linear equations : (1) 2x − y = 7, 3x − 2y = 11 (2) 7x + 3y = − 1, 2x + y = 0 (3) x + y + z = 9, 2x + 5y + 7z = 52, 2x + y − z = 0 (4) 2x − y + z = 7, 3x + y − 5z = 13, x+y+z=5 (5) x − 3y − 8z + 10 = 0, 3x + y = 4, 2x + 5y + 6z = 13
1.4 Rank of a Matrix : With each matrix, we can associate a non-negative integer, called its rank. The concept of rank plays an important role in solving a system of homogeneous and non-homogeneous equations. To define rank, we require the notions of submatrix and minor of a matrix. A matrix obtained by leaving some rows and columns from the matrix A is called a submatrix of A. In particular A itself is a submatrix of A, because it is obtained from A by leaving no rows or columns. The determinant of any square submatrix of the given matrix A is called a minor of A. If the square submatrix is of order r, then the minor is also said to be of order r. 13
Definition : The matrix A is said to be of rank r, if (i) A has atleast one minor of order r which does not vanish. (ii) Every minor of A of order (r + 1) and higher order vanishes. In other words, the rank of a matrix is the order of any highest order non vanishing minor of the matrix. The rank of A is denoted by the symbol ρ(A). The rank of a null matrix is defined to be zero. The rank of the unit matrix of order n is n. The rank of an m × n matrix A cannot exceed the minimum of m and n. i.e., ρ(A) ≤ min {m, n}. 7 − 1 Example 1.9 : Find the rank of the matrix 2 1 7 − 1 Solution : Let A = . This is a second order matrix. 2 1 ∴ The highest order of minor of A is also 2. 7 − 1 The minor is given by =9≠0 2 1 ∴ The highest order of non-vanishing minor of A is 2. Hence ρ(A) = 2. 2 − 4 Example 1.10 : Find the rank of the matrix − 1 2 2 − 4 Solution : Let A = . − 1 2 2 − 4 The highest order minor of A is given by = 0. Since the second − 1 2 order minor vanishes ρ(A) ≠ 2. We have to try for atleast one non-zero first order minor, i.e., atleast one non-zero element of A. This is possible because A has non-zero elements ∴ ρ(A) = 1. 1 −2 3 Example 1.11 : Find the rank of the matrix
5
1
Solution : Let A = − 2
−2 4 1
− 1 3
−6
14
− 2 5
4 1
− 1 −6
The highest order minor of A is 1 −2 3
− 2 5
1 − 6 = − 2 1 5 − 1
4 1
−2 −2 1
=0 − 1 3 3
Since the third order minor vanishes, ρ(A) ≠ 3 − 2 4 = − 22 ≠ 0 5 1 ∴ A has atleast one non-zero minor of order 2. ∴ ρ(A) = 2 1 1 1 3 Example 1.12 : Find the rank of the matrix
Solution : Let A = 2 5 1
1
1
−1 3
2 5
−1 3 −1 7
11 4
11 3
4
−1 7 This is a matrix of order 3 × 4 ∴ A has minors of highest order 3. They are given by 1 1 1 1 1 3
2 5
= 0 ; 2 − 1 4 = 0 ; 5 − 1 11 7 1 1 3 3 4 = 0 ; − 1 3 4 = 0 − 1 7 11 11
−1 3
−1 1 1
2 3 5 7
All the third order minors vanish. ∴ ρ(A) ≠ 3 Next, we have to try for atleast one non-zero minor of order 2. This is 1 1 possible, because A has a 2nd order minor = − 3 ≠ 0 ∴ ρ(A) = 2 2 − 1 Note : In the above examples, we have seen that the determination of the rank of a matrix involves the computation of determinants. The computation of determinants may be greatly reduced by means of certain elementary transformations of its rows and columns. These transformations will greatly facilitate our dealings with the problem of the determination of the rank and other allied problems.
15
1.4.1. Elementary transformations on a Matrix: (i) Interchange of any two rows (or columns) (ii) Multiplication of each element of a row (or column) by any non-zero scalar. (iii) Addition to the elements of any row (or column) the same scalar multiples of corresponding elements of any other row (or column) the above elementary transformations taken inorder can be represented by means of symbols as follows : (i) Ri ↔ Rj (Ci ↔ Cj) ; (ii) Ri → kRi (Ci → kCi) (ii) Ri → Ri + kRj (Ci → Ci + kCj) Two matrices A and B of the same order are said to be equivalent if one can be obtained from the other by the applications of a finite sequence of elementary transformation “The matrix A is equivalent to the matrix B” is symbolically denoted by A ∼ B. Result (without proof) : “Equivalent matrices have the same rank” Echelon form of a matrix : A matrix A (of order m × n) is said to be in echelon form (triangular form) if (i) Every row of A which has all its entries 0 occurs below every row which has a non-zero entry. (ii) The first non-zero entry in each non-zero row is 1. (iii) The number of zeros before the first non-zero element in a row is less than the number of such zeros in the next row. By elementary operations one can easily bring the given matrix to the echelon form. Result (without proof) : The rank of a matrix in echelon form is equal to the number of non-zero rows of the matrix. Note : (1) The above result will not be affected even if condition (ii) given in the echelon form is omitted. (i.e.) the result holds even if the non-zero entry in each non-zero row is other than 1. (2) The main advantage of echelon form is that the rank of the given matrix can be found easily. In this method we don’t have to compute determinants. It is enough, if we find the number of non-zero rows.
16
In the following examples we illustrate the method of finding the rank of matrices by reducing them to the echelon form. 1 1 −1 Example 1.13 : Find the rank of the matrix 1
−3 −2
3 4
−1
4 3 − 2 3 1 1 − 1 0 − 5 6 0 − 5 6 1 1 − 1 0 − 5 6 0 0 0 1
2 3
Solution : Let A = 2 − 3
R2 → R2 − 2R1 R3 → R3 − 3R1
R3 → R3 − R2
The last equivalent matrix is in echelon form. The number of non-zero rows is 2. ∴ ρ(A) = 2 1 2 3 −1 Example 1.14 : Find the rank of the matrix
1 Solution : Let A = 2 3
2 3
− 3
4 6 −2 6 9
2 3 −1
− 3
4 6 −2
6 9 1 2 3 − 1 R → R − 2R 2 1 0 0 0 0 2 R3 → R3 − 3R1 0 0 0 0
This equivalent matrix is in the echelon form. Since the number of non-zero rows of the matrix in this echelon form is 1, ρ(A) = 1. 4 2 1 3
Example 1.15 : Find the rank of the matrix 6 3 4 7 2 1 0 1
17
4 2 1 3 Solution : Let A = 6 3 4 7 2 1 0 1 1 2 4 3 4 3 6 7 C ↔ C 3 1 0 1 2 1 4 3 1 2 0 − 5 − 10 − 5 R → R − 4R 2 1 2 0 1 2 1 1 2 4 3 0 1 2 1 R → − 1 R 5 2 2 0 1 2 1 1 2 4 3 0 1 2 1 R → R − R 3 2 3 0 0 0 0 The last equivalent matrix is in the echelon form. The number of non-zero rows in this matrix is two. ∴ ρ(A) = 2 3 1 −5 −1 Example 1.16 : Find the rank of the matrix −5 −1
1 1 1 3 1
−2
1
5 −7 −2 1
2 −5 − 1 2
−5
1 − 5 1 −5 1 5 − 7 2 5 −7 1 −2 1 −5 R2 → R2 − 3R1 0 7 − 8 14 R3 → R3 − R1 0 7 − 8 7 1 − 2 1 − 5 0 7 − 8 14 R3 → R3 − R2 0 0 0 − 7 3
1
Solution : Let A = 1 − 2
The last equivalent matrix is in the echelon form. It has three non-zero rows. ∴ ρ(A) = 3
18
R1 ↔ R2
EXERCISE 1.3 Find the rank of the following matrices : (1)
(4)
1 3 2 0 2 1
1 −2 −3 1
−1
3 4 2
−3
0
1
−1
3 1 2 0 (3) 1 0 − 1 0 2 1 3 0
6 12 6 (2) 1 2 1 4 8 4
− 1 0 1
1 (5) 2 3
2 −1 4
1
6
3
− 2 − 7 3
1 (6) − 2 − 1
−2 4 2
3
6 4
−1 −3 7
1.5 Consistency of a system of linear equations : The system of linear equations arises naturally in many areas of Science, Engineering, Economics and Commerce. The analysis of electronic circuits, determination of the output of a chemical plant, finding the cost of chemical reaction are some of the problems which depend on the solutions of simultaneous linear equations. So, finding methods of solving such equations acquire considerable importance. In this connection methods using matrices and determinants play an important role. We have already seen the idea of solving a system of linear equations by the matrix inversion method. This method is applicable provided the number of equations is equal to the number of unknowns, and the coefficient matrix is non-singular. Also the solution obtained under this method is unique. But this is not so in all cases. For many of the problems the number of equations need not be equal to the number of unknowns. In such cases, we see that any one of the following three possibilities can occur. The system has (1) unique solution (2) more than one solution (3) no solution at all. Cases (1) and (3) have no significant role to play in higher studies. Although there exist many solutions, in some cases all the points in the solution are not attractive. Some provide greater significance than others. We have to select the best point among them. In this section we are going to discuss the following two methods. (1) Cramer’s rule method (or Determinant method) (2) Rank method These methods not only decide the existence of a solution but also help us to find the solution (if it exists) of the given system.
19
1.5.1 The Geometry of Solution sets : The solution set of a system of linear equations is the intersection of the solution sets of the individual equations. That is, any solution of a system must be a solution of each of the equations in that system. The equation ax = b (a ≠ 0) has only one solution, namely x = b/a and it represents a point on the line. Similarly, a single linear equation in two unknowns has a line in the plane as its solution set and a single linear equation in three unknowns has a plane in space as its solution set. Illustration I : (No. of unknowns ≥ No. of equations) Consider the solution of the following three different problems. (i) 2x = 10 (ii) 2x + y = 10 (iii) 2x + y − z = 10 Solution (i) 2x = 10 ⇒ x = 5 Solution (ii) 2x + y = 10 S We have to determine the values of two unknown from a single equation. To find the solution we can Fig. 1.1 assign arbitrary value to x and solve for y, or, choose an arbitrary value to y and solve for x. S Suppose we assign x an arbitrary value k, we obtain x = k and y = (10 − 2k) Fig. 1.2 These formulae give the solution set Y interms of the parameter ‘k’. Particular numerical solution can be obtained by S substituting values for ‘k’. For 1 example when k = 1, 2, 5, − 3, 2 , we O X get (1, 8), (2, 6), (5, 0), (− 3, 16) Z 1 and 2 , 9 as the respective solutions. Fig. 1.3 Solution (iii) 2x + y − z = 10 In this case, we have to determine three unknowns x, y and z from a single equation. We can assign arbitrary values to any two variables and solve for the third variable. We assign arbitrary values ‘s’ and ‘t’ to x and y respectively, and solve for z. We get x = s, y = t and z = 2s + t − 10 is the solution set. For different values of s and t we get different solutions.
20
1.5.2 Cramer’s Rule Method : (Determinant Method) Gabriel Cramer (1704 – 1752), a Swiss mathematician wrote on philosophy of law and government, and history of mathematics. He served in a public office, participated in artillery and fortifications activity for the government instructed workers on techniques of cathedral repair and undertook excavations of cathedral archives. Cramer, a bachelor, received numerous honours for his achievements. His theorem provides a useful formula for the solution of certain linear system of n equations in n unknowns. This formula, known as Cramer’s Rule, is of marginal interest for computational purposes, but it is useful for studying the mathematical properties of a solution without actually solving the system. Theorem 1.1 (without proof) : Cramer’s Rule : If AX = B is a system of n linear equations in n unknowns such that det(A) ≠ 0, then the system has a unique solution. This solution is det (A1) det (A2) det (An) x1 = det A , x2 = det (A) , … xn = det (A) Where Aj is the matrix obtained by replacing the entries in the jth column b1 of A by the entries in the matrix. B =
b 2 … bn
Cramer’s Rule for Non homogeneous equations of 2 unknowns : Let us start with the system of two linear equations in two unknowns ‘x’ and ‘y’. … (i) a11x + a12y = b1 … (ii)
a21x + a22y = b2 a11 a12 Let ∆ = a21 a22
a11 a12 = a21 a22
∴ x.∆ = x
a11x a12 a21x a22
b1 − a12y
a12
b2 − a22y
a22
=
(by equation (i) and (ii))
21
b1 a12 a12 a12 −y (by properties of determinants) b2 a22 a22 a22
=
b1 a12 − y . 0 (by properties of determinants) b2 a22
=
b1 a12 = ∆x (say) b2 a22
x.∆ =
Similarly
a11 b1 = ∆y (say) a21 b2
y. ∆ =
∆x, ∆y are the determinants which can also be obtained by replacing 1st and 2nd column respectively by the column of constants containing b1 and b2 i.e. by ∆x b1 b Thus, we have, x∆ = ∆x ⇒ x = ∆ 2 y∆ = ∆y ⇒ y =
∆y provided ∆ ≠ 0 ∆
Since ∆, ∆x, ∆y are unique, there exists a unique solution for the above system of equations. i.e., the system is consistent and has a unique solution. The method stated above to solve the system of equation is known as Cramer’s Rule. Cramer’s rule is applicable when ∆ ≠ 0. If ∆ = 0, then the given system may be consistent or inconsistent. Case 1 : If ∆ = 0 and ∆x = 0, ∆y = 0 and atleast one of the coefficients a11, a12, a21, a22 is non-zero, then the system is consistent and has infinitely many solutions. Case 2 : If ∆ = 0 and atleast one of the values ∆x, ∆y is non-zero, then the system is inconsistent i.e. it has no solution. To illustrate the possibilities that can occur in solving systems of linear equations with two unknowns, consider the following three examples. Solve : (1) x + 2y = 3 x+y=2
(2)
x + 2y = 3 2x + 4y = 6
22
(3)
x + 2y = 3 2x + 4y = 8
Solution (1) : We have
Unique solution
1 2
∆=
1 3 ∆x = 2 1 ∆y = 1
Y
= −1
1 2
= −1
1 3
S O
= −1 2
X X+2Y = 3 X+Y = 2
Fig. 1.4 Since ∆ ≠ 0, the system has unique solution. By Cramer’s rule ∆y ∆x =1 ; y= = 1 ∴ (x, y) = (1, 1) x= ∆ ∆ Solution (2) : Infinitely many solution Y 1 2 We have ∆ = = 0 2 4 3 2 ∆x = = 0 o 6 4 X S 1 3 X + 2Y = 3 ∆y = = 0 2X + 4Y = 6 2 6 Fig. 1.5 Since ∆ = 0 and ∆x = 0, ∆y = 0 and atleast one of a11, a12, a21, a22 is non zero, it has infinitely (case 1) many solutions. The above system is reduced to a single equation x + 2y = 3. To solve this equation, assign y = k ∴ x = 3 − 2y = 3 − 2k The solution is x = 3 − 2k, y = k ; k ∈ R For different value of k we get different solution. In particular (1, 1), (− 1, 2), (5 − 1) and (8, − 2.5) are some solutions for k = 1, 2, − 1 and − 2.5 respectively No Solution Solution (3) : Y 1 2 ∆= = 0 ; 2 4 3 2 1 3 ∆x = = − 4 ; ∆y = =2 8 4 2 8 o X Since ∆ = 0 and ∆x ≠ 0, ∆y ≠ 0 2X + 4Y = 8 X + 2Y = 3 (case 2 : atleast one of the value of ∆x, ∆y, non-zero), the system is Fig. 1.6 inconsistent. i.e. it has no solution.
23
1.5.3 Non homogeneous equations of three unknowns : Consider the system of linear equations a11x + a12y + a13z = b1 ; a21x + a22y + a23z = b2 ; a31x + a32y + a33z = b3 Let us define ∆, ∆x, ∆y and ∆z as already defined for two unknowns.
11 ∆ = a21 a31 a
11 ∆y = a21 a31 a
a12 a22 a32 b1 b2 b3
a23 , a33 a13
a23 , a33
1 ∆x = b2 b3 b
a13
a22 a32
11 a21 a31 a
∆z =
a12
a33 a13
a23
a12 a22 a32
b3 b1
b2
As we discussed earlier for two variables, we give the following rule for testing the consistency of the above system. Case 1 : If ∆ ≠ 0, then the system is consistent, and has a unique solution. Using Cramer’s Rule can solve this system. Case 2 : If ∆ = 0, we have three important possibilities. Subcase 2(a) : If ∆ = 0 and atleast one of the values of ∆x, ∆y and ∆z is non-zero, then the system has no solution i.e. Equations are inconsistent. Subcase 2(b) : If ∆ = 0 and ∆x = ∆y = ∆z = 0 and atleast one of the 2 × 2 minor of ∆ is non zero, then the system is consistent and has infinitely many solution. In this case, the system of three equations is reduced to two equations. It can be solved by taking two suitable equations and assigning an arbitrary value to one of the three unknowns and then solve for the other two unknowns. Subcase 2(c) : If ∆ = 0 and ∆x = ∆y = ∆z = 0 and all their (2 × 2) minors are zero but atleast one of the elements of ∆ is non zero (aij≠ 0) then the system is consistent and it has infinitely many solution. In this case, system is reduced to a single equation. To solve we can assign arbitrary values to any two variables and can determine the value of third variable. Subcase 2(d) : If ∆ = 0, ∆x = ∆y = ∆z = 0, all 2 × 2 minors of ∆ = 0 and atleast one 2 × 2 minor of ∆x or ∆y or ∆z is non zero then the system is inconsistent.
24
Theorem 1.2 (without proof) : If a non-homogeneous system of linear equations with more number of unknowns than the number of equations is consistent, then it has infinitely many solutions. To illustrate the different possibilities when we solve the above type of system of equations, consider the following examples. (1) 2x + y + z = 5 (2) x + 2y + 3z = 6 x+y+z=4 x+y+z=3 x − y + 2z = 1 2x + 3y + 4z = 9 (3) x + 2y + 3z = 6 (4) x + 2y + 3z = 6 2x + 4y + 6z = 12 x+y+z=3 3x + 6y + 9z = 18 2x + 3y + 4z = 10 (5) x + 2y + 3z = 6 2x + 4y + 6z = 12 3x + 6y + 9z = 24 Solution (1) : 2x + y + z = 5 ; x + y + z = 4 ; x − y + 2z = 1 Unique solution We have 2 1 1 ∆ = 1 1 1 = 3 1 − 1 2 5 1 1 ∆x = 4 1 1 = 3 1 − 1 2 Fig. 1.7
2 1 5 2 5 1 ∆ z = 1 1 4 = 3 ∆y = 1 4 1 = 6 ; 1 1 2 1 − 1 1 ∆ = 3, ∆x = 3, ∆y = 6, ∆z = 3 ‡ ∆ ≠ 0, The system has unique solution. By Cramer’s rule. ∆x 3 ∆y 6 ∆z x = = 3 = 1, y = = 3 = 2, z = =1 ∆ ∆ ∆ ∴ The solution is x = 1, y = 2, z = 1 (x, y, z) = (1, 2, 1)
25
Solution (2) : x + 2y + 3z = 6 ; x + y + z = 3 ; 2x + 3y + 4z = 9 1 2 3 6 2 3 ∆ x = 3 1 1 = 0 ∆ = 1 1 1 =0 ; 2 3 4 9 3 4 1 6 3 1 2 6 ∆y = 1 3 1 = 0 ; ∆z = 1 1 3 = 0 2 9 4 2 3 9 Since ∆ = 0 and ∆x = ∆y = ∆z = 0 but atleast one of the 2 × 2 minors of ∆ is 1 2 ≠ 0, the system is consistent (by case 2(b)) and has non-zero 1 1 infinitely many solution. The system is reduced to 2 equations. ∴ Assigning an arbitrary value to one of unknowns, say z = k, and taking first two equations. We get x + 2y + 3k = 6 Infinitely many solution x+y+k = 3 i.e., x + 2y = 6 − 3k x+y = 3−k
1 2 =−1 ∆= 1 1 Fig. 1.8 6 − 3k 2 ∆x = = 6 − 3k − 6 + 2k = − k 3 − k 1 1 6 − 3k ∆y = = 3 − k − 6 + 3k = 2k − 3 1 3 − k ∆x −k = = k x= −1 ∆ ∆y 2k − 3 y= = = 3 − 2k ∆ −1 The solution is x = k, y = 3 − 2k and z = k i.e. (x, y, z) = (k, 3 − 2k, k). k ∈ R Particularly, for k = 1, 2, 3, 4 we get (1, 1, 1), (2, − 1, 2), (3, − 3, 3), (4, − 5, 4) respectively as solution.
26
Solution (3) : x + 2y + 3z = 6 ; 2x + 4y + 6z = 12 ; 3x + 6y + 9z = 18 1 2 3 6 2 3 12 4 6 = 0 2 4 6 ∆= = 0 ; ∆x = 3 6 9 18 6 9 1 6 3 1 2 6 ∆y = 2 12 6 = 0 ; ∆z = 2 4 12 = 0 3 18 9 3 6 18 Here ∆ = 0 and ∆x = ∆y = ∆z = 0. Also all their 2 × 2 minors are zero, but atleast one of aij of ∆ is non- zero. ∴ It has infinitely many solution (by case 2(c)). The system given above is reduced to one equation i.e. x + 2y + 3z = 6
Infinitely many solution
Assigning arbitrary values to two of the three unknowns say y = s, z = t We get x = 6 − 2y − 3z = 6 − 2s − 3t
Fig. 1.9
∴ The solution is x = 6 − 2s − 3t, y = s, z = t i.e. (x, y, z) = (6 − 2s − 3t, s, t) s, t ∈ R For different value s, t we get different solution. Solution (4) : x + 2y + 3z = 6 ; x + y + z = 3 ; 1 2 3 ∆ = 1 1 1 = 0 2 3 4
2x + 3y + 4z = 10 No Solution
6 2 3 ∆x = 3 1 1 = − 1 10 3 4 Fig. 1.10 Since ∆ = 0, ∆x ≠ 0 (atleast one of the values of ∆x, ∆y, ∆z non-zero) The system is inconsistent (by case 2(a)). ∴ It has no solution.
27
Solution (5) : x + 2y + 3z = 6 1 ∆ = 2 3 1 ∆y = 2 3
; 2x + 4y + 6z = 12 ; 3x + 6y + 9z = 24 2 3 6 2 3 4 6 = 0 ; ∆x = 12 4 6 = 0 24 6 9 6 9 6 3 1 2 6 12 6 = 0 ; ∆z = 2 4 12 = 0 3 6 24 24 9
Here ∆ = 0 and ∆x = ∆y = ∆z = 0. All the 2 × 2 minors of ∆ are zero, but we see that atleast one of the 2 × 2 minors of ∆x or ∆y or ∆z is non zero. i.e. 12 4 ≠ 0 minor of 3 in ∆ x 24 6
No solution
∴ by case 2(d), the system is inconsistent and it has no solution.
Fig. 1.11
Example 1.17 : Solve the following system of linear equations by determinant method. (1)
x + y = 3,
(2) 2x + 3y = 8,
(3) x − y = 2,
2x + 3y = 7
4x + 6y = 16
3y = 3x − 7
Solution (1) : x + y = 3 ; 2x + 3y = 7 1 1 ∆= = 3 − 2 = 1, ; ‡ ∆ ≠ 0 2 3 3 1 1 ∆x = = 9 − 7 = 2 ; ∆y = 7 3 2 ∆ = 1,
∆x = 2,
It has unique solution 3
=7−6=1
7
∆y = 1
∴ By Cramer’s rule x=
∆x 2 = 1 =2 ∆
; y =
∆y 1 = 1 = 1 ∆
solution is (x, y) = (2, 1)
28
Solution (2) : 2x + 3y = 8 ; 4x + 6y = 16 2 3 ∆= = 12 − 12 = 0 4 6 8 3 ∆x = = 48 − 48 = 0 16 6 2 8 ∆y = = 32 − 32 = 0 4 16 Since ∆ = 0, and ∆x = ∆y = 0 and atleast one of the coefficients aij of ∆ ≠ 0, the system is consistent and has infinitely many solutions. All 2 × 2 minor are zero and atleast (1 × 1) minor is non zero. The system is reduced to a single equation. We assign arbitrary value to x (or y) and solve for y (or x). Suppose we assign x = t, from equation (1) 1 we get y = 3 (8 − 2t). 8 − 2t ∴The solution set is (x, y) = t, 3 , t ∈ R In particular
(x, y) = (1, 2) (x, y) = (− 2, 4) 1 (x, y) = − 2 , 3
for t = 1 for t = − 2 1 for t = − 2
Solution (3) : x − y = 2 ; 3y = 3x − 7 1 − 1 ∆= = 0, 3 − 3 2 − 1 ∆x = 7 − 3 = 1 Since ∆ = 0 and ∆x ≠ 0 (atleast one of the values ∆x or ∆y ≠ 0) the system is inconsistent. ∴ It has no solution. Example 1.18 : Solve the following non-homogeneous equations of three unknowns. (1) x + 2y + z = 7 (2) x + y + 2z = 6 (3) 2x + 2y + z = 5 2x − y + 2z = 4 3x + y − z = 2 x−y+z=1 x + y − 2z = − 1 4x + 2y + z = 8 3x + y + 2z = 4
29
(4)
x + y + 2z = 4 2x + 2y + 4z = 8 3x + 3y + 6z = 12
(5)
Solution (1) : x + 2y + z = 7, 1 2 1
x + y + 2z = 4 2x + 2y + 4z = 8 3x + 3y + 6z = 10
2x − y + 2z = 4,
2 − 1 2 = 15 1 1 − 2 7 2 1 ∆x = 4 − 1 2 = 15 − 1 1 − 2 1 2 7 ∆z = 2 − 1 4 = 30 1 1 − 1
x + y − 2z = − 1
∴ ∆ ≠ 0 it has unique solution.
∆=
1 7 1 ; ∆y = 2 4 2 = 30 1− 1 − 2
∆ = 15, ∆x = 15, ∆y = 30, ∆z = 30 By Cramer’s rule ∆y ∆z ∆x x= = 1, y = = 2, z = = 2 ∆ ∆ ∆ Solution is (x, y, z) = (1, 2, 2) 3x + y − z = 2, 4x + 2y + z = 8 Solution (2) : x + y + 2z = 6, 1 1 2 6 1 2 ∆ = 3 1 − 1 = 0,
4 1 ∆y = 3 4
2 6 2
1 2 − 1 = 0, 1
∆x = 2 1 − 1 = 0,
8 1 ∆z = 3 4
2 1 1 6 1 2 = 0
2 8 8 Since ∆ = 0 and ∆x = ∆y = ∆z = 0, also atleast one of the (2 × 2) minors of ∆ is not zero, the system is consistent and has infinitely many solution. Take two suitable equations and assign arbitrary value to one of the three unknowns. We solve for the other two unknowns. Let z = k ∈ R ∴ equation (1) and (2) becomes x + y =6 − 2k 3x + y = 2 + k
30
1 1 =1−3=−2 3 1
∆=
6 − 2k 1 = 6 − 2k − 2 − k = 4 − 3k 2 + k 1
∆x =
1 6 − 2k ∆y = = 2 + k − 18 + 6k = 7k − 16 3 2 + k ∴ By Cramer’s rule x=
∆x 4 − 3k 1 = = 2 (3k − 4) ∆ −2
y=
∆y 7k − 16 1 = = 2 (16 − 7k) −2 ∆
∴ The solution set is 3k − 4 16 − 7k (x, y, z) = 2 , , k 2
k∈R
Particular Numerical solutions for k = − 2 and 2 are (− 5, 15, − 2) and (1, 1, 2) respectively Solution (3) : 2x + 2y + z = 5, 2 2 1
∆ = 1 − 1 1 = 0
3
1
2
x − y + z = 1,
3x + y + 2z = 4 2 1
5 ; ∆ x = 1 − 1 1 ≠ 0 4 1 2
Since ∆ = 0 and ∆x ≠ 0 (atleast one of the values of ∆x, ∆y, ∆z non zero) the system is inconsistent. i.e. it has no solution. Solution (4) : x + y + 2z = 4, 1 1 2
∆ = 2 2 4 = 0 3 3 6 1 4 2
∆y = 2 8 4 = 0, 3 12 6
2x + 2y + 4z = 8, 4 1 2
3x + 3y + 6z = 12
∆x = 8 2 4 = 0 12 3 6 1 1 4 ∆ z = 2 2 8 = 0 3 3 12
31
Since ∆ = 0 and ∆x = ∆y = ∆z = 0 also all 2 × 2 minors of ∆, ∆x, ∆y and ∆z are zero, by case 2(c), it is consistent and has infinitely many solutions. (‡ all 2 × 2 minors zero and atleast one of aij of ∆ ≠ 0, the system is reduced to single equation). Let us take x = s and y = t, we get from equation (1) 1 z = 2 (4 − s − t) ∴ the solution set is 4 − s − t (x, y, z) = s, t, 2 , s, t ∈ R Particular numerical solution for (x, y, z) = (1, 1, 1)
when s = t = 1
3 (x, y, z) = − 1, 2, 2 Solution (5) : x + y + 2z = 4,
2x + 2y + 4z = 8,
when s = − 1, t = 2 3x + 3y + 6z = 10
1 1 2 ∆ = 2 2 4 = 0 3 3 6
1 4 2 ∆y = 2 8 4 = 0 3 10 6
4 1 2 ∆x = 8 2 4 = 0, 10 3 6
1 1 4 ∆ z = 2 2 8 = 0 3 3 10
∆ = 0 and ∆x = ∆y = ∆z = 0. Also all 2 × 2 minors of ∆ = 0, but not all the minors of ∆x, ∆y and ∆z are zero. Therefore the system is inconsistent. i.e. it has no solution. Example 1.19 : A bag contains 3 types of coins namely Re. 1, Rs. 2 and Rs. 5. There are 30 coins amounting to Rs. 100 in total. Find the number of coins in each category. Solution : Let x, y and z be the number of coins respectively in each category Re. 1, Rs. 2 and Rs. 5. From the given information x + y + z = 30 (i) x + 2y + 5z = 100 (ii) Here we have 3 unknowns but 2 equations. We assign arbitrary value k to z and solve for x and y.
32
(i) and (ii) become x + y = 30 − k k∈R
x + 2y = 100 − 5k
30 − k 1 1 30 − k 1 1 ∆= = 3k − 40, ∆y = = 70 − 4k = 1, ∆x = 1 2 100 − 5k 2 1 100 − 5k By Cramer’s Rule x =
∆x = 3k − 40, ∆
y=
∆y = 70 − 4k ∆
The solution is (x, y, z) = (3 k − 40, 70 − 4k, k) k ∈ R. Since the number of coins is a non-negative integer, k = 0, 1, 2 … Morever 3k − 40 ≥ 0,
and 70 − 4k ≥ 0 ⇒ 14 ≤ k ≤ 17
∴ The possible solutions are (2, 14, 14), (5, 10, 15), (8, 6, 16) and (11, 2, 17)
1.5.4 Homogeneous linear system : A system of linear equations is said to be homogeneous if the constant terms are all zero; that is, the system has the form a11 x1 + a12 x2 + ………… + a1n xn = 0 a21 x1 + a22 x2 + ………… + a2n xn = 0 …………………………………………. …………………………………………. am1x1 + am2 x2 + ……….. + amn xn = 0 Every homogeneous system of linear equations is always consistent, since all such systems have x1 = 0, x2 = 0 …… xn = 0 as a solution. This solution is called trivial solution. If there are other solution they are called non trivial solutions. Because a homogeneous linear system always has the trivial solution, there are only two possibilities. (i) (The system has only) the trivial solution (ii) (The system has) infinitely many solutions in addition to the trivial solution.
33
As an illustration, consider a homogeneous linear system of two equations in two unknowns. x+y = 0 x−y = 0 the graph of these equations are lines through the origin and the trivial solution corresponding to the point of intersection Fig. 1. 12 at the origin.
Y
X − Y =0
S o
X
X +Y =0
Y
For the following system X+ Y= 0 2X + 2Y = 0 x−y = 0 2x − 2y = 0 O the graph shows, that the system has X infinitely many solutions. There is one case in which a Fig. 1.13 homogeneous system is assured of having non-trivial solutions, namely, whenever the system involves more number of unknowns than the number of equations. Theorem 1.3 : (without proof) A homogeneous system of linear equations with more number of unknowns than the number of equations has infinitely many solutions. Example 1.20 : Solve : x + y + 2z = 0 2x + y − z = 0 2x + 2y + z = 0 Solution : 1 1 2 ∆ = 2 1 − 1 = 3
2 2
1 ‡ ∆ ≠ 0, the system has unique solution. ∴ The above system of homogeneous equation has only trivial solution. i.e., (x, y, z) = (0, 0, 0).
34
Example 1.21 : Solve : x + y + 2z = 0 3x + 2y + z = 0 2x + y − z = 0 Solution : 1 1 2 ∆ = 3 2
1 =0 2 1 − 1
Since ∆ = 0, it has infinitely many solutions. Also atleast one 2 × 2 minors of ∆ ≠ 0, the system is reduced to 2 equations. ∴ Assigning arbitrary value to one of the unknowns, say z = k and taking first and last equations. (Here we can take any two equations) we get x + y = − 2k 2x + y = k − 2k 1 1 − 2k 1 1 ∆y = ∴∆= = − 3k, = 5k = − 1, ∆x = 2 1 k 1 2 k By Cramer’s Rule x = 3k, y = − 5k ∴ Solution is (x, y, z) = (3k, − 5k, k) EXERCISE 1.4 Solve the following non-homogeneous system of linear equations by determinant method : (1) 3x + 2y = 5 (2) 2x + 3y = 5 x + 3y = 4 4x + 6y = 12 (3)
4x + 5y = 9 8x + 10y = 18
(4)
(5)
2x + y − z = 4 x + y − 2z = 0 3x + 2y − 3z = 4
(6)
(7)
x + 2y + z = 6 3x + 3y − z = 3 2x + y − 2z = −3
(8)
35
x+y+z=4 x−y+z=2 2x + y − z = 1 3x + y − z = 2 2x − y + 2z = 6 2x + y − 2z = − 2 2x − y + z = 2 6x − 3y + 3z = 6 4x − 2y + 2z = 4
2 4 1 1 2 1 (9) x + y − z = 1 ; x + y + z = 5 ;
3 2 2 x −y−z=0
(10) A small seminar hall can hold 100 chairs. Three different colours (red, blue and green) of chairs are available. The cost of a red chair is Rs.240, cost of a blue chair is Rs.260 and the cost of a green chair is Rs.300. The total cost of chair is Rs.25,000. Find atleast 3 different solution of the number of chairs in each colour to be purchased.
1.5.5 Rank method : Let us consider a system of “m” linear algebraic equation, in “n” unknowns x1, x2, x3, … xn as in section 1.2. The equations can be written in the form of matrix equation as AX = B Where the m × n matrix A is called the coefficient matrix. A set of values x1, x2, x3 … xn which satisfy the above system of equations is called a solution of the system. The system of equations is said to be consistent, if it has atleast one solution. A consistent system may have one or infinite number of solutions, when the system possesses only one solution then it is called a unique solution. The system of equations is said to be inconsistent if it has no solution. The m × (n + 1) matrix. a11 a12 a13 … a1n
a a … a
b is called the augmented matrix of the … b b1
21
a22 a23 … a2n b2
31
a32 a33
… a3n
3
… … … … a m1 m2 am3 … amn m system and it is denoted by [A, B]. The condition for the consistency of a system of simultaneous linear equations can be given interms of the coefficient and augmented matrices. The system of simultaneous linear equations AX = B is consistent if and only if the matrices A and [A, B] are of the same rank.
36
The solution of a given system of linear equations is not altered by interchanging any two equations or by multiplying any equation by a non-zero scalar or by adding a multiple of one equation to another equation. By applying elementary row operations to the augmented matrix the given system of equations can be reduced to an equivalent system and this reduced form is used to test for consistency and to find the solutions. Steps to be followed for testing consistency : (i) Write down the given system of equations in the form of a matrix equation AX = B. (ii) Find the augmented matrix [A, B] of the system of equations. (iii) Find the rank of A and rank of [A, B] by applying only elementary row operations. Column operations should not be applied. (iv) (a) If the rank of A ≠ rank of [A, B] then the system is inconsistent and has no solution. (b) If the rank of A = rank of [A, B] = n, where n is the number of unknowns in the system then A is a non-singular matrix and the system is consistent and it has a unique solution. (c) If the rank of A = rank of [A, B] < n, then also the system is consistent but has an infinite number of solutions. Example 1.22 : Verify whether the given system of equations is consistent. If it is consistent, solve them.
2x + 5y + 7z = 52,
x + y + z = 9,
2x + y − z = 0
Solution : The given system of equations is equivalent to the single matrix equation. 2 5 7 x 52
1 1 1 y = 9 2 1 − 1 z 0 AX = B
The augmented matrix is 2 5 [A, B] = 1 1
7
52
1 9 2 1 − 1 0
37
1 2 2 1 0 0
1
1
5
7
9
52 R1 ↔ R2
1 −1 0 1 1 9 R → R2 − 2R1 3 5 34 2 R3 → R3 − 2R1 − 1 − 3 − 18 9 1 1 1
0 − 1 − 3 − 18 R ↔ R 3 2 0 3 5 34
0 0 1
1
1
− 20 9
− 1 − 3 − 18 0
−4
R3 → R3 + 3R2
The last equivalent matrix is in the echelon form. It has three non-zero rows. ∴ ρ(A, B) = 3 1 Also A
0 0
1
− 4 1
−1 −3 0
Since there are three non-zero rows, ρ(A) = 3
ρ(A) = ρ[A, B] = 3 = number of unknowns. ∴ The given system is consistent and has a unique solution. To find the solution, we see that the given system of equations is equivalent to the matrix equation. 1 1 1 9 x
0 0
− 4
−1 −3 0
y = z
− 18 − 20
x+y+z = 9
… (1)
− y − 3z = − 18
(2)
− 4z = − 20
… (3)
38
(3) ⇒ z = 5 ; (2) ⇒ y = 18 − 3z = 3 ; (1) ⇒ x = 9 − y − z ⇒ x = 9−3−5 = 1 ∴ Solution is x = 1, y = 3, z = 5 Example 1.23 : Examine the consistency of the equations 2x − 3y + 7z = 5, 3x + y − 3z = 13, 2x + 19y − 47z = 32 Solution : The given system of equations can be written in the form of a matrix equation as 2 −3 7 x 5
3 2
1 19
− 47 −3
y = 13 z 32 AX = B
The augmented matrix is
2 [A, B] = 3 2
−3
7
1
−3
19 − 47
1 3 2
13 32 5
7 2
7 2
1
−3
19 − 47
R 0 112 − 272 112 R 0 22 − 54 27 3 7 5 1 −2 2 2 0 112 − 272 112 R 0 0 0 5 3 1 −2
3 −2
5 2
1 13 R1 → 2 32 5 2
2
→ R2 − 3R1
3
→ R3 − 2R1
3
R1
→ R3 − 4R2
The last equivalent matrix is in the echelon form. It has three non-zero rows. ∴ ρ[A, B] = 3 and ρ(A) = 2 ρ(A) ≠ ρ[A, B] ∴ The given system is inconsistent and hence has no solution. Note : This problem can be solved by not dividing R1 by 2 also. i.e., R2 → 2R2 − 3R1 Example 1.24 : Show that the equations x + y + z = 6, x + 2y + 3z = 14, x + 4y + 7z = 30 are consistent and solve them.
39
Solution : The matrix equation corresponding to the given system is 1 1 1 x 6
1 2 3 y = 14 1 4 7 z 30
AX = B The augmented matrix is 1 1 1 6
[A, B] = 1 2 3 14 1 4 7 30 1 1 1 6 R → R − R 2 1 0 1 2 8 2 R3 → R3 − R1 0 3 6 24 1 1 1 6
0 1 2 8 R → R − 3R 3 2 3 0 0 0 0
In the last equivalent matrix, there are two non-zero rows. ∴ ρ(A, B) = 2 and ρ(A) = 2 ρ(A) = ρ(A, B) ∴ The given system is consistent. But the value of the common rank is less than the number of unknowns. The given system has an infinite number of solutions. The given system is equivalent to the matrix equation 1 1 1 x 6
0 1 2 y = 8 0 0 0 z 0
x+y+z = 6 … (1) y + 2z = 8 … (2) (2) ⇒ y = 8 − 2z ; (1) ⇒ x = 6 − y − z = 6 − (8 − 2z) − z = z − 2 Taking z = k, we get x = k − 2, y = 8 − 2k ; k ∈ R Putting k = 1, we have one solution as x = − 1, y = 6, z = 1. Thus by giving different values for k we get different solutions. Hence the given system has infinite number of solutions. Example 1.25 : Verify whether the given system of equations is consistent. If it is consistent, solve them :
x− y + z = 5,
− x + y − z = − 5,
40
2x − 2y + 2z = 10
Solution : The matrix equation corresponding to the given system is 1 −1 1 x 5
− 1 2
1 −2
2
−1
y = −5 z 10 AX = B
The augmented matrix is
1 [A, B] = − 1 2
−1 1
1
10 5
−1 −5
−2 2 1 − 1 1 5 R → R + R 2 1 0 0 0 0 2 R3 → R3 − 2R1 0 0 0 0 In the last equivalent matrix, there is only one non-zero row ∴ ρ[A, B] = 1 and ρ(A) = 1 Thus ρ(A) = ρ[A, B] = 1. ∴ the given system is consistent. Since the common value of the rank is less than the number of unknowns, there are infinitely many solutions. The given system is equivalent to the matrix equation.
1 − 1 1 x 5 0 0 0 y = 0 0 0 0 z 0 x − y + z = 5 ; Taking y = k1, z = k2, we have x = 5 + k1 − k2. for various values of k1 and k2 we have infinitely many solutions. k1, k2 ∈ R Example 1.26 : Investigate for what values of λ, µ the simultaneous equations x + y + z = 6, x + 2y + 3z = 10, x + 2y + λz = µ have (i) no solution (ii) a unique solution and (iii) an infinite number of solutions. Solution : The matrix equations corresponding to the given system is 1 1 1 x 6
1 2 3 y = 10 1 2 λ z µ AX = B
41
The augmented matrix is 1 1 1
6
[A, B] = 1 2 3 10
1 2 1 0 0
λ µ 1 1
6 R → R2 − R1 1 2 4 2 R3 → R3 − R2 0 λ−3 µ−10 Case (i) : λ − 3 = 0 and µ − 10 ≠ 0 i.e. λ = 3 and µ ≠ 10. In this case ρ(A) = 2 while ρ[A, B] = 3 ∴ ρ(A) ≠ ρ[A, B] ∴ The given system is inconsistent and has no solution. Case (ii) : λ − 3 ≠ 0 i.e., λ ≠ 3 and µ can take any value in R. In this case ρ(A) = 3 and ρ[A, B] = 3 ρ(A) = ρ[A, B] = 3 = number of unknowns. ∴ The given system is consistent and has a unique solution. Case (iii) : λ − 3 = 0 and µ − 10 = 0 i.e., λ = 3 and µ = 10 In this case ρ(A) = ρ[A, B] = 2 < number of unknowns. ∴ The given system is consistent but has an infinite number of solutions.
1.5.6 Homogeneous linear Equations : A system of homogeneous linear equations is given by a11 x1 + a12 x2 + a13 x3 + ……...……+ a1n xn = 0 a21 x1 + a22 x2 + a23 x3 + …………. + a2n xn = 0 …………………………………………………… …………………………………………………… am1x1 + am2 x2 + am3 x3 + …………… + amn xn = 0 and the corresponding augmented matrix is a11 a12 … a1n 0
a … [A, B] = … a
21
m1
a22 … a2n …
…
…
…
…
…
am2 … amn
… = [A, O] … 0 0
42
Since rank of A = rank of [A, O] is always true, we see that the system of homogeneous equations is always consistent. Note that x1 = 0, x2 = 0, x3 = 0 … xn = 0 is always a solution of the system. This solution is called a trivial solution. If the rank of A = rank of [A, B] < n then the system has non trivial solutions including trivial solution. If ρ(A) = n then the system has only trivial solution. Example 1.27 : Solve the following homogeneous linear equations. x + 2y − 5z = 0, 3x + 4y + 6z = 0, x + y + z = 0 Solution : The given system of equations can be written in the form of matrix equation 1 2 − 5 x 0
3 4 1 1
6 y = 0 1 z 0 AX = B
The augmented matrix is 1 2 [A, B] = 3 4 1 1
1 0 0 1 0 0 1 0 0
−5
0
6
0
1 2
0 −5
−2
21
−1 2
6 −5
−1
6
−2 2
21 −5
−1
6
0
9
0 0 0 0 0 0 0 0
0
R2 → R2 − 3R1 R3 → R3 − R1
R2 ↔ R3
R3 → R3 − 2R2
This is in the echelon form. Clearly ρ[A, B] = 3. and. ρ(A) = 3 ∴ ρ(A) = ρ[A, B] = 3 = number of unknowns. ∴ The given system of equations is consistent and has a unique solution. i.e., trivial solution. ∴ x = 0, y = 0 and z = 0
43
Note : Since ρ(A) = 3, | A | ≠ 0 i.e. A is non-singular ; ∴ The given system has only trivial solution x = 0, y = 0, z = 0 Example 1.28 : For what value of µ the equations x + y + 3z = 0, 4x + 3y + µz = 0, 2x + y + 2z = 0 have a (i) trivial solution, (ii) non-trivial solution. Solution : The system of equations can be written as AX = B 1 1 3 x 0
4 3 µ 2 1 2 1 [A, B] = 4 2
y = 0 z 0 1 3 0 3 µ 0 1 2 0
0 − 1 0 − 1 1 1 0 − 1 0 0 1
1
3
0 0 0 0 0
µ−12 0 −4 3 µ−12
R2 → R2 − 4R1 R3 → R3 − 2R1 R3 → R3 − R2
8−µ Case (i) : If µ ≠ 8 then 8 − µ ≠ 0 and hence there are three non-zero rows. ∴ ρ[A] = ρ[A, B] = 3 = the number of unknowns. ∴ The system has the trivial solution x = 0, y = 0, z = 0 Case (ii) : If µ = 8 then. ρ[A, B] = 2 and ρ(A) = 2 ∴ ρ(A) = ρ[A, B] = 2 < number of unknowns. The given system is equivalent to x + y + 3z = 0 ; y + 4z = 0 ∴ y = − 4z ; x=z Taking z = k, we get x = k, y = − 4k, z = k [k ∈ R − {0}] which are non-trivial solutions. Thus the system is consistent and has infinitely many non-trivial solutions. Note : In case (ii) the system also has trivial solution. For only non-trivial solutions we removed k = 0.
44
EXERCISE 1.5 (1) Examine the consistency of the following system of equations. If it is consistent then solve the same. (i) 4x + 3y + 6z = 25 x + 5y + 7z = 13 2x + 9y + z = 1 (ii) x − 3y − 8z = − 10 3x + y − 4z = 0 2x + 5y + 6z − 13 = 0 (iii) x + y + z = 7 x + 2y + 3z = 18 y + 2z = 6 (iv) x − 4y + 7z = 14 3x + 8y − 2z = 13 7x − 8y + 26z = 5 (v) x + y − z = 1 2x + 2y − 2z = 2 − 3x − 3y + 3z = − 3 (2) Discuss the solutions of the system of equations for all values of λ. x + y + z = 2, 2x + y −2z = 2, λx + y + 4z = 2 (3) For what values of k, the system of equations kx + y + z = 1, x + ky + z = 1, x + y + kz = 1 have (i) unique solution (ii) more than one solution (iii) no solution
45
2. VECTOR ALGEBRA 2.1 Introduction : We have already studied two operations ‘addition’ and ‘subtraction’ on vectors in class XI. In this chapter we will study the notion of another operation, namely product of two vectors. The product of two vectors results in two different ways, viz., a scalar product and a vector product. Before defining these products we shall define the angle between two vectors.
2.2 Angle between two vectors : → → → → Let two vectors a and b be represented by OA and OB respectively. Then → → the angle between a and b is the angle between their directions when these directions both converge or both diverge from their point of intersection. B b
b
θ
O
a
A a
Fig. 2. 1 Fig. 2. 2 It is evident that if θ is the numerical measure of the angle between two vectors, then 0 ≤ θ ≤ π.
2.3 The Scalar product or Dot product → → Let a and b be two non zero vectors inclined at an angle θ. Then the → → → → scalar product of a and b is denoted by a . b and is defined as the scalar → → a b cos θ.
| || |
→ → → → Thus a . b = a b cos θ = ab cos θ Note : Clearly the scalar product of two vectors is a scalar quantity. Therefore → the product is called scalar product. Since we are putting dot between a and → b , it is also called dot product.
| || |
46
Geometrical Interpretation of Scalar Product → → → → Let OA = a , OB = b → → Let θ be the angle between a and b . From B draw BL ⊥r to OA. → → OL is called the projection of b on a . OL From ∆OLB, cos θ = OB O
B
b
θ a
L
A
Fig. 2.3 ⇒ OL = (OB) (cos θ) → ⇒ OL = b (cos θ)
| | → → → → Now by definition a . b = | a | | b | cos θ → = | a | (OL) → → → → → ∴ a . b = | a | [projection of b on a ] → → a . b → → Projection of b on a = = → a
… (1)
[‡ using (1)]
→ a → ∧ → . b =a. b → a
| | | |
→ → a . b → → → Projection of a on b = = a . → b
| |
→ b → ∧ = a . b → b
| |
2.3.1 Properties of Scalar Product : Property 1 : The scalar product of two vectors is commutative → → → → → → (i.e.,) a . b = b . a for any two vectors a and b Proof : → → Let a and b be two vectors and θ the angle between them. → → → a . b = a
| | |→b | cos θ → → → → ∴ b . a = | b | | a | cos θ 47
… (1)
→ → → b . a = a
| | |→b | cos θ
… (2)
From (1) and (2) → → → → a . b = b . a Thus dot product is commutative. Property 2 : Scalar Product of Collinear Vectors : → → (i) When the vectors a and b are collinear and are in the same direction, then θ = 0 → → → → → → Thus a . b = a b cos θ = a b (1) = ab … (1) → → (ii) When the vectors a and b are collinear and are in the opposite direction, then θ = π Thus → → → → → → a . b = a b cos θ = a b (cos π) … (1) → → = a b (− 1) = − ab Property 3 : Sign of Dot Product → → The dot product a . b may be positive or negative or zero. (i) If the angle between the two vectors is acute (i.e., 0 < θ < 90°) then cos θ is positive. In this case dot product is positive. (ii) If the angle between the two vectors is obtuse (i.e., 90 < θ < 180) then cos θ is negative. In this case dot product is negative. (iii) If the angle between the two vectors is 90° (i.e., θ = 90°) then cos θ = cos 90° = 0. In this case dot product is zero. → → Note : If a . b = 0, we have the following three possibilities → → → → a . b = 0 ⇒ a b cos θ = 0
| || |
| || |
| || | | || |
| || |
| || |
(i) (ii)
|→a | = 0 (i.e.,) →a is a zero vector and →b any vector. |→b | = 0 (i.e.,) →b is a zero vector and →a any vector.
→ → (iii) cos θ = 0 (i.e.,) θ = 90° (i.e.,) a ⊥ b Important Result : → → → → → → Let a and b be two non-zero vectors, then a . b = 0 ⇔ a ⊥ b
48
Property 4 : Dot product of equal vectors : 2
→ → → a . a = a Convention :
| | |→a | cos 0 = |→a | |→a | = |→a | 2 2 (→a ) = →a . →a = |→a | = →a 2 = a2
= a2
Property 5 : → → → → → → (i) i . i = j . j = k . k = 1 → → → → → → → → → → → → (ii) i . j = j . i = j . k = k . j = k . i = i . k =0 → → → i . i = i
| | |→i | cos 0 = (1) (1) (1) = 1 → → → → i . j = | i | | j | cos 90 = (1) (1) (0) = 0
Property 6 : → → If m is any scalar and a , b are any two vectors, then
(m→a ) . →b = m(→a . →b ) = →a . (m→b ) Property 7 : → → If m, n are scalars and a , b are two vectors then → → → → → → → → m a . n b = mn a . b = mn a . b = a . mn b
(
) (
)
(
)
Property 8 : The scalar product is distributive over addition. → → → → → → → → → → a . b + c = a . b + a . c , for any three vectors a , b , c B Proof : c → → C Let OA = a b c → → + b OB = b → → BC = c O L M → → → a Then OC = OB + BC Fig. 2.4 → → = b + c Draw BL ⊥ OA and CM ⊥ OA
(
)
49
A
→ → ∴ OL = Projection of b on a → → LM = Projection of c on a → → → OM = Projection of b + c on a
(
)
→ → → We have a . b = a
| | (Projection of →b on →a ) → → → ⇒ a . b = | a | (OL) → → → → → a . c = | a | (Projection of c on a ) → → → ⇒ a . c = | a | (LM) → → → → → → → a .( b + c ) = | a | Projection of ( b + c ) on a → → = | a | (OM) = | a | (OL + LM) → → = | a | (OL) + | a | (LM)
Also
Now
→ → → → = a . b + a . c
… (1)
… (2)
[by using (1) and (2)]
→ → → → → → → a . b + c = a . b + a . c
( ) → → → → → → → Corollary : a .( b − c ) = a . b − a . c
Hence
Property 9 : → → (i) For any two vectors a and b , 2
Proof :
2
(→a + →b ) = (→a ) + 2→a . →b + (→b ) 2 (→a + →b ) = (→a + →b ) . (→a + →b )
2
→ → = a2 + 2 a . b + b2
→→ →→ →→ →→ = a . a + a . b + b . a + b . b (by distribution law) 2
2
(→a ) + →a .→b + →a .→b + (→b ) (‡ →a .→b = →b .→a ) →→ → 2 →→ → 2 = ( a ) + 2 a . b + ( b ) = a2 + 2 a . b + b2 2 2 2 → → ( a − b ) = (→a ) − 2→a . →b + (→b ) = a2 − 2→a . →b + b2 =
(ii)
50
2
(iii) Proof :
2
(→a + →b ) . (→a − →b ) = (→a ) − (→b ) = a2 − b2 (→a + →b ) . (→a − →b ) = →a . →a − →a . →b + →b . →a − →b . →b → 2 → 2 = ( a ) − ( b ) = a 2 − b2
Property 10 : Scalar product in terms of components :
→ → → → → → → → b = b1 i + b2 j + b3 k Let a = a1 i + a2 j + a3 k → → → → → → → → a . b = a1 i + a2 j + a3 k . b1 i + b2 j + b3 k →→ →→ →→ →→ = a1b1 i . i + a1b2 i . j + a1b3 i . k + a2b1 j . i →→ →→ →→ →→ + a2b2 j . j + a2b3 j . k + a3b1 k . i + a3b2 k . j + a3b3 →→ k.k
(
(
)
)
(
(
)
(
)
)
(
(
)
(
)
) (
)
= a1b1(1) + a1b2(0) + a1b3(0) + a2b1(0) + a2b2(1) + a2b3(0) + a3b1(0) + a3b2(0) + a3b3(1) = a1b1 + a2b2 + a3b3 Thus, the scalar product of two vectors is equal to the sum of the products of their corresponding components. Property 11 : Angle between two vectors : → → a , b be two vectors inclined at an angle θ. Let → → → → Then a . b = a b cos θ
| || |
⇒ cos θ =
→ → a . b → → a b
| || |
→ → a . b → → a b
⇒ θ = cos−1
| || |
→ → → → → → → → If a = a1 i + a2 j + a3 k and b = b1 i + b2 j + b3 k → → Then a . b = a1b1 + a2b2 + a3b3
|→a | =
a12 + a22 + a32 ;
∴ θ = cos−1
|→b | =
b12 + b22 + b32
a1b1 + a2b2 + a3b3 2
2
a1 + a2 + a3
51
2
b12
+
b22
+
b32
→ → Property 12 : For any two vectors a and b
|→a + →b | ≤ |→a | + |→b | (Triangle inequality) 2 → → →2 →2 → → We have | a + b | = | a | + | b | + 2( a . b ) 2 → → →2 →2 → → | | ⇒ a + b = | a | + | b | + 2| a | | b | cos θ →2 →2 → → ≤ | a | + | b | + 2| a | | b | 2 → → → 2 → ⇒ | a + b | ≤ | a | + | b | → → → → ⇒ |a + b| ≤ |a|+|b|
[‡ cosθ ≤ 1]
→ → Example 2.1 : Find a . b when → → → → → → → → (i) a = i − 2 j + k and b = 4 i − 4 j + 7 k → → → → → → (ii) a = j + 2 k and b = 2 i + k → → → → → → → (iii) a = j − 2 k and b = 2 i + 3 j − 2 k Solution : (i)
(ii) (iii)
→ → → → → → → → a . b = i −2J + k . 4 i −4 j +7k
(
) (
)
= (1) (4) + (− 2) (− 4) + (1) (7) = 19 → → → → → → a . b = j + 2 k . 2 i + k = (0) (2) + (1) (0) + (2) (1) = 2
( ) ( ) → → → → → → → a . b = ( j − 2 k ) . (2 i + 3 j − 2 k )
= (0) (2) + (1) (3) + (− 2) (− 2) = 7 → → Example 2.2 : For what value of m the vectors a and b are perpendicular to each other → → → → → → → → (i) a = m i + 2 j + k and b = 4 i − 9 j + 2 k → → → → → → → → (ii) a = 5 i − 9 j + 2 k and b = m i + 2 j + k Solution : → → (i) Given : a ⊥ b
52
→ → → → → → → → ∴ a . b =0 ⇒ m i +2 j + k . 4 i −9J +2k =0 ⇒ 4m − 18 + 2 = 0 ⇒ m = 4 → → → → → → (ii) 5 i −9J +2k . m i +2 j + k =0 16 ⇒ 5m − 18 + 2 = 0 ⇒ m = 5 → → → → Example 2.3 : If a and b are two vectors such that a = 4, b = 3 and → → → → a . b = 6. Find the angle between a and b Solution : → → 6 1 π a . b = (4) (3) = 2 ⇒ θ=3 cos θ = → → a b Example 2.4 : Find the angle between the vectors → → → → → → 3 i − 2 j − 6 k and 4 i − j + 8 k → → → → → → → → Solution : Let a = 3 i − 2 j − 6 k ; b = 4 i − j + 8 k Let ‘θ’ be the angle between the vectors → → a . b = 12 + 2 − 48 = − 34 → → a = 7, b = 9 → → a . b − 34 = cos θ = 7×9 → → a b 34 θ = cos−1 − 63 → → Example 2.5 : Find the angle between the vectors a and b → → → → → → where a = i − j and b = j − k → → → → → → i − j . j − k a . b Solution : cos θ = = → → → → → → a b i − j j − k
(
(
) (
) (
)
)
| |
| || |
| |
| |
| || |
( | || | |
) ( ||
(1) (0) + (− 1) (1) + (0) (− 1) 2× 2 2π 1 ⇒ cos θ = − 2 ⇒ θ = 3 ⇒ cos θ =
53
) |
| |
→ Example 2.6 : For any vector r → → → → → → → → → → i + r . j j+ r . k k prove that r = r . i
(
)
(
)
(
)
→ → → → Solution : Let r = x i + y j + z k be an arbitrary vector. → → → → →→ r. i = x i +y j +zk . i =x
( ) →→ → → → → r . j = (x i + y j + z k ) . j = y →→ → → → → r . k = (x i + y j + z k ) . k = z (→r . →i ) →i + (→r . →j ) →j + (→r . →k ) →k = x→i + y→j + z→k = →r Example 2.7 : Find the projection of the vector → → → → → → 7 i + j − 4 k on 2 i + 6 j + 3 k → → → → → → → → Solution : Let a = 7 i + j − 4 k ; b = 2 i + 6 j + 3 k → → → → → → → → 7 i + j −4k . 2 i +6 j +3k → → a . b Projection of a on b = = → → → → b 2 i +6 j +3k
(
) (
| |
|
)
|
14 + 6 − 12 8 = 7 4 + 36 + 9 → → Example 2.8 : For any two vectors a and b =
|
Solution :
2
2
| + |→a − →b | = 2 |→a | + |→b | 2 2 2 2 |→a + →b | = (→a + →b ) = |→a | + |→b | + 2→a . →b 2 2 2 2 |→a − →b | = (→a − →b ) = |→a | + |→b | − 2→a . →b
→ → prove that a + b
2
2
… (1) … (2)
Adding (1) and (2) 2
2
2
2
|→a + →b | + |→a − →b | = |→a | + |→b |
→ → →2 +2a . b + a
| | → → →2 +|b| −2a . b → 2 → 2 →2 →2 = 2| a | + 2| b | = 2| a | + | b | 54
∧ ∧ Example 2.9 : If a and b are unit vectors inclined at an angle θ, then prove that 1 ∧ ∧ θ sin 2 = 2 a − b Solution :
2 ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ a − b = a 2 + b 2 − 2a . b = 1 + 1 − 2 a b cos θ θ = 2 − 2 cos θ = 2 (1 − cos θ) = 2 2 sin2 2 θ θ 1 ∧ ∧ ∧ ∧ ∴ a − b = 2 sin 2 ⇒ sin 2 = 2 a − b
→ → → → → Example 2.10 : If a + b + c = 0 , a = 3, → → angle between a and b → → → a + b + c Solution : → → a + b
| |
|→b |= 5 and |→c | = 7, find the
2
⇒
→ = 0 → =− c
(→a + →b ) = (− →c ) → → → 2 → 2 → 2 ⇒ (a) + (b) +2a . b = (c) 2 2 2 |→a | + |→b | + 2|→a | |→b |cos θ = |→c |
2
⇒ 32 + 52 + 2(3) (5) cos θ = 72 1 π cos θ = 2 ⇒ θ = 3 Example 2.11 : Show that the vectors → → → → → → → → → 2 i − j + k , i − 3 j − 5 k , −3 i + 4 j + 4 k form the sides of a right angled triangle. → → → → → → → → → → → → Solution : Let a = 2 i − j + k ; b = i −3 j −5 k ; c = −3 i + 4 j + 4 k → → → → We see that a + b + c = 0 → → → ∴ a , b , c forms a triangle → → → → → → → → Further a . b = 2 i − j + k . i −3 j −5k = 2+3−5=0 → → ∴ a ⊥ b ∴ The vectors form the sides of a right angled triangle.
(
) (
55
)
(1) (2)
EXERCISE 2.1 → → → → → → → → → → Find a . b when a = 2 i + 2 j − k and b = 6 i − 3 j + 2 k → → → → → → → → If a = i + j + 2 k and b = 3 i + 2 j − k find → → → → a +3b . 2a − b → → → → → → Find λ so that the vectors 2 i + λ j + k and i − 2 j + k are perpendicular to each other. → → → → Find the value of m for which the vectors a = 3 i + 2 j + 9 k and → → → → b = i + m j + 3 k are (i) perpendicular (ii) parallel → → → Find the angles which the vector i − j + 2 k makes with the coordinate axes. → → → Show that the vector i + j + k is equally inclined with the coordinate axes. ∧ ∧ If a and b are unit vectors inclined at an angle θ, then prove that ∧ ∧ a − b θ θ 1 ∧ ∧ (ii) tan 2 = (i) cos 2 = 2 a + b ∧ ∧ a + b If the sum of two unit vectors is a unit vector prove that the magnitude of their difference is 3 . → → → If a , b , c are three mutually perpendicular unit vectors, then prove → → → that a + b + c = 3
(
(3) (4)
(5)
(6) (7)
(8) (9)
) (
)
|
(10)
| → → → → → → If | a + b | = 60, | a − b | = 40 and | b | = 46 find | a |.
→ → → → → → → (11) Let u , v and w be vector such that u + v + w = 0 . → → → →→ →→ →→ If u = 3, v = 4 and w = 5 then find u . v + v . w + w . u → → → → → → (12) Show that the vectors 3 i − 2 j + k , i − 3 j + 5 k and → → → 2 i + j − 4 k form a right angled triangle. (13) Show that the points whose position vectors → → → → → → → → 4 i − 3 j + k , 2 i − 4 j + 5 k , i − j form a right angled triangle.
| |
| |
| |
56
(14) Find the projection of → → → → → → → → (i) i − j on z-axis (ii) i + 2 j − 2 k on 2 i − j + 5 k → → → → → → (iii) 3 i + j − k on 4 i − j + 2 k
2.3.2 Geometrical Applicaton of dot product Cosine formulae : Example 2.12 : With usual notations : c2 + a2 − b2 a2 + b2 − c2 b2 + c2 − a2 ; (ii) cos B = (iii) cos C = (i) cos A= 2bc 2ac 2ab Solution (i) : From the diagram π-A A → → → → → → → → b AB + BC + CA = 0 ⇒ a + b + c = 0 c → → → a =− b + c
(
)
B
→ 2 → → 2 a = b + c → → ⇒ a2 = b2 + c2 + 2 b . c
( ) ( ⇒
)
C
a
Fig. 2.5
a2 = b2 + c2 + 2bc cos(π − A) a2 = b2 + c2 − 2bc cos A 2bc cosA = b2 + c2 − a2 cos A =
b2 + c2 − a2 2bc
Similarly we can prove the results (ii) & (iii) Projection Formulae : Example 2.13 : With usual notations (i) a = b cos C+c cos B (ii) b = a cos C+c cos A (iii) c = a cos B+b cos A Solution (i) : A From the diagram → → → → c b AB + BC + CA = 0 B → → → → C π-B ⇒ a+b+c = 0 a → → → a =− b − c Fig. 2.6 →→ →→ → → a . a = −a.b − a.c
57
We have
⇒ ⇒
a2 = − ab cos (π − C) − ac cos (π − B) a2 = − ab (− cos C) − ac (− cos B) a2 = ab cos C + ac cos B a = b cos C + c cos B
Similarly (ii) and (iii) can be proved. Example 2.14 : Angle in a semi-circle is a right angle. Prove by vector method. Solution : Let AB be the diameter of the circle with centre O. Let P be any point on the semi-circle. P
To prove APB = 90°
We have OA = OB = OP (radii) → → → Now PA = PO + OA → → → A O Also PB = PO + OB → → Fig. 2.7 = PO − OA → → → → → → ∴ PA . PB = PO + OA . PO − OA 2
→ → = PO − OA
B
2
= PO2 − OA2 = 0 → → π ∴ PA ⊥ PB ⇒ APB = 2 Hence angle in a semi-circle is a right angle. Example 2.15 : Diagonals of a rhombus are at right angles. Prove by vector methods. → → → → Solution : Let ABCD be a rhombus. Let AB = a and AD = b We have AB = BC = CD = DA i.e.,
Also
→ → a = b
| | | |
D
… (1)
→ → → → → AC = AB + BC = a + b → → → BD = BC + CD → → → → = AD − AB = b − a
C
b A
a
Fig. 2.8
58
B
→ → → → → → ∴ AC . BD = a + b . b − a
( ) ( ) → → → → = (b + a).(b − a) → 2 → 2 → → = ( b ) − ( a ) = 0 ‡| a | = | b |
→ → → → Thus AC . BD = 0 ⇒ AC ⊥ BD Hence the diagonals of a rhombus are at right angles. Example 2.16 : Altitudes of a triangle are concurrent – prove by vector method. Solution : Let ABC be a triangle and let AD, BE be its two altitudes intersecting at O. In order to prove that the altitudes are concurrent it is sufficient to prove that CO is perpendicular to AB. → → → Taking O as the origin, let the position vectors of A, B, C be a , b , c respectively. A → → → → → → Then OA = a ; OB = b ; OC = c E F Now AD ⊥ BC → → O ⇒ OA ⊥ BC B
D
C
Fig. 2.9 → → OA . BC = 0
⇒
→ → → a . c − b =0 → → → → ⇒ a . c − a . b = 0 …(1) → → BE ⊥ CA ⇒ OB ⊥ CA → → → → → ⇒ OB . CA = 0 ⇒ b . a − c = 0 ⇒
(
)
(
)
→ → → → ⇒ b . a − b . c =0 Adding (1) and (2), we get → → → → a . c − b . c =0 ⇒
59
… (2)
(→a − →b ) . →c = 0
→ → → → ⇒ BA . OC = 0 ⇒ OC ⊥ AB Hence the three altitudes are concurrent. Example 2.17 : Prove that cos (A − B) = cos A cos B + sin A sin B Solution : Y Take the points P and Q on the unit P (Cos A, Sin A) circle with centre at the origin O. Q (Cos B, Sin B ) Assume that OP and OQ make angles A B A and B with x-axis respectively. X O M L ∴ POQ = POx − QOx = A − B Clearly the coordinates of P and Q Fig. 2.10 are (cos A, sin A) and (cos B, sin B) . → → Take the unit vectors i and j along x and y axes. → → → → → ∴ OP = OM + MP = cos A i + sin A j → → → → → OQ = OL + LQ = cos B i + sin B j → → → → → → By value, OP . OQ = cos A i + sin A j . cos B i + sin B j ..(1) = cos A cos B + sin A sin B → → → → By definition , OP . OQ = OP OQ cos (A − B) = cos (A − B) .. (2) From (1) and (2) cos (A − B) = cos A cos B + sin A sin B
(
) (
)
2.3.4 Application of Scalar Product in Physics Work done by force : The work done by a force is a scalar quantity and its measure is equal to the product of the magnitude of the force and the resolved part of the A displacement in the direction of the force. Let a particle be placed at O and a d → → force F represented by OB be acting on the particle at O. Due to the application → θ of force F , the particle is displaced in O L B F → → the direction of OA. Here OA is the displacement and OL is the displacement Fig. 2.11 → in the direction of F In right angled ∆ OLA
60
OL = OA cos θ
where θ is the angle → → between F and d
→ = d cos θ
| |
The work done by a force = (Magnitude of force) (displacement in the direction of force) → → → = F OL = F d cos θ → → Work done by the force = F . d Note : If a number of forces are acting on a particle, then the sum of the works done by the separate forces is equal to the work done by the resultant force. Example 2.18 : Find the work done in moving a particle from the point A, → → → with position vector 2 i − 6 j + 7 k , to the point B, with position vector → → → → → → → 3 i − j − 5 k , by a force F = i + 3 j − k Solution : → → → → → → → → → → → → F = i + 3 j − k ; OA = 2 i − 6 j + 7 k ; OB = 3 i − j − 5 k → → → → → → → d = AB = OB − OA = i + 5 j − 12 k → → Work done = F . d
| |
=
→ (→i + 3→j − →k ) . (→i + 5→j − 12k )
= (1) (1) + 3(5) + 12 = 28 → → → → Example 2.19 : The work done by the force F = a i + j + k in moving the point of application from (1, 1, 1) to (2, 2, 2) along a straight line is given to be 5 units. Find the value of a. → → → → → → → → → → → → Solution : F = a i + j + k ; OA = i + j + k ; OB = 2 i + 2 j + 2 k Work done = 5 units → → → → → → → d = AB = OB − OA = i + j + k → → Work done = F . d → → → → → → 5= ai + j + k . i + j + k
(
) (
5 = a+1+1 ⇒ a = 3
61
)
EXERCISE 2.2 Prove by vector method (1) If the diagonals of a parallelogram are equal then it is a rectangle. (2) The mid point of the hypotenuse of a right angled triangle is equidistant from its vertices (3) The sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of the sides. (4) cos (A + B) = cos A cos B − sin A sin B → → → → (5) Find the work done by the force F = 2 i + j + k acting on a particle, if the particle is displaced from the point with position vector → → → → → → 2 i + 2 j + 2 k to the point with position vector 3 i + 4 j + 5 k . → → → (6) A force of magnitude 5 units acting parallel to 2 i − 2 j + k displaces the point of application from (1, 2, 3) to (5, 3, 7). Find the work done. → → → → → → → → (7) The constant forces 2 i − 5 j + 6 k , − i + 2 j − k and 2 i + 7 j act → → → on a particle which is displaced from position 4 i − 3 j − 2 k to position → → → 6 i + j − 3 k . Find the work done. (8) Forces of magnitudes 3 and 4 units acting in the directions → → → → → → 6 i + 2 j + 3 k and 3 i − 2 j + 6 k respectively act on a particle which is displaced from the point (2, 2, − 1) to (4, 3, 1). Find the work done by the forces.
2.4 Vector product : 2.4.1 Right-handed and left handed systems : → → → Consider a set of three linearly independent vectors a , b , c through the origin O. As they are linearly independent no two of them have parallel directions and not all of them lie on the same plane. Let θ be the smaller angle → → → → (i.e. 0 < θ < π) between a and b . Let an observer walk from a to b through the angle θ keeping O always to his left. If the observer’s head is on the → → → → → → same side of the plane of a and b as the vector c , we say a , b , c is a right handed system or right handed triple (or) triad. → → → → If c has the opposite direction, a , b , c is a left handed system.
62
z
z b
c
b O y
O
a
a
y
c
x
x
Fig. 2. 12 →
→
→
→
Definition : The vector product of two vectors a and b is denoted as a × b
→ → and it is defined as a vector whose magnitude is a b sin θ where θ is the → → angle between a and b , 0 ≤ θ ≤ π and whose direction is perpendicular to → → → → both a and b in such a way that a , b and this direction constitute a right handed system. In other words,
| || |
a x b
<
| | |→b | sin θ ∧n, where
n
<
→ → → a × b = a
n
→ → ∧ b θ is the angle between a and b and n is a unit vector perpendicular to both → → → → ∧ a and b such that a , b , n form a a Fig. 2.13 right handed system. Note : → → ∧ → (1) a , b , n form a right handed system means that if we rotate a into → ∧ b , then n will point in the direction perpendicular to the plane → → containing a and b in which a right handed screw will move if it is turned in the same manner. → → → → → (2) a × b is read as a cross b since we are putting cross between a → and b .
2.4.2 Geometrical interpretation of Vector product : → → → → Let OA = a , OB = b
63
→ → Let θ be the angle between a and b Complete the parallelogram OACB →
<
→
B n
with OA and OB as adjacent sides. Draw BN ⊥ OA. In right angled triangle ONB
C
b θ
O
→ BN = b sin θ
| |
N A a Fig. 2.14
→ → → a × b = a
| | |→b | sin θ ∧n |→a × →b | = |→a | |→b | sin θ
Now
= (OA) (BN) = Base × height = Area of parallelogram OACB Area of parallelogram with → → ∴ a × b = → → a and b as adjacent sides 1 Also, area of ∆OAB = 2 area of a parallelogram OACB
|
|
→ 1 → 1 → → = 2 OA × OB = 2 a × b
| | 1 → → Vector area of ∆OAB = 2 ( a × b )
|
|
Some important results : → → Result : (1) The area of a parallelogram with adjacent sides a and b is → → a × b
|
|
→ → (2) The vector area of a parallelogram with adjacent sides is a × b → → 1 → → (3) The area of a triangle with sides a and b is 2 a × b 1 → → 1 → → (4) The area of a triangle ABC is 2 AB × AC (or) 2 BC × BA 1 → → (or) 2 CA × CB
|
64
|
2.4.3 Properties of Vector Product : Property (1) : Non-Commutativity of Vector product : → → Vector product is not commutative (i.e.) if a and b are any two vectors, → → → → → → → → then a × b ≠ b × a however a × b = − b × a .
(
)
→ → → a × b = a
B
<
→ → Let a and b be two non-zero, non parallel vectors and let θ be the angle between them. Then
b
n
→ ∧ b sin θ n where
| || |
θ
2π - θ
a
<
∧ n is a unit vector perpendicular to the → → plane of a and b
−θ A
-n
Fig. 2.15
→ → → → ∧ → b × a = b a sin(θ) (− n ) = − a
| | |→b | sin θ ∧n = − (→a × →b )
| || |
→ → ∧ Note that b , a and − n form a right handed system. → → → → Hence a × b ≠ b × a → → → → But a × b = − b × a
(
)
Property (2) : Vector product of Collinear (Parallel) Vectors : → → → → → If the vectors a and b are collinear or parallel then a × b = 0 → → The vectors a and b are collinear or parallel, then θ = 0, π sin θ = 0 for θ = 0, π → → ∧ → → Thus a × b = a b sin θ n
| || | ∧ → → → = | a | | b | (O) n = 0
Result : The vector product of two non-zero vectors is zero vector if and only if they are parallel (collinear) → → → → → → → i.e., a × b = o ⇔ a is parallel to b , where a , b are non zero vectors. Proof (i) : → → → Suppose a × b = O
65
then
|→a | |→b | sin θ ∧n = →O
→ → ∧ → But a ≠ 0 & b ≠ 0, n ≠ O ⇒ θ = 0 or π
| |
| |
⇒ sin θ = 0 → → ⇒ a and b are collinear (parallel) → → conversely if a || b then θ = O or π ⇒ sin θ = O → → → → ∧ → ⇒ a × b = a b sin θ n = o → → → ⇒ a × b = o → → → Note : If a × b = o , we have the following three possibilities. → → (i) a is a zero vector and b is any vector. → → (ii) b is a zero vector and a is any vector → → (iii) a and b are parallel (collinear) Property (3) : Cross Product of Equal Vectors : → → → → ∧ a × a = a a sin θ n → → ∧ = a a (0) n → = o → → → → ∴ a × a = o for every non-zero vector a Property (4) : → → → Cross product of Unit Vectors i , j , k By the above property → → → → → → → i × i = j × j = k× k = O
| || |
| || | | || |
Z
k k O
j
j Y
i i
X
Fig. 2. 16
66
→ → → i × j = i
| | |→j | sin 90°→k = (1) (1) (1) →k = →k
Also
→ → → → → → j × k = i, k × i = j → → → → → → → → → j × i = − k, k × j =− i , i × k =− j
Similarly and Property (5) :
→ → If m is any scalar and a , b are two vectors inclined at angle θ, then → → → → → → ma × b = m a × b = a ×mb Property (6) : Distributivity of vector product over vector addition → → → Let a , b , c be any three vectors. then
(
(i)
)
→ → → → → → → a × b + c = a × b + a × c (Left distributivity)
(
)
→ → → → → → → b+c × a = b × a + c × a (ii) Result : Vector Product in the determinant form → → → → Let a = a1 i + a2 j + a3 k and
(
)
(
) (
) (Right distributivity)
→ → → → b = b1 i + b2 j + b3 k be the two vectors Then
→ → → → → → → → a × b = a1 i + a2 j + a3 k × b1 i + b2 j + b3 k → → → → → → = a1b1 i × i + a1b2 i × j + a1b3 i × k
(
)
(
)
(
)
→ → → → → → + a2b1 j × i + a2b2 j × j + a2b3 j × k
( ) ( ) ( ) → → → → → → + a3b1( k × i ) + a3b2( k × j ) + a3b3( k × k ) → → → → = a1b2 k + a1b3(− j ) + a2b1(− k ) + a2b3 i → → + a3b1 j + a3b2(− i ) → → → = ( a2b3 − a3b2) i − (a1b3 − a3b1) j + ( a1b2 − a2b1) k
67
→i = a1 b 1
→ → j k a2 b2
a3 b3
Property (7) : Angle between two vectors : → → Let a , b be two vectors inclined at an angle θ. → → → Then a × b = a
| | |→b | sin θ ∧n ∧ → → → → ⇒ | a × b | = | a | | b | sin θ n → → → → ⇒ | a × b | = | a | | b |sin θ |→a × →b | ⇒ θ = sin−1 |→a × →b | ⇒ sin θ = → → |→a | |→b | | a | | b |
Note : In this case θ is always acute. Thus if we try to find the angle using vector product, we get only the acute angle. Hence in problems of finding the angle, the use of dot product is preferable since it specifies the position of the angle θ. Property (8) : Unit vectors perpendicular to two given vectors (i.e.) Unit vectors normal to the plane of two given vectors. → → Let a , b be two non-zero, non-parallel vectors and θ be the angle between them. → → → a × b = a
| | |→b | sin θ ∧n
… (1)
∧ → → Where n is a unit vector perpendicular to the both of a and b
|→a × →b | = |→a | |→b | sin θ From (1) and (2)
→ → a × b ∧ n = → → a × b
|
68
|
… (2)
→ → a × b → → is also a unit vector perpendicular to a and b Note that − → → a × b
|
|
→ → Unit vectors perpendicular to a and b are → → a × b ∧ ∴±n = ± → → a × b
|
|
→ → Vectors of magnitude µ normal to the plane containing a and b is given → → µ a ×b by ± → → a × b Example 2.20 :
( |
) |
2
→ → → → → → If a , b are any two vectors, then a × b + a . b Solution : → → Let θ be the angle between a and b → → ∧ → → ∴ a × b = a b sin θ n
|
| (
2
2
2
) = |→a | |→b |
| || | |→a × →b | = |→a | |→b | sin θ 2 2 2 |→a × →b | = |→a | |→b | sin2 θ 2 2 2 (→a . →b ) = |→a | |→b | cos2θ 2 2 2 2 2 2 |→a × →b | + (→a . →b ) = |→a | |→b | (sin2θ + cos2θ) = |→a | |→b |
Example 2.21 : Find the vectors of magnitude 6 which are perpendicular to → → → → → → both the vectors 4 i − j + 3 k and − 2 i + j − 2 k Solution : → → → → → → → → Let a = 4 i − j +3k ; b =−2 i + j −2k → → → i j k → → → → → Then a × b = 4 −1 3 =− i +2 j +2k
− 2
1
69
− 2
|→a × →b | =
(− 1)2 + (2)2 + (2)2 = 3 → → a × b Required vectors = 6 ± → → a × b
|
|
→ → → = ± −2 i +4 j +4k
( ) → → →→ → → Example 2.22 : If | a | = 13, | b | = 5 and a . b = 60 then find | a × b | Solution : 2
2
2
2
|→a × →b | + (→a . →b ) = |→a | |→b | 2 2 2 2 |→a × →b | = |→a | |→b | − (→a . →b ) = (13)2 (5)2 − (60)2 = 625 ⇒
|→a × →b | = 25
→ → → Example 2.23 : Find the angle between the vectors 2 i + j − k and → → → i + 2 j + k by using cross product. Solution : → → → → → → → → Let a = 2 i + j − k ; b = i + 2 j + k → → Let θ be the angle between a and b
→ → a × b → → a b
| | | || |
∴ θ = sin−1
→ → a × b =
|→a × →b | = |→a | = |→b | =
→i 2 1
→ → j k 1 2
32 + (− 3)2 + 32 = 3 3 22 + 12 + (− 1)2 = 6 12 + 2 2 + 1 2 = 6
70
→ → → − 1 = 3 i − 3 j + 3 k 1
→ → a × b ∴ sin θ = → → a b
| | = 3 | | | | 6
3 3 = 6 2
π θ=3 → → → → → → → → Example 2.24 : If p = − 3 i + 4 j − 7 k and q = 6 i + 2 j − 3 k then find → → → → → p × q . Verify that p and p × q are perpendicular to each other and also → → → verify that q and p × q are perpendicular to each other. Solution : → → → i j k → → p × q = −3 4 −7
6
− 3
2 → → → = 2 i − 51 j − 30 k Now
→ → → → → → → → → p . p × q = − 3 i + 4 j − 7 k . 2 i − 51 j − 30 k
(
) (
) (
)
= − 6 − 204 + 210 = 0 → → → Hence p and p × q are perpendicular to each other. Now
→ → → → → → → → → q . p × q = 6 i + 2 j − 3 k . 2 i − 51 j − 30 k
(
) (
) (
)
= 12 − 102 + 90 = 0 → → → Hence q and p × q are perpendicular to each other. Example 2.25 : If the position vectors of three points A, B and C are → → → → → → → → → → respectively i + 2 j + 3 k , 4 i + j + 5 k and 7 i + k . Find AB × AC . Interpret the result geometrically. Solution : → → → → → → → → → → → OA = i + 2 j + 3 k , OB = 4 i + j + 5 k ; OC = 7 i + 7 k → → → → → → → → → AB = OB − OA = 4 i + j + 5 k − i + 2 j + 3 k
(
(
) (
→ → → → AB = 3 i − j + 2 k
71
)
)
→ → → → → → AC = OC − OA = 6 i − 2 j + 4 k → → → i j k → → → AB × AC = 3 − 1 2 = 0
6
4
−2 → → The vectors AB and AC are parallel. But they have the point A as a common point. → → ∴ AB and AC are along the same line. ∴ A, B, C are collinear. EXERCISE 2.3 (1) Find the magnitude of → → → → → → → → → a × b if a = 2 i + k , b = i + j + k → → →→ → → (2) If a = 3, b = 4 and a . b = 9 then find a × b (3) Find the unit vectors perpendicular to the plane containing the vectors → → → → → → 2 i + j + k and i + 2 j + k (4) Find the vectors whose length 5 and which are perpendicular to the → → → → → → → → vectors a = 3 i + j − 4 k and b = 6 i + 5 j − 2 k → → → →→ → (5) Find the angle between two vectors a and b if a × b = a . b
| |
| |
|
|
|
(6) If
|
|→a | = 2, |→b | = 7 and →a × →b = 3→i − 2→j + 6→k find the angle
→ → between a and b . → → → → → → → → → (7) If a = i + 3 j − 2 k and b = − i + 3 k then find a × b . Verify → → → → that a and b are perpendicular to a × b separately. → → → (8) For any three vectors a , b , c show that → → → → → → → → → → a × b + c + b × c + a + c × a + b = 0 → → → → → → → (9) Let a , b , c be unit vectors such that a . b = a . c = 0 and the angle → → → → → π between b and c is 6 . Prove that a = ± 2 b × c
(
)
(
)
72
(
)
(
)
→ → → → → → → → (10) If a × b = c × d and a × c = b × d , → → → → show that a − d and b − c are parallel.
2.4.4 Geometrical applications of cross product : 1 → → Example 2.26 : Prove that the area of a quadrilateral ABCD is 2 AC × BD where AC and BD are its diagonals. Solution : Vector Area of quadrilateral ABCD = Vector area of ∆ABC + Vector area of ∆ACD 1 → → 1 → → = 2 AB × AC + 2 AC × AD
C D
1 → → 1 → → = − 2 AC × AB + 2 AC × AD 1 → → → = 2 AC × − AB + AD 1 → → → = 2 AC × BA + AD 1 → → = 2 AC × BD
A
B
Fig. 2.17
1 → → The area of the quadrilateral ABCD = 2 AC × BD Deduction : 1 → → → → Area of a parallelogram = 2 d 1 × d 2, where d 1 and d 2 are the diagonals. Example 2.27 : → → → If a , b , c are the position vectors of the vertices A, B, C of a triangle ABC, then prove that the area of triangle ABC is 1→ → → → → → → → → 2 a × b + b × c + c × a Deduce the condition for points a , b , c to be collinear.
|
|
73
1 → → Solution : Area of ∆ ABC =2 AB × AC → → → → → Now AB = OB − OA = b − a → → → → → and AC = OC − OA = c − a 1 → → 1 → → → → Hence, area of ∆ABC = 2 AB × AC = 2 b − a × c − a 1 → → → → → → → → =2 b × c − b × a − a × c + a × a 1 → → → → → → =2 b × c + a × b + c × a 1 → → → → → → Area of ∆ABC = 2 a × b + b × c + c × a If the points A, B, C are collinear, then the area of ∆ABC = 0 1 → → → → → → ⇒ 2 a × b + b × c + c × a =0
(
) (
| | |
) |
| |
| | → → → → → → ⇒ |a × b + b × c + c × a| = 0
→ → → → → → → (or) a × b + b × c + c × a = 0 → → → → → → → Thus a × b + b × c + c × a = 0 is the required condition of → → → collinearity of the points with positions a , b , c . a b c Example 2.28 : With usual notation prove that sin A = sin B = sin C → → → → → → Solution :Let BC = a , CA = b , AB = c 1 → → 1 → → 1 → → By the area property of triangles 2 a × b = 2 b × c = 2 c × a
|
|
|
|
→ → → → → → ⇒ a × b = b × c = c × a ab sin (π−C) = bc sin(π−A) = ca sin (π−B)
|
| |
| |
|
|
|
⇒ ab sinC = bc sinA = ca sinB
π-A
Divide by abc sin B sinC sinA c = a = b
c B
π-B
A b π-C a
Fig. 2.18
74
C
a b c sin A =sin B = sin C Example 2.29 : Prove that sin(A + B) = sinA cosB + cosA sinB Solution : Y (Co sA Take the points P and Q on the P , S in A) unit circle with centre at the origin O. Assume that OP and OQ make angles A and B with x-axis A M N O B respectively. Take the reciprocals,
in B
)
X
sB
(Co Q
, -S
∴ POQ = POx + QOx = A + B Fig. 2.19
Clearly the coordinates of P and Q are (cosA, sinA) and (cosB, − sinB). → → Take the unit vectors i and j along x and y axes respectively. → → → → → OP = OM + MP = cos A i + sin A j → → → → → OQ = ON + NQ = cos B i + sin B − j
( )
→ ‡ NQ = sinB
→ → = cos B i −sin B j → → → → → → OQ × OP = OQ OP sin (A + B) k = sin (A + B) k → → OQ × OP =
→i cosB cosA
→ j − sinB sinA
… (1)
→ k
→ 0 = k [sinA cosB + cosA sinB] 0
…(2)
From (1) and (2) sin (A + B) = sinA cosB + cosA sinB Example 2.30 : Show that the area of a parallelogram having diagonals → → → → → → 3 i + j − 2 k and i − 3 j + 4 k is 5 3. → → → → → → → → Solution : Let d 1= 3 i + j − 2 k and d 2 = i − 3 j + 4 k 1 → → Area of the parallelogram = 2 d 1 × d 2
75
→ i → → d1× d2 = 3 1 → → ⇒ d 1 × d 2 = =
→ → j k 1 −3
→ → → − 2 = − 2 i − 14 j − 10 k 4
(− 2)2 + (− 14)2 + (− 10)2 300 = 10 3
1 → → 1 Area of the parallelogram = 2 d 1 × d 2= 2 10 3 = 5 3 sq. units
2.4.5 Applications of Vector Product in Physics The moment of a force about a point : → Let a force F be applied at a point r x F P of a rigid body. Then the moment of O → force F about a point O measures the → r tendency (amount) of F to turn the body about point O. If this tendency of θ F P N rotation about O is in anti-clockwise θ Fig. 2.20 direction the moment is positive, otherwise it is negative. → → → Let F be the force and P be a point on the line of action of F . Let r be the position vector of P relative to O. → The magnitude of the moment of the force F about O is the product of the → magnitude of F and the length of the perpendicular from O to the line of action of the force. → ∴ Magnitude of the moment = F (ON) In right angled triangle ONP ON ON sinθ = OP = → r
| |
|→r | sin θ = ON 76
→ ∴ Magnitude of the moment = F (ON) → → = r F sin θ
| |
→ → = r × F → ∴ Moment (or) Torque of force F about the point O is defined as the → → → vector M = r × F Example 2.31 : → → → A force given by 3 i + 2 j − 4 k is applied at the point (1, − 1, 2). Find the moment of the force about the point (2, − 1, 3). Solution : We have A(2,-1,3) → → → → r F = 3 i +2 j −4k → → → → P(1,-1,2) OP = i − j + 2 k → → → → F OA = 2 i − j + 3 k Fig. 2.21 → → → → r = AP = OP − OA =
(→i − →j + 2→k ) − (2→i − →j + 3→k )
→ → → r =−i − k → → The moment M of the force F about the point A is given by → → → i j k → → → → → → M = r × F = −1 0 −1 =2 i −7 j −2k
3
2
− 4
EXERCISE 2.4 (1) Find the area of parallelogram ABCD whose vertices are A(− 5, 2, 5), B(− 3, 6, 7), C(4, − 1, 5) and D(2, − 5, 3) (2) Find the area of the parallelogram whose diagonals are represented by → → → → → → 2 i + 3 j + 6 k and 3 i − 6 j + 2 k 77
(3) Find the area of the parallelogram determined by the sides → → → → → → i + 2 j + 3 k and − 3 i − 2 j + k (4) Find the area of the triangle whose vertices are (3, − 1, 2), (1, − 1, − 3) and (4, − 3, 1) (5) Prove by vector method that the parallelograms on the same base and between the same parallels are equal in area. (6) Prove that twice the area of a parallelogram is equal to the area of another parallelogram formed by taking as its adjacent sides the diagonals of the former parallelogram. (7) Prove that sin (A − B) = sin A cos B − cos A sin B. → → → → → → → → (8) Forces 2 i + 7 j , 2 i − 5 j + 6 k , − i + 2 j − k act at a point P → → → whose position vector is 4 i − 3 j − 2 k . Find the moment of the resultant of three forces acting at P about the point Q whose position → → → vector is 6 i + j − 3 k . → → → (9) Show that torque about the point A(3, − 1, 3) of a force 4 i + 2 j + k → → → through the point B(5, 2, 4) is i + 2 j − 8 k . (10) Find the magnitude and direction cosines of the moment about the point → → → (1, − 2, 3) of a force 2 i + 3 j + 6 k whose line of action passes through the origin.
2.5 Product of three vectors : → → → Let a , b , c be three vectors. By inserting dot and cross between → → → a , b , c in the same alphabetical order we introduce the following :
(→a . →b ) . →c , (→a . →b ) × →c , (→a × →b ) . →c and (→a × →b ) × →c → → → Consider ( a . b ) . c → → Here a . b is a scalar quantity and dot product is not defined between a → → → scalar and vector quantity. Therefore a . b . c is not meaningful.
(
Similary
)
(→a . →b ) × →c is not meaningful. 78
→ → → → → But a × b . c is meaningful, because a × b is a vector and → → → → → → a × b . c is the dot product between the vectors a × b and c .
(
( )
Similarly
)
(→a × →b ) × →c is meaningful.
2.5.1 Scalar Triple Product : → → → → → → Let a , b , c be three vectors. Then the product a × b . c is called a scalar triple product.
(
)
Geometrical Interpretation of Scalar Triple Product : → → → Let a , b , c be three non-coplanar vectors. Consider a parallelopiped → → → → having co-terminus edges OA, OB and OC such that OA = a , OB = b and → → OC = c . G F a x b → → Then a × b is a vector perpendicular to the plane E C → → B c ϕ containing a and b . D ϕ b L Let φ be the angle between O A → → → a c and a × b . Fig. 2.22 Let CL be perpendicular to the base OADB. Here CL is the height of the parallelopiped. → → Here CL and a × b are perpendicular to the same plane → → ⇒ CL is parallel to a × b . ⇒ OCL = φ → In right angled triangle OLC, CL = c cos φ
| | → ∴ Height of the paralleopiped CL = | c | cos φ
Area of the parallelogram → → with a and b as adjacent sides
Base area of the parallelopiped =
→ → Base area of the parallelopiped = a × b
|
79
|
(→a × →b ) . →c = |→a × →b | |→c | cos φ
Now,
= [base area] [height] → → → Volume of the parallelopiped with a × b . c = → → → co-terminous edges a , b , c
(
)
Thus, the scalar triple product
(→a × →b ) . →c represents the volume of the
→ → → paralleopiped whose co-terminous edges a , b , c form a right handed system of vectors.
2.5.2 Properties of Scalar Triple Product : Property (1) :
(→a × →b ) . →c = (→b × →c ) . →a = (→c × →a ) . →b
[Cyclic order]
Proof : → → → Let a , b , c represent the co-terminous edges of a parallelopiped such that they form a right handed system. Then the volume V of the parallelopiped → → → is given by V = a × b . c
(
)
→ → → → → → Clearly b , c , a as well as c , a , b form a right handed system of vectors and represent the co-terminous edges of the same parallelopiped. → → → → → → ∴ V = b × c . a and V = c × a . b
( ) ( ) → → → → → → → → → ∴V = ( a × b ) . c = ( b × c ) . a = ( c × a ) . b
… (1) Since dot product is commutative (1) gives → → → → → → → → → V= c . a × b = a. b × c = b . c × a … (2) From (1) and (2) → → → → → → a × b . c = a . b × c The dot and cross are interchangeable in a scalar triple product. In view of this property, the scalar triple product is written in the following notation. → → → → → → → →→ a × b . c = a . b × c = a b c
(
(
)
)
(
)
(
)
(
)
( ) ( ) [ ] → → → →→ → → → → ∴[a , b , c] = [b c a]=[c a b] 80
Property (2) : The change of cyclic order of vectors in scalar triple product changes the sign of the scalar triple product but not the magnitude. → →→ → → → → → → → → → (i.e.) a b c = − b a c = − c b a = − a c b Proof : → →→ → → → a b c = a × b . c We have
[
] [ [
] [
] (
] [
]
)
(→b × →a ) . →c Q →a × →b = − →b × →a [→a , →b , →c ] = − [→b →a →c ] … (1) =−
Similarly we can prove other results. Property (3) : The scalar triple product of three vector is zero if any two of them are equal. → → → Proof : Let a , b , c be three vectors. → → When a = b ,
[→a →b →c ] = (→a × →b ) . →c = (→b × →b ).→c → → → → → = o . c =0 Q b × b= o → → → → Similarly we can prove for b = c and for c = a Property (4) : → →→ → →→ → → → For any three vectors a , b , c and scalar λ a b c = λ a b c
[
Proof :
] [ ] [λ→a →b →c ] = (λ→a × →b ) . →c = λ (→a × →b ) . →c → →→ = λ[a b c]
Property (5) : The scalar triple product of three vectors is zero if any two of them are parallel or collinear. → → → → Proof : Let a , b , c be three vectors such that a is parallel or collinear to → → → b . Then a = λ b for some scalar λ.
[→a →b →c ] = [λ→b →b →c ] = λ (0) = 0 81
Property (6) (without proof) : The necessary and sufficient condition for three non-zero, non-collinear → →→ → → → vectors a , b , c to be coplanar is a b c = 0
[
]
[→a →b →c ] = 0 → →→ Note : Three possibilities for [ a b c ] to be zero are → → → i.e., a , b , c are coplanar ⇔
→ → → (i) atleast one of the vectors a , b , c is a zero vector. → → → (ii) any two of the vectors a , b , c are parallel. → → → (iii) The vector a , b , c are co-planar. But for cases (i) and (ii), the case (iii) is trivially true. Result : Scalar Triple Product in terms of components : → → → → → → → → Let a = a1 i + a2 j + a3 k , b = b1 i + b2 j + b3 k , → → → → c = c1 i + c2 j + c3 k , a1 a2 a3 → →→ Then a b c = b1 b2 b3
[
]
c1 → i → → Proof : We have a × b = a1 b 1
c2 → j a2 b2
c3 → k a3 b3
→ → → = (a2b3 − a3b2) i −(a1b3 − a3b1) j + (a1b2 − a2b1) k ∴
[→a , →b , →c ] = (→a × →b ) . →c = (→a × →b ) . c1→i + c2→j + c3→k = (a2b3 − a3b2) c1 − (a1b3 − a3b1)c2 + (a1b2 − a2b1)c3 a1 a2 a3 =
b1 c1
c3
b2 b3 c2
82
Distributivity of Cross product over Vector addition : Result : → → → For any three vectors a , b , c → → → → → → → we have a × b + c = a × b + a × c This can be proved by determinant form of cross product. → → → → → → → → Example 2.32 : If the edges a = − 3 i + 7 j + 5 k , b = − 5 i + 7 j − 3 k → → → → c = 7 i − 5 j − 3 k meet at a vertex, find the volume of the parallelopiped. Solution :
(
)
Volume of the parallelopiped =
[→a , →b , →c ]
− 3 = − 5 7
7 7
= − 264 − 3 5
−3
−5 The volume cannot be negative ∴ Volume of parallelopiped = 264 cu. units. Note : Box product may be negative. → →→ Example 2.33 : For any three vectors a b c prove that
[→a + →b , →b + →c , →c + →a ] = 2 [→a →b →c ] → → → → → → Solution : [ a + b , b + c , c + a ] → → → → → → = ( a + b ) × ( b + c ) . ( c + a ) → → → → → → → → → → = ( a × b ) + ( a × c ) + ( b × b ) + ( b × c ) . ( c + a ) → → → → → → → → = {a × b + a × c + b × c}.(c + a) → → → → → → → → → = ( a × b ) . c + ( a × c ) . c + ( b × c ). c → → → → → → → → → + ( a × b ) . a + ( a × c ) . a + ( b × c ). a → →→ →→→ → →→ = [a b c]+[b c a] = 2[a b c] 83
→ → → → →→ → → Example 2.34 : If x . a = 0, x . b = 0, x . c = 0 and x ≠ 0 then show → → → that a , b , c are coplanar. Solution : → → → → → → → x . a = 0 and x . b = 0 implies a and b are ⊥r to x → → → ∴ a × b is parallel to x → → → ∴ x =λ a × b
(
)
→→ → → → Now x . c = 0 ⇒ λ a × b . c = 0 ⇒ → → → ⇒ a , b , c are coplanar
(
)
[→a →b →c ] = 0
2.5.3 Vector Triple Product : Definition : → → → → → → Let a , b , c be any three vectors, then the product a × b × c and → → → → → → a × b × c are called vector triple products of a , b , c Result : → → → For any three vectors a , b , c → → → → → → → → → a × b × c= a . c b − b . c a → → → → This result can be proved by taking a = a1 i + a2 j + a3 k ;
(
(
)
)
(
)
(
)
(
)
→ → → → → → → → b = b1 i + b2 j + b3 k ; c = c1 i + c2 j + c3 k Property (1) : → → → The vector triple product a × b × c is a linear combination of those two vectors which are within brackets. Property (2) : → → → → The vector triple product a × b × c is perpendicular to c and lies in → → the plane which contains a and b .
(
)
(
)
Property (3) : → a ×
(→b × →c ) = (→a . →c )→b − (→a . →b )→c 84
→ → → → → → Example 2.35 : If a × b × c = a × b × c then
(
) ( ) → → → → Prove that ( c × a ) × b = o → → → → → → Proof : Given : a ×(b × c) = (a × b)× c (→a . →c ) →b − (→a . →b ) →c = (→a . →c ) →b − (→b .→c ) →a → → → →→ → ⇒ (a . b)c = (b.c) a → → → → → → → ⇒ (c . b)a −(a . b)c = 0 (→c × →a ) × →b = →0 ⇒ → → → → → → → → Example 2.36 : If a = 3 i + 2 j − 4 k , b = 5 i − 3 j + 6 k , → → → → → → → → → → (ii) a × b × c c = 5 i − j + 2 k , find (i) a × b × c and show that they are not equal. Solution : → → → i j k → → → → b × c = 5 − 3 6 = 20 j + 10 k (i)
(
(ii)
∴
)
(
)
5 −1 2 → → → i j k → → → → → → ( ) ∴ a × b × c = 3 2 − 4 = 100 i − 30 j + 60 k 0 20 10 → → → i j k → → → → a × b = 3 2 − 4 = − 38 j − 19 k 5 −3 6 → → → i j k (→a × →b ) × →c = 0 − 38 − 19 = − 95→i − 95→j + 190→k 5 −1 2
→ → → From (i) and (ii) a × b × c ≠
(
) (→a × →b ) × →c 85
2.5.4 Vector product of four vectors ; → → → For the four vectors a , b , c , → → → → a × b and c × d namely product of four vectors. → → → → Example 2.37 : Let a , b , c and d
(
(i)
)
(
)
→ d the vector product of the two vectors → → → → a × b × c × d is called vector
(
) (
)
be any four vectors then
(→a × →b ) × (→c × →d ) = [→a →b →d ] →c − [→a →b →c ] →d (→a × →b ) × (→c × →d ) = [→a →c →d ] →b − [→b →c →d ] →a
(ii) Solution : (i)
(→a × →b ) × (→c × →d ) = →x × (→c × →d ) where →x = →a × →b → → → → → → = (x . d) c −(x . c) d → → → → → → → → = ( a × b ) . d c − ( a × b ) . c d → →→ → → →→ → = [a b d] c −[a b c] d
→ → → Similarly we can prove other result by taking x = c × d → → → → Note : (1) If the four vectors a , b , c , d are coplanar then → → → → → a × b × c × d = o . → → → → (2) Let a , b be lie on one plane and c , d lie on another plane. → → → → → These planes are perpendicular then a × b . c × d = o Example 2.38 :
(
) (
)
(
Prove that
[→a × →b ,
→ → → → b × c, c × a =
) (
)
2
] [→a , →b , →c ]
Solution :
[→a × →b , →b × →c , →c × →a ] = (→a × →b ) × (→b × →c ) . (→c × →a ) → →→ → → → → → → → = [ a b c ] b − [ a b b ] c . ( c × a ) → → → → → → →→→ = [ a b c ] b . ( c × a ) since [ a b b ] = 0 86
[→a →b →c ] [→b →c →a ] 2 → → → →→→ → → → = [ a b c ] since [ b c a ] = [ a b c ] =
2.5.5 Scalar product of four vectors : → → → → For four vectors a , b , c , d the scalar product of the two vectors → → → → namely a × b and c × d is called scalar product of four vectors. → → → → i.e. a × b . c × d → → → → Result : Determinant form of a × b . c × d → → → → a.c a.d → → → → i.e. a × b . c × d = → → → → b.c b.d Proof : → → → → → → → → → → a × b . c × d = a × b . x where x = c × d → → → = a. b × x (interchange dot and cross) → → → → = a .b × c × d → → → → → → → = a . b . d c − b . c d → → → → → → → → = b . d a . c − b . c a . d → → → → a.c a.d = → → → → b.c b.d
(
) (
(
(
) (
) (
) (
) (
)
)
) (
)
(
)
(
) ( ) ( ( )( ) (
)
)(
)
EXERCISE 2.5 → → → (1) Show that vectors a , b , c are coplanar if and only if → → → → → → a + b , b + c , c + a are coplanar. (2) The volume of a parallelopiped whose edges are represented by → → → → → → → − 12 i + λ k , 3 j − k , 2 i + j − 15 k is 546. Find the value of λ. → → → → →→ (3) Prove that a b c = abc if and only if a , b , c are mutually perpendicular.
[
]
87
(4) Show that the points (1, 3, 1), (1, 1, − 1), (− 1, 1, 1) (2, 2, − 1) are lying on the same plane. (Hint : It is enough to prove any three vectors formed by these four points are coplanar). → → → → → → → → → → (5) If a = 2 i + 3 j − k , b =−2 i +5k, c = j −3k → → → → → → → → → Verify that a × b × c = a . c b − a . b c → → → → → → → → → → (6) Prove that a × b × c + b × c × a + c × a × b = o → → → → → → → → (7) If a = 2 i + 3 j − 5 k , b = − i + j + 2 k and → → → → → → → → → → c = 4 i − 2 j + 3 k , show that a × b × c ≠ a × b × c → → → → → → → → (8) Prove that a × b × c = a × b × c iff a and c are collinear. (where vector triple product is non-zero). → (9) For any vector a → → → → → → → → → → prove that i × a × i + j × a × j + k × a × k = 2 a
( (
(
)( )
)
(
)
) (
(
)
( (
) )
(
)
)
(
)
(
(
)
)
(→ →) (→ →) (→ →) (→ →) (→ →) (→ →) → → → → → → → → Find ( a × b ) . ( c × d ) if a = i + j + k
(10) Prove that a × b . c × d + b × c . a × d + c × a . b × d = 0 (11)
→ → → → → → → → → → → b =2 i + k, c =2 i + j + k, d = i + j +2k → → → → → → → → → → → (12) Verify a × b × c × d = a b d c − a b c → → → → for a , b , c and d in problem 11.
(
) (
) [
]
[
] →d
2.6 Lines : 2.6.1 Equation of a line : Parametric and non parametric vector equations : → Let P be an any point with position vector r on the given line. A relation → satisfied by r for all points on the line is then found using certain conditions. This relation is called the vector equation of the line. Parametric vector equations : → If r is expressed in terms of some fixed vectors and variable scalars, (parameters) the relation is then called a parametric vector equation.
88
Non-parametric vector equation : If no parameter is involved, the equation is called a non-parametric vector equation. Vector and Cartesian Equations of Straight lines : A straight line is uniquely determined in space if (i) a point on it and its direction are given (ii) two points on it are given. Note : Eventhough the syllabus does not require the derivations (2.6.2, 2.6.3) and it needs only the results, the equations are derived for better understanding the results.
2.6.2 Equation of a straight line passing through a given point and parallel to a given vector : Vector form : → Let the line pass through a given point A whose position vector is a w.r.to → O and parallel to the given vector v . Let P be any point on the line and its → position vector w.r.to O be r . → → → → z We have OA = a , OP = r v A → → AP and v are parallel. P a → → r ∴ AP = t v for some scalar t y O → → → OP = OA + AP x Fig. 2.23 → → → … (1) r = a +tv This represents the vector equation of the given straight line. → → → Note : r = a + t v , where t is a variable scalar (i.e., a parameter) is the parametric vector equation of the line. Corollary : If the straight line is given to be passing through the origin, then → → the equation (1) becomes r = t v Cartesian form : Let the co-ordinates of the fixed point A be (x1, y1, z1) and the direction ratios of the parallel vector be l, m, n. Then → → → → → → → → a = x1 i + y1 j + z1 k ; v = l i + m j + n k → → → → r = x i +y j +zk
89
→ → → → → → → → → → → → r = a +t v ⇒x i + y j + z k =x1 i + y1 j + z1 k + t l i + m j + n k → → → Equating the coefficients of i , j , k we get x = x1 + tl These are the y = y1 + tm parametric equations of the line z = z + tn
(
)
1
x − x1 = t, ⇒ l
y − y1 z − z1 = t, m n = t y − y1 z − z1 x − x1 = = Eliminating t, we get l m n This is the cartesian equation of the line passing through a point (x1, y1, z1) and parallel to a vector whose drs are l, m, n. Non-parametric vector equation : → → → → → AP = OP − OA = r − a → → → → → But AP | | v ⇒ AP × v = 0 → → → → r − a × v = 0 ⇒ → → → → → ⇒ r × v − a × v = 0 → → → → ⇒ r × v = a × v This is the non-parametric vector equation of the line.
(
)
2.6.3 Equation of a straight line passing through two given points: Vector Form : Let the line pass through two given points A and B whose position vectors → → are a and b respectively. z Let P be any point on the line and its A → position vector be r B b P a We have r → → → → → → y O OA = a , OB = b and OP = r → → x AP and AB are parallel vectors. → → Fig. 2.24 ∴ AP = t AB for some scalar t
90
→ → → → = t OB − OA = t b − a
(
)
→ → → OP = OA + AP → → → → (or) … (1) r = a +t b − a → → → r = (1 − t) a + t b This represents the vector equation of the given straight line. → → → Note : r = (1 − t) a + t b where t is a variable scalar (i.e., a parameter) is the parametric vector equation of the required line. Cartesian form : Let the co-ordinates of the fixed points A be (x1, y1, z1) and B be (x2, y2, z2)
(
)
→ → → → → → → → → → → → a = x1 i + y1 j + z1 k ; b = x2 i + y2 j + z2 k ; r = x i + y j + z k Substituting these values in equation (1) we get → → → → → → x i + y j + z k = x1 i + y1 j + z1 k → → → → → → + tx2 i + y2 j + z2 k − x1 i + y1 j + z1 k → → → Equating the coefficients of i , j , k x = x1 + t(x2 − x1) These are the parametric equations y = y1 + t(y2 − y1) of the line z = z1 + t(z2 − z1) ⇒
x − x1 = t, x2 − x1
y − y1 = t, y2 − y1
z − z1 = t z2 − z 1
Eliminating t, we get y − y1 z − z1 x − x1 = = x2 − x1 y2 − y1 z2 − z 1 This is the cartesian equation of the required line. Note : x2 − x1, y2 − y1, z2 − z1 are the d.r.s of the line joining the points (x1, y1, z1) and (x2, y2, z2) Non-parametric vector equation : → → → → → AB = OB − OA = b − a
91
→ → → → → AP = OP − OA = r − a → → Since AP and AB are collinear vectors → → → ⇒ AP × AB = 0 → → → → → r − a × b − a = 0 ⇒ This is the non parametric vector equation.
(
) (
)
2.6.4 Angle between two lines : → → → → → → Let r = a 1 + t u and r = a 2 + s v be the two lines in space. These two lines are in the direction of
→ → u and v . “Angle between the two lines is defined as the angle between their directions”.
a1 + r =
t u u
θ
v
r =a 2 + s v
Fig. 2.25
→ → u . v If θ is the angle between the given lines then θ = cos → → u v −1
| || |
Cartesian form : If the equations of the lines are in Cartesian form y − y1 z − z1 x − x1 y − y1 z − z1 x − x1 = = and = a1 b1 c1 a2 b2 = c2 Where a1, b1, c1 and a2, b2, c2 are the direction ratios of two lines, then angle between them is a1a2 + b1b2 + c1c2 θ = cos−1 2 2 2 2 2 2 a2 + b2 + c2 a1 + b1 + c1 Note : When two lines are perpendicular then a1a2 + b1b2 + c1c2 = 0 Example 2.39 : Find the vector and cartesian equations of the straight line → → → passing through the point A with position vector 3 i − j + 4 k and parallel to → → → the vector − 5 i + 7 j + 3 k Solution : We know that vector equation of the line through the point with → → → → → position vector a and parallel to v is given by r = a + t v where t is a scalar.
92
→ → → → Here a = 3 i − j + 4 k → → → → v = − 5 i +7 j +3k ∴ Vector equation of the line is → → → → → → → r = 3 i − j +4k +t − 5 i +7 j +3k … (1) The cartesian equation of the line passing through (x1, y1, z1) and parallel to a vector whose d.r.s are l, m, n is x − x1 y − y1 z − z1 = l m = n Here (x1, y1, z1) = (3, − 1, 4)
(
) (
)
(l, m, n) = (− 5, 7, 3) x−3 y+1 z−4 = 7 = 3 ∴ The required equation is −5 Example 2.40 : Find the vector and cartesian equations of the straight line passing through the points (− 5, 2, 3) and (4, − 3, 6) Solution : Vector equation of the straight line passing through two points with → → position vectors a and b is given by → → → → r = a +t b − a → → → → a = −5 i +2 j +3k Here → → → → b = 4 i −3 j +6k → → → → → b − a = 9 i −5 j +3k ∴ Vector equation of the line is → → → → → → → r = − 5 i + 2 j + 3 k + t 9 i − 5 j + 3 k or → → → → → → → r = (1 − t) − 5 i + 2 j + 3 k + t 4 i − 3 j + 6 k Cartesian Form : The required equation is y − y1 z − z1 x − x1 = = x2 − x1 y2 − y1 z2 − z 1
(
(
)
) (
(
)
) (
)
(x1, y1, z1) = (− 5, 2, 3) ; (x2, y2, z2) = (4, − 3, 6) x+5 y−2 z−3 ∴ 9 = = 3 is the cartesian equation of the line −5
Here
93
Example 2.41 : Find the angle between the lines → → → → → → → r = 3 i + 2 j − k + t i + 2 j + 2 k and
(
)
→ → → → → → r =5 j +2k +s 3 i +2 j +6k
(
)
→ → Solution : Let the given lines be in the direction of u and v → → → → → → → → Then u = i + 2 j + 2 k , v = 3 i + 2 j + 6 k Let θ be the angle between the given lines → → u . v ∴ cos θ = → → u v
| || | → → → → u . v = 19 ; | u | = 3 ; | v | = 7 19 cos θ = 21
19 ⇒ θ = cos−1 21
EXERCISE 2.6 (1) Find the d.c.s of a vector whose direction ratios are 2, 3, − 6. (2) (i) Can a vector have direction angles 30°, 45°, 60°. (ii) Can a vector have direction angles 45°, 60°, 120°? (3) What are the d.c.s of the vector equally inclined to the axes? → (4) A vector r has length 35 2 and direction ratios (3, 4, 5) , find the → direction cosines and components of r . (5) Find direction cosines of the line joining (2, − 3, 1) and (3, 1, − 2). (6) Find the vector and cartesian equation of the line through the point → → → (3, − 4, − 2) and parallel to the vector 9 i + 6 j + 2 k . (7) Find the vector and cartesian equation of the line joining the points (1, − 2, 1) and (0, − 2, 3). (8) Find the angle between the following lines. x−1 y+1 z−4 y+2 z−4 2 = 3 = 6 and x + 1 = 2 = 2 (9) Find the angle between the lines → → → → → → r =5 i −7 j +µ − i +4 j +2k
( → → → → → r = − 2 i + k + λ( 3 i + 4 k )
94
)
2.6.5 Skew lines Consider two straight lines in the space. There are three possibilities. (i) either they are intersecting (ii) (or) parallel L1 (iii) (or) neither intersecting nor A parallel Two lines in space which are d either intersecting or parallel are L2 called coplanar lines. B i.e., a plane can be defined passing through (the two lines completely lie on the plane) two Fig. 2.26 intersecting lines or through two parallel lines. Therefore, two lines lie on the same plane are called coplanar lines. Two lines L1 and L2 in space, which are neither intersecting nor parallel are called skew lines. (See Fig. 2.26) i.e., two lines in space which are not coplanar are called skew lines. Shortest distance between two lines (i) Trivially the shortest distance between two intersecting lines is zero. (ii) Parallel lines Theorem : (without proof) The distance between two parallel lines → → → → → → r = a 1 + t u ; r = a 2 + s u is given by d=
→ → → u × a 2 − a 1
|→u |
(iii) Skew lines : Theorem : (without proof) The distance between the skew lines → → → → → → r = a 1 + t u ; r = a 2 + s v is given by d=
→ → → → a 2 − a 1 u v
|→u × →v | 95
Condition for two lines to intersect : The shortest distance between the intersecting lines → → → → → → r = a 1 + t u ; r = a 2 + s v is 0 → → →→ The condition for intersecting is d = 0 ⇒ a 2 − a 1 u v = 0 (or) x2 − x1 y2 − y1 z2 − z1
l1
m1
n1
l2
m2
n2
=0
if
→ (x1, y1, z1)and (x2, y2, z2) are the points whose position vectors are a 1 and → → → a 2 and l1, m1, n1 ; l2, m2, n2 are the d.rs of the vectors u and v → → respectively. ( u and v are not parallel) Example 2.42 : Find the shortest distance between the parallel lines → → → → → → r = i − j + t 2 i − j + k and
( ) ( ) → → → → → → → r = (2 i + j + k ) + s(2 i − j + k ) → → → → → → Solution : Compare the given equations with r = a 1 + t u and r = a 2 + s u , → → → → → → → → → → → a 1 = i − j ; a 2 = 2 i + j + k and u = 2 i − j + k → → → → → a2− a1 = i +2 j + k → → → u × a 2 − a 1 =
→i 2 1
→ → k j −1 2
1= 1
→ → → u × a 2 − a 1 = 9 + 1 + 25 =
|→u | =
→ → → −3 i − j +5k
35
4+1+1 = 6
→ → → The distance between u × a 2 − a 1 35 = = the parallel lines 6 → u
| |
96
Note : If the equations are in the Cartesian form, write in the vector form and find the distance between them. → → → → → Example 2.43 : Show that the two lines r = i − j + t 2 i + k and
(
) (
)
→ → → → → → r = 2 i − j +s i + j − k are skew lines and find the distance between them. → → → → → → Solution : Compare the given equations with r = a 1 + t u and r = a 2 + s v
(
) (
)
→ → → → → → → → → → → → → a 1 = i − j ; a 2 = 2 i − j and u = 2 i + k ; v = i + j − k → → → a2− a1 = i 1 0 → → → 2 0 → a 2 − a 1 u v =
0 1 =−1≠0
1 1 − 1
∴ They are skew lines. → → u × v =
→i 2 1
|→u × →v | = Shortest distance between the lines =
→ → j k 0 1
1 = − 1
→ → → − i +3 j +2k
14 → → → → a 2 − a 1 u v
|→u × →v |
… (1)
1 14 x−4 y z+1 x−1 y−1 z+1 = 0 and 2 = 0 = 3 Example 2.44 : Show that the lines 3 = −1 intersect and hence find the point of intersection. Solution : The condition for intersecting is x2 − x1 y2 − y1 z2 − z1 From (1) shortest distance between them is
l1
m1
n1
l2
m2
n2
=0
97
Compare with
y − y1 z− z1 x − x2 y − y2 z − z2 x − x1 = = and l1 m1 n1 l2 = m2 = n2 , we
get (x1, y1, z1) = (1, 1, − 1) ; (x2, y2, z2) = (4, 0, − 1) (l1, m1, n1) = (3, − 1, 0) ; (l2, m2, n2) = (2, 0, 3) The determinant becomes 3 −1 0 → → 3 − 1 0 = 0. Note that u and v are not parallel.
2
3
0 ∴ The lines are intersecting lines. Point of intersection : x−1 y−1 z+1 Take 3 = = 0 =λ −1 ∴ Any point on the line is of the form (3λ + 1, − λ + 1, − 1) x−4 y z+1 Take 2 = 0 = 3 = µ Any point on this line is of the form (2µ + 4, 0, 3µ − 1) Since they are intersecting, for some λ, µ (3λ + 1, − λ + 1, − 1) = (2µ + 4, 0, 3µ − 1) ⇒ λ = 1 and µ = 0 To find the point of intersection either take λ = 1 or µ = 0 ∴ The point of intersection is (4, 0, − 1). Note : If the two lines are in the vector form convert into cartesian form and do it. Example 2.45 : Find the shortest distance between the skew lines → → → → → → r = i − j +λ 2 i + j + k
( ) ( ) and → → → → → → → r = ( i + j − k ) + µ(2 i − j − k )
Solution : → → → → → → Compare the given equation with r = a 1 + t u and r = a 2 + s v , → → → → → → → → → → → a1= i − j ; a2= i + j − k ; u =2 i + j + k ; → → → → v=2 i − j − k
98
→ → → → → → → → a 2 − a 1 = 2 j − k and u × v = 4 j − 4 k → → → → a 2 − a 1 u v =
0 2 2
|→u × →v | = 4 distance =
→ → → → a 2 − a 1 u v
|→u × →v |
=
2 1 −1
−1
= 12 − 1 1
2
12 3 = 4 2 2
2.6.7 Collinearity of three points : Theorem (without proof) : → → → Three points A, B and C with position vectors a , b and c respectively are collinear if and only if there exists scalars λ1, λ2, λ3, not all zeros such that → → → → λ1 a + λ2 b + λ3 c = o and λ1 + λ2 + λ3 = 0 Working rule to find the collinearity : Write the equation of the line in cartesian form using any two points and verify the third point. Note : If the three points are collinear then their position vectors are coplanar, but the converse need not be true. Example 2.46 : Show that the points (3, − 1, − 1), (1, 0, − 1) and (5, − 2, − 1) are collinear. Solution : The equation of the line passing through (3, − 1, − 1) and (1, 0, − 1) is x−3 y+1 z+1 2 = − 1 = 0 = λ (say) Any point on the line is of the form (2λ + 3, − λ − 1, − 1) The point (5, − 2, − 1) is obtained by putting λ = 1. ∴ The third point lies on the same line. Hence the three points are collinear. Note : If the position vectors of the points are given then take the points and do the problem.
99
Example 2.47 : Find the value of λ if the points (3, 2, − 4), (9, 8, − 10) and (λ, 4, − 6) are collinear. Solution : Since the three points are collinear, the position vectors of the points are coplanar. → → → → → → → → → → → → Let a = 3 i + 2 j − 4 k ; b = 9 i + 8 j − 10 k ; c = λ i + 4 j − 6 k 3 2 −4 → → → a b c = 9 8 − 10 = 0
]
[
λ
−6
4 ⇒ 12λ = 60 ⇒ λ = 5
EXERCISE 2.7 (1) Find the shortest distance between the parallel lines (i)
→ → → → r = 2 i − j −k
(ii)
y z+3 x−3 y+1 z−1 x−1 = 3 = 2 and = 3 = 2 −1 −1
( ) + t(→i − 2→j + 3→k ) → → → → → → → r = ( i − 2 j + k ) + s( i − 2 j + 3 k )
and
(2) Show that the following two lines are skew lines : → → → → → → → r = 3 i + 5 j + 7 k + t i − 2 j + k and
( ) ( ) → → → → → → → r = ( i + j + k ) + s( 7 i + 6 j + 7 k )
x−1 y+1 z x−2 y−1 −z−1 1 = − 1 = 3 and 1 = 2 = 1 intersect and find their point of intersection.
(3) Show that the lines
(4) Find the shortest distance between the skew lines and
x−6 y−7 z−4 3 = −1 = 1
x y+9 z−2 = 2 = 4 −3
(5) Show that (2, − 1, 3), (1, − 1, 0) and (3, − 1, 6) are collinear. (6) If the points (λ, 0, 3), (1, 3, − 1) and (− 5, − 3, 7) are collinear then find λ.
100
2.7 Planes : A plane is defined as a surface such that the line joining of any two points on it lies completely on the surface. Vector and Cartesian Equations of the planes in parametric and nonparametric form : A plane is determined uniquely in the following cases : (i) Given a point on the plane and a normal to the plane. (ii) Given a normal to the plane and distance of the plane from the origin. (iii) Given a point and two parallel vectors to the plane. (iv) Given two points on it and a line parallel to the plane. (v) Given three non-collinear points. (vi) Equation of a plane that contains two given lines. (vii) Equation of a plane passing through the line of intersection of two given planes and a given point. Note : Eventhough the syllabus does not require the derivations (2.7.1 to 2.7.5) and it needs only the results, the equations are derived for better understanding the results.
2.7.1 Equation of a plane passing through a given point and perpendicular to a vector.
<
Vector Form : Let the plane pass through the point A whose position vector be → → a w.r.to O and perpendicular to the given vector n . Let P be any point on the plane and its z n → → position vector be r . Join AP A → → Here AP is perpendicular to n P a →→ → → → r ∴ AP. n = 0 ⇒ OP − OA. n = 0 O y → → → →→ →→ Fig. 2.27 r − a .n = 0⇒ r.n = a.n x This is the vector equation of the required plane (non parametric form.) Cartesian Form : → If (x1, y1, z1) are the coordinates of A and a, b, c are the direction ratios of n
(
)
→ → → → → → → → → → → → then a = x1 i + y1 j + z1 k ; n = a i + b j + c k ; r = x i + y j + z k
101
Now,
(→r − →a ) .→n = 0
→ → → → → → ⇒ (x − x1) i + (y − y1) j + (z − z1) k . a i + b j + c k = 0
(
)
⇒ a(x − x1) + b(y − y1) + c(z − z1) = 0 This is the cartesian equation of the plane (in non-parametric form). Corollary : The vector equation of the plane passing through the origin and → →→ perpendicular to the vector n is r . n = 0
2.7.2 Equation of the plane when distance from the origin and unit normal is given : z
Let p be the length of the perpendicular ON from the origin O ∧
<
to the given plane. Let n be the unit N vector normal to the plane in the direction O to N. P pn r → ∧ Then ON = pn. O Let P be any point on the plane → x and let its position vector be r Fig. 2.28 → → (i.e.,) OP = r . Join NP. → → NP lies on the plane and ON is perpendicular to the plane → → → → → ⇒ NP . ON = 0 ⇒ OP − ON. ON = 0
(→r − p∧n) . pn∧ = 0 i.e.,
⇒
∧ ∧ → ∧ r . n − pn . n = 0
→ ∧ r .n = p
(‡ ∧n . n∧ = 1)
This is the vector equation of the plane (in non-parametric form). Cartesian form : ∧ → → → → If l, m, n are the direction cosines of n then n = l i + m j + n k
→ → → ∧ → → → → r .n=p ⇒ x i +y j +zk . l i +m j +nk =p
(
)(
)
lx + my + nz = p This is the cartesian equation of the plane (in non-parametric form).
102
y
→ Corollary : If n is a normal vector but not a unit vector, → n ∧ then n = → n → → n → → → = p ⇒ r . n = p n = q (say) r . → n
| |
| |
| |
→ → r . n =q → This is the vector equation of the plane perpendicular to the vector n . q The length of the perpendicular from origin to this plane is → n
| |
2.7.3 Equation of the plane passing through a given point and parallel to two given vectors : → Let a be the position vector of the given point A referred to the → → origin O. Let u and v be the given vectors, which are parallel to the plane. Let P be any point on the plane → and let its position vector be r → → (i.e.,) OP = r .
z
v
u C
A
P B
a O
r
y
x
Fig. 2.29
→ → Through A, draw a lines AB and AC parallel to u and v lying in the → → → → planes such that AB = u and AC = v . → → → Now AP is coplanar with AB and AC → → → ∴ AP = s AB + t AC where s and t are scalars → → = su +tv → → → OP = OA + AP → → → ⇒ r = a+s u +t v … (1) This is the vector equation of the plane (in parametric form).
103
Cartesian form : → → → → Let a = x1 i + y1 j + z1 k → → → → → → → → u = l1 i + m1 j + n1 k ; v = l2 i + m2 j + n2 k From (1)
→ → → → r = a +su +tv
→ → → → → → → → → x i + y j + z k = x1 i + y1 j + z1 k + sl1 i + m1 j + n1 k → → → + t l2 i + m2 j + n2 k → → → Equating the coefficients i , j , k x = x1 + sl1 + tl2 y = y1 + sm1 + tm2
These are the parametric equations in cartesian form
z = z1 + sn1 + tn2 ⇒ x − x1 = sl1 + tl2 y − y1 = sm1 + tm2 z − z1 = sn1 + tn2
x − x1 Eliminating s and t, we get l1 l2
y − y1 z − z1 m1
n1
m2
n2
=0
This is the cartesian equation of the required plane (in non-parametric form). Non-parametric vector equation → → → → → → → AP, AB and AC are coplanar i.e., the vectors r − a , u , v are coplanar ∴
[→r − →a , →u , →v ] = 0
or
[→r →u →v ] = [→a , →u , →v ]
This is the vector equation of plane in non-parametric form.
104
2.7.4 Equation of the plane passing through two given points and parallel to a given vector : Vector Form : → → Let a and b be the position vectors of the points A and B (respectively) referred to the origin
z
v C
→ O. Let v be the given vector.
P
A a
The required plane passes through the points A and B and is parallel to the vector v. Let P be any point on the plane → and let its position vector be r → → (i.e.,) OP = r .
r
b
B y
O x
Fig. 2.30
→ → Through A, draw a line AC lying in the plane such that AC = v . → → → Now AP is coplanar with AB and AC → → → ∴ AP = s AB + t AC where s and t are scalars → → → → → → = s OB − OA + t v = s b − a + t v
(
)
→ → → OP = OA + AP ⇒
→ → → → → r = a +s b − a +t v
(
)
… (1)
→ → → → r = (1 − s) a + s b + t v This is the vector equation of the plane (in parametric form). Non-parametric vector equation → → → → → → → → AP, AB and AC are coplanar i.e., the vectors r − a , b − a and v are coplanar ∴
[→r − →a , →b − →a , →v ] = 0
This is the required vector equation of plane in non-parametric form.
105
Cartesian form : → → → → → → → → Let a = x1 i + y1 j + z1 k ; b = x2 i + y2 j + z2 k → → → → → → → → v = l i +m j +nk ; r =x i +y j +zk From (1) → → → → → → x i + y j + z k = x1 i + y1 j + z1 k → → → → → → + s (x2 − x1) i + (y2 − y1) j + (z2 − z1) k + t l i + m j + n k → → → Equating the coefficients of i , j , k x = x1 + s(x2 − x1) + tl These are the parametric equations y = y1 + s(y2 − y1) + tm in cartesian form z = z1 + s(z2 − z1) + tn ⇒ (x − x1) = s(x2 − x1) + tl (y − y1) = s(y2 − y1) + tm (z − z1) = s(z2 − z1) + tn x − x1 y − y1 z − z1
(
Eliminating s and t we get x2 − x1 l
y2 − y1
)
z2 − z 1 = 0 n
m This is the (non-parametric) equation of the plane in cartesian form. 2.7.5 Vector and cartesian equations of the plane passing through three given non-collinear points. → → → z Let a , b and c be the position C vectors of the points A, B and C A referred to the origin O. P The required plane passes a r c B through the points A, B and C. b y O Let P be any point on the plane → x and let its position vector be r → → (i.e.,) OP = r . Fig. 2.31 Now join AB, AC and AP. → → → AP is coplanar with AB and AC
106
→ → → ∴ AP = s AB + t AC where s and t are scalars → → → → = s OB − OA + tOC − OA =s
(→b − →a ) + t (→c − →a )
→ → → OP = OA + AP ⇒
→ → → → → → r = a +s b − a +t c − a
(
) (
) (or)
… (1)
→ → → → r = (1 − s − t) a + s b + t c This is the vector equation of the plane (in parametric form). Non-parametric vector equation : → → → AP, AB and AC are coplanar. → → → AP, AB, AC = 0 (i.e.,) → → → → → → ∴ r − a, b − a, c − a = 0 This is the required vector equation of plane in non-parametric form. Cartesian form : → → → → → → → → → → → → Let a = x1 i + y1 j + z1 k ; b = x2 i + y2 j + z3 k ; c = x3 i + y3 j + z3 k
[
]
→ → → → r =x i +y j +zk From (1) → → → → → → x i + y j + z k = x1 i + y1 j + z1 k → → → + s (x2 − x1) i + (y2 − y1) j + (z2 − z1) k → → → + t (x3 − x1) i + (y3 − y1) j + (z3 − z1) k → → → Equating the coefficients of i , j and k , we get x = x1 + s(x2 − x1) + t(x3 − x1) These are the parametric equations y = y1 + s(y2 − y1) + t(y3 − y1) in cartesian form z = z1 + s(z2 − z1) + t(z3 − z1)
107
x − x1 Eliminating s and t we get x2 − x1 x − x 3 1
y − y1
z − z1
=0 z3 − z 1
y2 − y1 z2 − z1 y3 − y1
This is the (non-parametric) equation of the plane in cartesian form. Example 2.48 : Find the vector and cartesian equation of a plane which is at a distance of 8 units from the origin and which is normal to the vector → → → 3 i +2 j −2k → → → → Solution : Here p = 8 and n = 3 i + 2 j − 2 k → → → → → → → n 3 i +2 j −2k 3 i +2 j −2k ∧ ∴n = = = 9+4+4 17 → n
| |
Hence the required vector equation of the plane is → ∧ r .n = p → → → → 3 i +2 j −2k r . =8 17 → → → → r . 3 i +2 j −2k
(
)= 8
17
Cartesian form is (xi +yj + zk). (3i + 2j − 2k) = 8 17 3x + 2y − 2z = 8 17 Example 2.49 : The foot of perpendicular drawn from the origin to the plane is (4, − 2, − 5), find the equation of the plane. Solution : The required plane passes through the point A(4, − 2, − 5) and is → perpendicular to OA. → → → → → → → → → ∴ a = 4 i − 2 j − 5 k and n = OA = 4 i − 2 j − 5 k →→ →→ ∴ The required equation of the plane is r . n = a . n → → → → → → → → → → r . 4 i −2 j −5k = 4 i −2 j −5k . 4 i −2 j −5k = 16 + 4 + 25
(
) (
→ → → → r . 4 i − 2 j − 5 k = 45
(
)
108
) (
)
… (1)
Cartesian form :
(x→i + y→j + z→k ) . (4→i − 2→j − 5→k ) = 45 4x − 2y − 5z = 45 Example 2.50 : Find the vector and cartesian equations of the plane through the point (2, − 1, − 3) and parallel to the lines x−2 y−1 z−3 x−1 y+1 z− 2 3 = 2 = − 4 and 2 = − 3 = 2 . Solution : The required plane passes through A(2, − 1, − 3) and parallel to → → → → → → → → u = 3 i + 2 j − 4 k and v = 2 i − 3 j + 2 k → → → → The required equation is r = a + s u + t v
→ → → → → → → → → → r = 2 i − j − 3 k +s 3 i + 2 j − 4 k + t 2 i − 3 j + 2 k Cartesian form : (x1, y1, z1) is (2, − 1, −3) ; (l1, m1, n1) is (3, 2, −4) ; (l2, m2, n2) is (2, −3, 2)
(
) (
) (
The equation of the plane is
i.e.,
x − 2 3 2
y+1 z+3 2
−4
−3
2
x − x1 l1 l2
)
y − y1 z − z1 m1
n1
m2
n2
=0
=0
⇒ 8x + 14y + 13z + 37 = 0 This is the required equation in cartesian form. Example 2.51 : Find the vector and cartesian equations of the plane passing through the points (− 1, 1, 1) and (1, − 1, 1) and perpendicular to the plane x + 2y + 2z = 5 → → → Solution : The normal vector to the plane x + 2y + 2z = 5 is i + 2 j + 2 k . This vector is parallel to the required plane. ∴ The required plane passes through the points (− 1, 1, 1) and (1, − 1, 1) → → → and parallel to the vector i + 2 j + 2 k .
109
Vector equation of the plane : The vector equation of the plane passing through two given points and parallel to a vector is → → → → r = (1 − s) a + s b + t v where s and t are scalars.
→ → → → → → → → → → → → Here a = − i + j + k ; b = i − j + k ; v = i + 2 j + 2 k → → → → → → → → → → ∴ r = (1 − s) − i + j + k + s i − j + k + t i + 2 j + 2 k This is the required vector equation of the plane. Cartesian form : (x1, y1, z1) is (− 1, 1, 1) ; (x2, y2, z2) is (1, − 1, 1) ; (l1, m1, n1) is (1, 2, 2)
(
) (
The equation of the plane is
i.e.,
x + 1 2 1
y−1 z−1 −2
0
x − x1 x2 − x1 l1
) (
y − y1
)
z − z1
y2 − y1 z2 − z1 m1
n1
=0
=0
2 2 ⇒ 2x + 2y − 3z + 3 = 0 Example 2.52 : Find the vector and cartesian equations of the plane passing through the points (2, 2, − 1), (3, 4, 2) and (7, 0, 6) Solution : Vector equation of the plane passing through three given noncollinear points is → → → → r = (1 − s − t) a + s b + t c where s and t are scalars. → → → → → → → → → → → Here a = 2 i + 2 j − k ; b = 3 i + 4 j + 2 k ; c = 7 i + 6 k → → → → → → → → → ∴ r = (1−s −t) 2 i + 2 j − k + s 3 i + 4 j + 2 k + t 7 i + 6 k Cartesian equation of the plane : Here (x1, y1, z1) is (2, 2, − 1) ; (x2, y2, z2) is (3, 4, 2) ; (x3, y3, z3) is (7, 0, 6)
(
) (
The equation of the plane is
x − x1 x2 − x1 x − x 3 1 110
) (
y − y1
z − z1
=0 z3 − z 1
y2 − y1 z2 − z1 y3 − y1
)
i.e.,
x − 2 1 5
y−2 z+1 2
3
=0
−2 7 5x + 2y − 3z = 17 This is the Cartesian equation of the plane. EXERCISE 2.8 (1) Find the vector and cartesian equations of a plane which is at a → distance of 18 units from the origin and which is normal to the vector 2 i → → +7 j +8k (2) Find the unit normal vectors to the plane 2x − y + 2z = 5. (3) Find the length of the perpendicular from the origin to the plane → → → → r . 3 i + 4 j + 12 k = 26. (4) The foot of the perpendicular drawn from the origin to a plane is (8, − 4, 3). Find the equation of the plane. (5) Find the equation of the plane through the point whose p.v. is → → → → → → 2 i − j + k and perpendicular to the vector 4 i + 2 j − 3 k . (6) Find the vector and cartesian equations of the plane through the point → → → → (2, − 1, 4) and parallel to the plane r . 4 i − 12 j − 3 k = 7. (7) Find the vector and cartesian equation of the plane containing the line z+ 1 x−2 y−2 z−1 x+1 y−1 2 = 3 = 3 and parallel to the line 3 = 2 = 1 . (8) Find the vector and cartesian equation of the plane through the point (1, 3, 2) and parallel to the lines x+1 y+2 z+3 x−2 y+1 z+2 = = 3 and 1 = 2 = 2 2 −1 (9) Find the vector and cartesian equation to the plane through the point (−1, 3, 2) and perpendicular to the planes x+2y+2z = 5 and 3x+y+2z = 8. (10) Find the vector and cartesian equation of the plane passing through the points A(1, − 2, 3) and B (− 1, 2, − 1) and is parallel to the line x−2 y+1 z−1 2 = 3 = 4 (11) Find the vector and cartesian equation of the plane through the points (1, 2, 3) and (2, 3, 1) perpendicular to the plane 3x− 2y + 4z − 5 = 0
(
)
(
111
)
(12) Find the vector and cartesian equation of the plane containing the line y−2 z−1 x−2 2 = 3 = − 2 and passing through the point (− 1, 1, − 1). (13) Find the vector and cartesian equation of the plane passing through the → → → → → → points with position vectors 3 i + 4 j + 2 k , 2 i − 2 j − k and → → 7 i + k. (14) Derive the equation of the plane in the intercept form. (15) Find the cartesian form of the following planes : → → → → (i) r = (s − 2t) i + (3 − t) j + (2s + t) k → → → → (ii) r = (1 + s + t) i + (2 − s + t) j + (3 − 2s +2 t) k
2.7.6 Equation of a plane passing through the line of intersection of two given planes : Vector form : The vector equation of the plane passing through the line of intersection of → → → → the planes r . n 1 = q1 and r . n 2 = q2 is → → → → r . n 1 − q 1 + λ r . n 2 − q2 = 0 → → → i.e. r . n 1 + λ n 2 = q1 + λq2 Cartesian form : The cartesian equation of the plane passing through the line of intersection of the planes a1x + b1y + c1z+ d1 = 0 and a2x + b2y + c2z+ d2 = 0 is (a1x + b1y + c1z + d1) + λ(a2x + b2y + c2z+ d2 ) = 0 Example 2.53 : Find the equation of the plane passing through the line of intersection of the plane 2x − 3y + 4z = 1 and x − y = − 4 and passing through the point (1, 1, 1). Solution : Any plane through the line of intersection of the given two planes is of the form (2x − 3y + 4z − 1) + λ(x − y + 4) = 0 1 But it passes through the point (1, 1, 1). ∴ λ = − 2 1 ∴ The equation of the required plane is (2x − 3y + 4z − 1)− 2 (x − y + 4)= 0 i.e., 3x − 5y + 8z − 6 = 0
112
Example 2.54 : Find the equation of the plane passing through the intersection of the planes 2x − 8y + 4z = 3 and 3x − 5y + 4z + 10 = 0 and perpendicular to the plane 3x − y − 2z − 4 = 0 Solution : The equation of the plane passing through the line of intersection of the given two planes is of the form (2x − 8y + 4z − 3) + λ (3x − 5y + 4z + 10) = 0 i.e., (2 + 3λ) x + (− 8 − 5λ)y + (4 + 4λ)z + (− 3 + 10λ) = 0 . But the required plane is perpendicular to the plane 3x − y − 2z − 4 = 0 ∴ Their normals are perpendicular. i.e., (2 + 3λ) 3 + (− 8 − 5λ) (− 1) + (4 + 4λ) (− 2) = 0 6λ + 6 = 0
⇒ λ=−1
∴ The required equation is (2x − 8y + 4z − 3) −1(3x − 5y + 4z + 10) = 0 − x − 3y − 13 = 0 x + 3y + 13 = 0
2.7.7 The distance between a point and a plane : Let (x1, y1, z1) be a point and ax + by + cz + d = 0 be the equation of the ax1 + by1 + cz1 + d plane. The distance between the point and the plane is a2 + b2 + c2 Corollary (1) : The distance between the origin and the plane ax + by + cz + d = 0 is d 2 2 2 a +b +c Corollary (2) : The distance between the two parallel planes ax + by + cz + d1 = 0 and d1 − d2 ax + by + cz + d2 = 0 is 2 a + b2 + c2 Note : If the given equation is in vector form, convert into cartesian form and find the distance. Example 2.55 : Find the distance from the point (1, − 1, 2) to the plane → → → → → → → → r = i + j + k +s i − j +t j − k
(
)
(
) (
113
)
Solution : The given plane is passing through the point (1, 1, 1) and parallel to two → → → → vectors i − j and j − k .
(
)
(
)
∴ The corresponding cartesian equation is of the form x − x1 y − y1 z − z1 (x1, y1, z1) = (1, 1, 1) l1 m1 n1 (l1, m1, n1) = (1 − 1, 0) =0 (l2, m2, n2) = (0, 1, − 1) m n l
2 x − 1 i.e., 1 0
2
y−1 −1
2 z−1 0 =0 −1
i.e., x + y + z − 3 = 0
1 Here (x1, y1, z1) = (1, −1, 2) 1 ax1 + by1 + cz1 + d 1 − 1 + 2 − 3 ∴ The distance = = = 3 2 2 2 1 + 1 + 1 a +b +c Find the distance between the parallel Example 2.56 : → → → → → → → → r . − i − j + k = 3 and r . i + j − k = 5 Solution : The corresponding cartesian equations of the planes are − x − y + z − 3 = 0 and x + y − z − 5 = 0 i.e., x + y − z + 3 = 0 and x + y − z − 5 = 0 d 1 − d2 3 + 5 8 = distance = 2 = 3 a + b2 + c2 1 + 1 + 1
(
)
(
planes
)
2.7.8 Equation of the plane which contain two given lines (i.e. passing through two given lines) → → → → → → Let r = a 1 + t u and r = a 2 + s v be the lines, lie on the plane. → → → → → → → → Clearly r − a 1, u , v are coplanar and r − a 2, u , v are also coplanar → → → → → → → → Thus r − a 1, u , v = 0 and r − a 2, u , v = 0 Note that the above two equations represent the same required plane. The cartesian form is
114
x − x1 l1 l2
y − y1 z − z1 m1
n1
m2
n2
=0
x − x2 or l1 l2
y − y2 z − z2 m1
n1
m2
n2
=0
→ → → → → → → → Where a 1 = x1 i + y1 j + z1 k ; a 2 = x2 i + y2 j + z2 k → → → → → → → → u = l1 i + m1 j + n1 k ; v = l2 i + m2 j + n2 k Note : (1) If the two lines are parallel then take the two trivial points from the lines and the parallel vector. Now find the equation of the plane passing through two points and parallel to a vector. (2) Through two skew lines, we can’t draw a plane. Example 2.57 : Find the equation of the plane which contains the two lines x−4 y−1 z x−1 y−2 z−3 2 = 3 = 4 and 5 = 2 = 1 Solution : Take the trivial point from the first line and the two parallel vectors i.e. (x1, y1, z1) = (1, 2, 3). (l1, m1, n1) = (2, 3, 4) and (l2, m2, n2) = (5, 2, 1) The required equation is x − x1 y − y1 z − z1 x−1 y−2 z−3
l1
m1
n1
l2
m2
n2
=0 ⇒ 2 5
3 2
4 =0 1
⇒ 5x − 18y + 11z − 2 = 0 Example 2.58 : Find the point of intersection of the line passing through the two points (1, 1, − 1) ; (− 1, 0, 1) and the xy-plane. Solution : The equation of the line passing through (1, 1, − 1) and (− 1, 0, 1) is y−1 z+1 x− 1 2 = 1 = −2 It meets the xy-plane i.e. z = 0 y−1 1 1 x−1 ∴ 2 = 1 = ⇒ x = 0, y = 2 −2 1 The required point is 0, 2, 0
115
Example 2.59 : Find the co-ordinates of the point where the line → → → → r = i +2 j −5k
( ) + t(2→i − 3→j + 4→k ) meets the plane → → → → r .(2 i + 4 j − k ) = 3 Solution : The equation of the straight line in the cartesian form is y−2 z+5 x−1 2 = − 3 = 4 = λ (say) ∴ Any point on this line is of the form (2λ + 1, − 3λ + 2, 4λ − 5). The cartesian equation of the plane is 2x + 4y − z − 3 = 0 But the required point lies on this plane. ∴ 2(2λ + 1) + 4(− 3λ + 2) − (4λ − 5) − 3 = 0 ⇒ λ = 1 ∴ The required point is (3, − 1, − 1) EXERCISE 2.9 (1) Find the equation of the plane which contains the two lines x−4 y−1 x+1 y−2 z−3 2 = − 3 = 4 and 3 = 2 = z − 8 (2) Can you draw a plane through the given two lines? Justify your answer. → → → → → → → r = i +2 j −4 k +t 2 i +3 j +6 k
( ) ( ) and → → → → → → → r = (3 i + 3 j − 5 k ) + s(−2 i + 3 j + 8 k ) (3) Find the point of intersection of the line → → → r = j − k
(
) + s(2→i − →j + →k ) and xz – plane
(4) Find the meeting point of the line → → → → r = 2 i + j −3k
(
) + t(2→i − →j − →k ) and the plane
x − 2y + 3z + 7 = 0 (5) Find the distance from the origin to the plane → → → → r . 2 i − j +5k
(
) =7
(6) Find the distance between the parallel planes x − y + 3z + 5 = 0 ; 2x − 2y + 6z + 7 = 0
116
2.7.9 Angle between two given planes : The angle between two planes is defined as the angle between their normals. → → → → Let the r . n 1 = q1 and r . n 2 = q2 the equations of the given two planes → → (where n 1 and n 2 are normals to the planes.) Now if θ be the angle between the two planes (i.e., between their normals) then
n2
θ
n1
Fig. 2.32
→ → n 1 . n 2 θ = cos → → n 1 n 2 −1
Note :
→ → (i) If the two planes are perpendicular then n 1 . n 2 = 0 → → (ii) If the two planes are parallel then n 1 = t n 2 where t is a scalar.
2.7.10 Angle between a line and a plane The angle between a line and a plane is the complement angle between the line and the normal to the plane. → → → Let r = a + t b be the line and → → r . n = q be the plane. If θ is the angle between the line and the plane then (90 − θ) is the angle between the line and the normal to the plane. i.e., (90 − θ) is the angle between → → b . n ∴ cos (90° − θ) = ⇒ sin θ = → → b n
| || |
n
b
90-θ θ
Fig. 2.33
→ → b and n → → b . n ⇒ θ = sin−1 → → b n
| || |
117
→ → b . n → → b n
| || |
Note : If the line is parallel to the plane i.e., the normal to the plane is → → perpendicular to the line then b . n = 0 Example 2.60 : Find the angle between 2x − y + z = 4 and x + y + 2z = 4 Solution : The normals to the given planes are → → → → → → → → n 1 = 2 i − j + k and n 2 = i + j + 2 k Let θ be the angle between the planes then → → n1. n2 3 1 π cos θ = = = ⇒θ=3 → 6 6 2 → n 1 n 2 Example 2.61 : Find the angle between the line → → → → r = i +2 j − k
( ) + µ (2→i + →j + 2→k ) and the plane → → → → r . (3 i − 2 j + 6 k ) = 0 Solution : Let θ be the angle between the line and the plane. → → b . n sinθ = → → b n
| || |
→ → → → → → → → b = 2 i + j +2k ; n =3 i −2 j +6k 16 16 sin θ = ⇒ θ = sin−1 21 3 ×7 EXERCISE 2.10 (1) Find the angle between the following planes : (i) 2x + y − z = 9 and x + 2y + z = 7 (ii) 2x − 3y + 4z = 1 and − x + y = 4 → → → → → → → → (iii) r . 3 i + j − k = 7 and r . i + 4 j − 2 k = 10 (2) Show that the following planes are at right angles. → → → → → → → → r . 2 i − j + k = 15 and r . i − j − 3 k = 3
(
(
(3)
)
(
)
)
( ) → → → → → → → → The planes r . (2 i + λ j − 3 k ) = 10 and r . (λ i + 3 j + k ) = 5
are perpendicular. Find λ.
118
(4) Find the angle between the line
x−2 y+1 z−3 3 = − 1 = − 2 and the plane
3x + 4y + z + 5 = 0 → → → → → → → (5) Find the angle between the line r = i + j + 3 k + λ 2 i + j − k → → → and the plane r . i + j = 1.
(
(
)
)
2.8 Sphere : A sphere is the locus of a point which moves in space in such a way that its distance from a fixed point remains constant. The fixed point is called the centre and the constant distance is called the radius of the sphere. Note : Eventhough the syllabus does not require the derivations (2.8.1, 2.8.2) and it needs only the results, the equations are derived for better understanding the results.
2.8.1 Vector equation of the sphere whose position vector of centre is → c and radius is a. Let O be the point of reference (origin) and C be the centre of the → sphere having position vector c → → (i.e.,) OC = c Let P be any point on the sphere → whose position vector be r → → (i.e.,) OP = r
C
a P
c
r
O
Fig. 2.34
→ → The radius of the sphere is given as a. (i.e.,) CP = a → → → From the figure (2.34) OP = OC + CP → → → r = c + a → → → r − c = a → → → ⇒ r − c = a … (1) This is the vector equation of the sphere. Corollary : Vector equation of a sphere whose centre is origin and radius is a.
|
| | |
119
→ → When O coincides with the centre C then c = o and the vector equation → → of the sphere (1) becomes r = a Cartesian form : → → → → Let r = x i + y j + z k → → → → c = c1 i + c2 j + c3 k
| | | |
→ → → → → ⇒ r − c = (x − c1) i + (y − c2) j + (z − c3) k 2
But 2
|→r − →c |
= a2
2
2
… (2) 2
From (2) (x − c1) + (y − c2) + (z − c3) = a … (3) This is the cartesian equation of the sphere whose centre is (c1, c2, c3) and raidus is a. Corollary : If the centre is at the origin, then the equation (3) takes the form x2 + y2 + z2 = a2. This is known as the standard form of the equation of the sphere. Note : General Equation of a Sphere : The equation x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 represents a sphere with centre (− u, − v, − w) and the radius = Note :
u2 + v2 + w2 − d
(i) the coefficients of x2, y2, z2 are equal. (ii) The equation does not contain the terms of xy, yz and zx.
2.8.2 Vector and Cartesian equations of the sphere when the extremities of the diameter being given : P
Let C be the centre of the sphere. Let A and B be the end points of the diameter AB. → Let a be the position vector of → the point A and b be the position vector of the point B with reference to the origin O.
A
B
C
a
r b O
Fig. 2.35
120
→ → → → (i.e.,) OA = a and OB = b → Let P be any point on the surface of the sphere. Let r be the position → → vector of P. (i.e.,) OP = r → → → → → AP = OP − OA = r − a → → → → → BP = OP − OB = r − b We know that the diameter AB subtends a right angle at P. → → ⇒ AP ⊥ BP → → ⇒ AP . BP = 0 ⇒
(→r − →a ) . (→r − →b ) = 0
… (1)
which is the required equation of the sphere. Cartesian Form : Let A(x1, y1, z1) and B(x2, y2, z2) be the end points of the diameter AB. Let P (x, y, z) be any point on the surface of the sphere. → → → → → → → → → → Now a = OA = x1 i + y1 j + z1 k ; b = OB = x2 i + y2 j + z2 k → → → → → r = OP = x i + y j + z k
(→r − →a ) . (→r − →b ) = 0 ( → → →) → → → ( → → →) → → → x i +y j + z k −(x1 i +y1 j +z1 k ) . x i +y j +z k −(x2 i +y2 j + z2 k ) = 0 From (1)
→ → → → → (x − x )→ 1 i + (y − y1) j + (z − z1) k . (x − x2) i + (y − y2) j + (z − z2) k =0 ⇒ (x − x1) (x − x2) + (y − y1) (y − y2) + (z − z1) (z − z2) = 0 This is required cartesian form of the equation of the sphere in terms of the end points of the diameter. Example 2.62 : Find the vector and cartesian equations of the sphere whose → → → centre is 2 i − j + 2 k and radius is 3.
121
Solution : We know that the vector equation of the sphere with centre and radius is → → r − c =a → → → → Here c = 2 i − j + 2 k and a = 3 → → → → ∴ The required vector equation is r − 2 i − j + 2 k = 3 Cartesian equation : → → → → Putting r = x i + y j + z k we get → → → → → → x i +y j +zk − 2 i − j +2k = 3
|
|
(
(
⇒
) (
)
)
|(x − 2)→i + (y + 1)→j + (z − 2)→k | = 3 2 |(x − 2)→i + (y + 1)→j + (z − 2)→k | = 32
(x − 2)2 + (y + 1)2 + (z − 2)2 = 9 ⇒ x2 + y2 + z2 − 4x + 2y − 4z = 0 Example 2.63 : Find the vector and cartesian equation of the sphere whose centre is (1, 2, 3) and which passes through the point (5, 5, 3). (5 − 1)2 + (5 − 2)2 + (3 − 3)2 16 + 9 = 25 = 5 → → → → Here a = 5 and c = i + 2 j + 3 k ∴ Vector equation of the sphere is → → r − c =a
Solution :
Radius = =
|
|
→ → → → r − i +2 j +3k = 5
(
)
→ → → → Cartesian Equation : Let r = x i + y j + z k From (1) → → → → → → x i +y j +zk − i +2 j +3k = 5
(
) (
)
|(x − 1)→i + (y − 2)→j + (z − 3)→k | = 5 (x − 1)2 + (y − 2)2 + (z − 3)2 = 25 x2 + y2 + z2 − 2x − 4y − 6z − 11 = 0
122
… (1)
Example 2.64 : Find the equation of the sphere on the join of the points A and B → → → → → → having position vectors 2 i + 6 j − 7 k and 2 i − 4 j + 3 k respectively as a diameter. Solution : Vector equation of the sphere is
(→r − →a ) . (→r − →b ) = 0
→ → → → → → → → a = 2 i + 6 j − 7 k and b = 2 i − 4 j + 3 k → → → → Let r = x i +y j +zk The required equation is → → → → → → → → → → → → x i +y j + z k − 2 i +6 j −7 k . x i +y j + z k − 2 i −4 j +3 k = 0 Here
( )( ) ( )( ) [(x − 2)→i + (y − 6)→j + (z+7)→k ] . [(x − 2)→i +(y+4)→j +(z−3)→k ] = 0 …(1) Cartesian Equation : From (1) (x − 2) (x − 2) + (y − 6) (y + 4) + (z + 7) (z − 3) = 0 ⇒ x2 + y2 + z2 − 4x − 2y + 4z − 41 = 0 Example 2.65 : Find the coordinates of the centre and the radius of the sphere → → → →2 → whose vector equation is r − r . 8 i − 6 j + 10 k − 50 = 0 → → → → Solution : Let r = x i + y j + z k → → → →2 → r − r . 8 i − 6 j + 10 k − 50 = 0 →2 r = x2 + y2 + z2 ⇒ x2 + y2 + z2 − 8x + 6y − 10z − 50 = 0 Here 2u = coefficient of x = − 8 ⇒ u = − 4 2v = coefficient of y = 6 ⇒ v = 3 2w = coefficient of z = − 10 ⇒ w = − 5 Centre : (− u, − v, − w) = (4, − 3, 5)
(
(
Radius : u2 + v2 + w2 − d = Example 2.66 :
)
)
16 + 9 + 25 + 50 = 100 = 10 uts.
→ → → Chord AB is a diameter of the sphere r − 2 i + j − 6 k = 18 with coordinate of A as (3, 2, − 2) Find the coordinates B.
(
123
)
→ → → Solution : The equation of the sphere is r − 2 i + j − 6 k = 18 ⇒ Centre of the sphere is (2, 1, − 6) → → → (i.e.,) Position vector of the centre is 2 i + j − 6 k We know that Centre is the mid point of diameter AB The co-ordinates of A are (3, 2, − 2) and let the coordinates of B be α, β, γ) α+3 β+2 γ−2 ∴ (2, 1, − 6) = 2 , 2 , 2 ⇒ α = 1, β = 0, γ = − 10 ∴ Coordinates of B are (1, 0, − 10)
(
)
EXERCISE 2.11 (1) Find the vector equation of a sphere with centre having position vector → → → 2 i − j + 3 k and radius 4 units. Also find the equation in cartesian form. (2) Find the vector and cartesian equation of the sphere on the → → → join of the points A and B having position vectors 2 i + 6 j − 7 k and → → → −2 i + 4 j − 3 k respectively as a diameter. Find also the centre and radius of the sphere. (3) Obtain the vector and cartesian equation of the sphere whose centre is (1, −1, 1) and radius is the same as that of the sphere → → → → r − i + j + 2 k = 5. (4) If A (− 1, 4, − 3) is one end of a diameter AB of the sphere
(
)
x2 + y2 + z2 − 3x − 2y + 2z − 15 = 0, then find the coordinates of B. (5) Find the centre and radius of each of the following spheres. → → → → (i) r − 2 i − j + 4 k = 5
( ) → → → → (ii) 2 r + (3 i − j + 4 k ) = 4
(iii) x2 + y2 + z2 + 4x − 8y + 2z = 5 → → → → → (iv) r 2 − r . 4 i + 2 j − 6 k − 11 = 0 (6) Show that diameter of a sphere subtends a right angle at a point on the surface.
(
)
124
3. COMPLEX NUMBERS 3.1 Introduction : The number system that we are aware of today is the gradual development from natural numbers to integers, from integers to rational numbers and from rational numbers to the real numbers. If we consider the following polynomial equations (i) x − 1 = 0, (ii) x + 1 = 0, (iii) x + 1 = 1, (iv) 2x + 1 = 0 and (v) x2 − 3 = 0, we see that all of them have solutions in the real number system. However this real number system is not sufficient to solve equations of the form x2 + 9 = 0 i.e., there does not exist any real number which satisfies x2 = − 9. The mathematical need to have solutions for equations of the above form led us to extend the real number system to a new kind of number system that allows the square root of negative numbers. Let us consider solution of a simple quadratic equation x2 + 16 = 0. Its solutions are x = ± 4 − 1 . We assume that square root of − 1 is denoted by the symbol i, called the imaginary unit. Thus for any two real numbers a and b, we can form a new number a + ib. This number a + ib is called a complex number. The set of all complex numbers is denoted by ÷ and the nomenclature of a complex number was introduced by C.F. Gauss, a German mathematician. Hence the extension of the concept of numbers from real numbers enables one to solve any polynomial equation. The symbol i was first introduced in mathematics by the famous Swiss mathematician, Leonhard Euler (1707 – 1783) in 1748. ‘i’ is the first letter of the Latin word “imaginarius” and it is also referred to as ‘iota’, a Greek alphabet. Later on the subject was enriched by the original work of A.L. Cauchy, B. Riemann, K. Weierstrass and others.
3.2 The complex number system : A complex number is of the form a + ib where ‘a’ and ‘b’ are real numbers and i is called the imaginary unit, having the property that i2 = − 1. If z = a + ib then a is called the real part of z, denoted by Re(z) and b is called the imaginary part of z and is denoted by Im(z) . 2 Some examples of complex numbers are 3 − i2, 2 + i3, − 5 + i.
125
Note that 3 is the real part and − 2 is the imaginary part of 3 − i2 and so on. Two complex numbers a + ib and c + id are equal if and only if a = c and b = d. i.e., the corresponding real parts are equal and the corresponding imaginary parts are equal. The real numbers can be considered as a subset of the set of complex numbers with b = 0. Hence the complex numbers 0 + i0 and − 2 + i0 represents the real numbers 0 and − 2 respectively. If a = 0 the complex number 0 + ib or ib is called a pure imaginary number. Negative of a complex number : If z = a + ib is a complex number then the negative of z is denoted by − z and it is defined as − z = − a + i(− b)
Basic Algebraic operations : Addition : (a + ib) + (c + id) = (a + c) + i (b + d) (a + ib) − (c + id) = (a − c)+ i(b − d) Subtraction : To perform the operations with complex numbers we can proceed as in the algebra of real numbers replacing i2 by − 1 whenever it occurs. Multiplication :
(a + ib) (c + id) = ac + iad + ibc + i2bd = (ac − bd) + i (ad + bc)
3.3 Conjugate of a complex number :
If z = a + ib, then the conjugate of z is denoted by z and is defined by z = a − ib c − id a + ib a + ib Division : c + id = c + id × c − id Multiplying the numerator and denominator by the conjugate of the denominator, we get a + ib ac + bd bc − ad c + id = c2 + d2 + i c2 + d2
3.3.1 Properties :
(i) zz = (a + ib) (a − ib) = a2 + b2 which is a non-negative real number. = (ii) Conjugate of z is z i.e., z = z (iii) If z is real, i.e., b = 0 then z = z . Conversely, if z = z , i.e., if a + ib = a − ib then b = − b ⇒ 2b = 0 ⇒ b = 0 (‡ 2 ≠ 0 in the real number system). ∴ b = 0 ⇒ z is real. Thus z is real ⇔ the imaginary part is 0
126
(iv) Let z = a + ib. Then z = a − ib
∴ z + z = (a + ib) + (a − ib) = 2a
z+ z ⇒ a = Re(z) = 2
z− z Similarly, b = Im(z) = 2i (v) The conjugate of the sum of two complex numbers z1, z2 is the sum of
their conjugates i.e., z1 + z2 = z1 + z2
z1 = a + ib and z2 = c + id
Proof : Let Then
z1 + z2 = (a + ib) + (c + id) = (a + c) + i (b + d)
z1 + z2 = (a + c) − i (b + d) z1 z1
= a − ib, z2 = c − id
+ z2 = (a − ib) + (c − id) = (a + c) − i(b + d)
= z1 + z2 Similarly it can be proved that the conjugate of the difference of two complex numbers z1, z2 is the difference of their conjugates.
i.e., z1 − z2 = z1 − z2
(vi) The conjugate of the product of two complex numbers z1, z2 is the product of their conjugates. z1z2
= z 1 z2 Proof : Let z1 = a + ib and z2 = c + id. Then i.e.,
z1 z2 = (a + ib) (c + id) = (ac − bd) + i(ad + bc) z1z2 z1 z1 z 2
= (ac − bd) − i(ad + bc)
= a − ib, z2 = c − id = (a − ib) (c − id) = (ac − bd) − i(ad + bc)
= z1z2
127
(vii) The conjugate of the quotient of two complex numbers z1, z2, (z2 ≠ 0) is the quotient of their conjugates.
z1 z 1 i.e., z = (without proof) 2 z 2 n (viii) zn = ( z ) Example 3.1 : Write the following as complex numbers
(i) − 35 Solution : (i)
− 35 =
(ii) 3 − (− 1) × (35) =
−7 −1 .
35 = i 35
(ii) 3 − − 7 = 3 − (− 1) × 7 = 3 − − 1 7 = 3 − i 7 Example 3.2 : Write the real and imaginary parts of the following numbers : 3 (i) 4 − i 3 (ii) 2 i Solution : (i) Let z = 4 − i 3 ; Re(z) = 4, Im(z) = − 3 3 3 (ii) Let z = 2 i ; Re(z) = 0, Im(z) = 2 Example 3.3 : Find the complex conjugate of (i) 2 + i 7, (ii) − 4 − i9 (iii) 5 Solution : By definition, the complex conjugate is obtained by reversing the sign of the imaginary part of the complex number. Hence the required conjugates are (i) 2 − i 7, (ii) − 4 + i9 and (iii) 5 (‡ the conjugate of any real number is itself). Example 3.4 : Express the following in the standard form of a + ib (i) (3 + 2i) + (− 7 − i) (ii) (8 − 6i) − (2i − 7) 5 + 5i (iii) (2 − 3i) (4 + 2i) (iv) 3 − 4i Solution : (i) (3 + 2i) + (− 7 − i) = 3 + 2i − 7 − i = − 4 + i (ii) (8 − 6i) − (2i − 7) = 8 − 6i − 2i + 7 = 15 − 8i
128
(iii) (2 − 3i) (4 + 2i)=8 + 4i − 12i − 6i2 = 14 − 8i (iv)
15 + 20i + 15i + 20i2 5 + 5i 5 + 5i 3 + 4i = × 3 + 4i = 3 − 4i 3 − 4i 32 + 42 =
Note :
−1 7 − 5 + 35i = 5 +5i 25
i4 = 1 i3 = − i i2 = − 1 (i)4n = 1
(i)4n − 1 = − i (i)4n − 2 = − 1 ; n ∈ z Example 3.5 : Find the real and imaginary parts of the complex number 3i20 − i19 2i − 1 Solution : z=
(i)
3i20 − i19 z= = 2i − 1 = = =
10
9
3(i2) − (i2) i 2i − 1 3(− 1)10 − (− 1)9i − 1 + 2i 3+i − 1 + 2i − 1 − 2i 3+i × − 1 + 2i − 1 − 2i − 3 − 6i − i − 2i2 (− 1)2 + 22 −1 7 − 1 − 7i = 5 − 5 i = 5 1 −7 − 5 and Im(z) = 5 =
Re(z) =
3 −1 Example 3.6 : If z1 = 2 + i, z2 = 3 − 2i and z3 = 2 + 2 i find the conjugate of (i) z1z2
(ii) (z3)4 129
Solution :
(i) Conjugate of z1z2 is z1 z2 i.e. (2 + i) (3 − 2i) = = =
(ii)
(2 + i) (3 − 2i) (2 − i) (3 + 2i) (2 − i) (3 + 2i)
= 6 + 4i − 3i − 2i2 = 6 + 4i − 3i + 2 = 8+i 4 4 1 3 z34 = z3 = − 2 + 2 i 2
2
2 1 3 3 3 1 = − 2 − 2 i = 4 + 2 i + 4 i2 2
1 3 3 3 −1 = 2 + 2 i = 4 − 2 i + 4 i2 1 3 = −2 − 2 i EXERCISE 3.1 (1) Express the following in the standard form a + ib (1 + i) (1 − 2i) 2(i − 3) (ii) (i) 2 1 + 3i (1 + i) i4 + i9 + i16 3 − 2i8 − i10 − i15 (2) Find the real and imaginary parts of the following complex numbers: 1 2 + 5i (i) 1 + i (ii) (iii) (2 + i) (3 − 2i) 4 − 3i (iii) (− 3 + i) (4 − 2i) (iv)
n
1 + i =1 (3) Find the least positive integer n such that 1 − i (4) Find the real values of x and y for which the following equations are satisfied. (i) (1 − i)x + (1 + i)y = 1 − 3i (1 + i)x − 2i (2 − 3i)y + i + =i (ii) 3+i 3−i (iii)
x2 + 3x + 8 + (x + 4)i = y(2 + i)
130
(5) For what values of x and y, the numbers − 3 + ix2y and x2 + y + 4i are complex conjugate of each other?
3.4 Ordered pair Representation : In view of the representation of complex numbers, it is desirable to define a complex number a + ib as an ordered pair (a, b) of real numbers a and b subject to certain operational definitions. These definitions are as follows: (i) Equality : (a, b) = (c, d) if and only if a = c, b = d (ii) Sum : (a, b) + (c, d) = (a + c, b + d) (iii) Product : (a, b) . (c, d) = (ac − bd, ad + bc) m(a, b) = (ma, mb)
Result : The imaginary unit i is defined as i = (0, 1). We have (a, b) = (a, 0) + (0, b) = a(1, 0) + b(0, 1) and (0, 1) (0, 1) = (0 − 1, 0 + 0) = (− 1, 0). By identifying (1, 0) with 1 and (0, 1) with i we see that (a, b) = a + ib. Thus we associate the complex number a + ib with the ordered pair (a, b). The ordered pair (0, 0) corresponds to the real number 0. Remark : Though the set of real numbers is ordered, the set of complex numbers is not ordered. i.e., order relation does not exist in ÷. Given two complex numbers z1 and z2 we cannot say z1 < z2 or z1 > z2. We can only say that z1 = z2 or z1 ≠ z2, since the points are represented in a plane. Thus the order relations ‘greater than’ and ‘less than’ are not definied for complex numbers. i.e., the inequalities like 1` + i > 3 − 2i, i > 0, (3 + i) < 2 etc. are meaningless.
3.5 Modulus of a complex number : Let z = a + ib be a complex number. The modulus or absolute value of z denoted by | z | is defined by |z|=
a2 + b2
From the definition, it is obvious that | z | = | z |. Since a2 + b2 = zz ,
| z | = zz (Taking the positive square root) Result : Let z = x + iy Now, x < x2 + y2, then Re(z) < | z | for y ≠ 0 If y = 0 then x = | z | then Re(z) = | z |
131
… (1) … (2)
Re(z) ≤ | z |
Combining (1) and (2) Similarly Im(z) ≤ | z |
Example 3.7 : Find the modulus of the following complex numbers: (i) − 2 + 4i (ii) 2 − 3i (iii) − 3 − 2i (iv) 4 + 3i Solution : (i)
| − 2 + 4i | =
(− 2)2 + 42 =
20 = 2 5
(ii)
|2−3i| =
22 + (− 3)2 =
13
(iii)
| − 3 − 2i | =
(iv)
| 4 + 3i | =
2
2
(− 3) + (− 2) = 42 + 32 =
13
25 = 5
3.5.1 Properties : If z1, z2, … zm are complex numbers, then the following properties hold. (i) The modulus of a product of two complex numbers is equal to the product of their moduli. i.e. |z1 z2 | = | z1 | | z2 | Proof :
|z1 z2 |2 = (z1z2) z1 z2
[‡ zz = | z |2]
= (z1 z2) z1 z2 =
(z1 z1 )( z2 z2 )
= | z1 | 2 | z 2 | 2 Taking the positive square root on both sides, we get |z1 z2 | = | z1 | | z2 | Note : This result can be extended to any finite number of complex numbers i.e., | z1 z2 … zn | = | z1 | | z2 | … | zn | (ii) The modulus of a quotient of two complex numbers is equal to the quotient of their moduli. z1 | z1 | i.e., z = | z | where z2 ≠ 0. 2 2
132
Proof : Since z2 ≠ 0,
z1 z1 z1 = z . z2 and so | z1 | = z | z2 | (by the previous 2 2
result) Therefore Hence
| z1 | z1 = z | z2 | 2 z1 | z1 | z = | z | 2 2
(iii) Triangle inequality : The modules of sum of two complex numbers is always less than or equal to the sum of their moduli. i.e., | z1 + z2 | ≤ | z1 | + | z2 | Proof : Let z1 and z2 be two complex numbers.
We know that | z1 + z2 |2 = (z1 + z2) (z1 + z2)
[‡ | z |2 = zz ]
( )
= (z1 + z2) z1 + z2
= z1 z1 + z1 z2 + z2 z1 + z2 z2
= z1 z1 + z2 z2 + z1 z2 + z1 z2
( )
= | z1 |2 + | z2 |2 + 2 Re z1 z2
≤ | z1 |2 + | z2 |2 + 2 | z1 z 2 | = | z1 |2 + | z2 |2 + 2 | z1| | z2|
[‡ Re (z) ≤ | z |]
‡| z | = |z|
= [ | z11 + | z2 |]2 ∴ | z1 + z2 |2 ≤ [| z1 | + | z2 |]2 Thus taking positive square root we get | z 1 + z2 | ≤ | z 1 | + | z2 | Note : 1 Writing − z2 for z2 in this result We also have | z1 − z2 | ≤ | z1 | + | − z2 | ⇒ | z1 − z2 | ≤ | z1 | + | z2 |
133
Note : 2 The above inequality can be immediately extended by induction to any finite number of complex numbers i.e., for any n complex numbers z1, z2, z3, …, zn |z1 + z2 + z3 + … + zn | ≤ | z1 | + | z2 | + … + | zn | (iv) The modulus of the difference of two complex numbers is always greater than or equal to the difference of their moduli. Proof : Let z1 and z2 be two complex numbers.
We know that | z1 − z2 |2 = (z1 − z2) (z1 − z2)
[Q| z |2 = z z ]
( )
= (z1 − z2) z1 − z2
= z1 z1 − z1 z2 − z2 z1 + z2 z2
(
= z1 z1 + z2 z2 − 2 Re z1 z2
)
≥ | z1 |2 + | z2 |2 − 2 | z1 z 2 |
[Q Re (z) ≤ | z | − Re (z) ≥ − | z |]
2
z2
2
= | z1 | + | z2 | − 2 | z1 | | | = | z1 |2 + | z2 |2 − 2 | z1 | | z2 | = [ | z1| − | z2 |]2 ∴ | z1 − z2 |2 ≥ [| z1 | − | z2 |]2 Thus taking positive square root we get | z1 − z 2 | ≥ | z 1 | − | z 2 | Example 3.8 : Find the modulus or the absolute value of
(1 + 3i) (1 − 2i) (3 + 4i)
Solution :
(1 + 3i) (1 − 2i) = | 1 + 3i | | 1 − 2i | | 3 + 4i | (3 + 4i) =
12 + 32 2
12 + (2)2
3 +4 =
10 5 = 5
134
2
2
=
10 5 25
3.6. Geometrical Representation 3.6.1 Geometrical meaning of a Complex Number If real scales are chosen on two mutually perpendicular axes X′OX and Y′OY (called the x axis and y axis respectively), We can locate any point in the plane determined by these lines, by the ordered pair of real numbers (a, b) called rectangular co-ordinates of the point. Since every complex number a + ib can be considered as an ordered pair (a, b) of real numbers, we can represent such number by a point P in the xy plane, called the complex plane. Such a representation is also known as the Argand diagram. The complex number represented by P can therefore be read as either (a, b) or a + ib.
y P(z)
|
z| x
Fig. 3.1
With this representation the modulus of the complex number z = a + ib a2 + b2. The → complex number z = a + ib can also be represented by the vector OP (Fig. 3.1) where P = (a, b) and pictured as an arrow from the origin to the point (a, b). To each complex number there corresponds one and only one point in the plane, and conversely to each point in the plane there corresponds one and only one complex number. Because of this we often refer to the complex number z as the point z. represents the distance between z and the origin i.e., | z | =
Clearly, the set of real numbers (x, 0) corresponds to the x-axis called real axis. The set of all purely imaginary number (0, y) corresponds to the y-axis called the imaginary number axis. The origin identifies the complex number 0 = 0 + i0.
3.6.2 Polar form of a Complex Number : Let (r, θ) be the polar co-ordinates of the point P = P(x, y) in the complex plane corresponding to the complex number z = x + iy.
135
Then we get from the figure (Fig. 3.2), OM x PM y cos θ = OP = r and sinθ = OP = r x = r cos θ ; y = r sin θ 2
y P(x,y)
2
r
where r = x + y = | x + iy | is called y the modulus or the absolute value of )θ z = x + iy denoted by mod z or | z | x x M O (i.e., the distance from the origin to the Fig. 3.2 point z) y −1 y tan θ = x , ∴ θ = tan x is called the amplitude or argument of z = x + iy denoted by amp z or arg z and is measured as the angle which the line OP makes with the positive x-axis (in the anti clockwise sense). Thus z = x + iy = r(cosθ + i sin θ) is called the polar form or the modulus amplitude form of the complex number. It is sometimes convenient to use the abbreviation cis θ for cos θ + i sin θ. y θ = tan−1 x is applicable only for first quadrant numbers i.e., x & y are positive.
3.6.3 Principal Value : The argument of z is not unique. Any two distinct arguments of z differ from each other by an integral multiple of 2π. In order to specify a unique value of arg z, we may restrict its value to some interval of length 2π. For this purpose, we introduce the concept of “principal value” for arg z as follows : For an arbitrary z ≠ 0 the principal value of arg z is defined to be the unique value of z that satisfies − π < arg z ≤ π. Note : For z = 0, the argument is indeterminate.
Results : (1) For any two complex numbers z1 and z2 (ii) arg (z1 . z2) = arg z1 + arg z2 (i) | z1 z2 | = | z1 | . | z2 | Proof : Let z1 = r1 (cos θ1 + i sin θ1) and z2 = r2 (cos θ2 + i sin θ2) then
| z1 | = r1, arg z1 = θ1 ; | z2 | = r2, arg z2 = θ2 z1.z2 = r1r2 (cos θ1 + i sin θ1) . (cos θ2 + i sin θ2) = r1r2 [(cosθ1 . cos θ2 − sinθ1 . sin θ2) + i (sin θ1 . cos θ2 + cos θ1 . sin θ2)]
136
= r1r2 [cos (θ1 + θ2) + i sin (θ1 + θ2)] ∴ | z1 z2 | = r1r2 = | z1 | . | z2 | and arg (z1 z2) = θ1 + θ2 = arg z1 + arg z2 Note : This result can be extended to any finite number of complex numbers i.e., (i) | z1 . z2 … zn | = | z1 | . | z2 | … | zn | (ii) arg (z1 z2 … zn) = arg z1 + arg z2 + … + arg zn (2) For any two complex numbers z1 and z2
z1 | z1 | z1 (i) z = | z | , (z2 ≠ 0) (ii) arg z = arg z1 − arg z2 2 2 2 Proof : Let z1 = r1 (cos θ1 + i sin θ1) and z2 = r2 (cos θ2 + i sin θ2) Then Now
| z1 | = r1, arg z1 = θ1 and | z2 | = r2, arg z2 = θ2 z1 r1 (cos θ1 + i sin θ1) z2 = r2 (cos θ2 + i sin θ2) r1 (cos θ1 + i sin θ1) (cos θ2 − i sin θ2) =r 2 (cosθ2 + i sin θ2) (cos θ2 − i sin θ2) r1 (cosθ1 cosθ2 + sinθ1 sinθ2) + i (sinθ1 cosθ2−cosθ1 sin θ2) =r 2 cos2θ + sin2θ 2
2
r1 = r [cos (θ1 − θ2) + i sin (θ1 − θ2)] 2
z1 r1 | z1 | ∴ z = r = | z | and 2 2 2 z 1 arg z = θ1 − θ2 = arg z1 − arg z2 2 Exponential form of a Complex Number : The symbol eiθ or exp (iθ) (called exponential of iθ) is defined by eiθ = cos θ + i sin θ This relation is known as Euler’s formula.
137
If z ≠ 0 then z = r (cos θ + i sin θ) = reiθ. This is called the exponential form of the complex number z. By straight forward multiplication of iθ iθ e 1 = (cos θ1 + i sin θ1) and e 2 = cos θ2 + i sin θ2 iθ
i(θ + θ )
iθ
we have e 1.e 2 = e 1 2 Remarks : (1) If θ1 = θ and θ2 = − θ in the above definition then we have eiθ . ei(− θ) = ei(θ − θ) = ei0 = 1 1 ⇒ ei(− θ) = iθ . Thus writing ei(− θ) as e− iθ e 1 we observe that e−iθ = iθ e 2
(2) If θ1 = θ2 = θ then (eiθ) = e2iθ . By mathematical induction it can be n
shown that (eiθ) = einθ where n = 0, 1, 2 …
(3) Since eiθ = e− iθ, we see that, if z = reiθ then z = re− iθ (4) Two complex numbers z1 = r1eiθ1 and z2 = r2eiθ2 are equal if and only if r1 = r2 and θ1 = θ2 + 2nπ, n ∈ Z (the set of all integers) General rule for determining the argument θ. Let z = x + iy θ=π−α where x, y ∈ R θ=−π+α | y | Take α = tan−1 | x | (i) Both cos θ and sin θ are + νe. z lies in the first quadrant. (ii) Sin θ is + νe, cos θ is − νe z lies in the second quadrant. (iii) Both sin θ and cos θ are − ve z lies in the third quadrant. (iv) Sin θ is − νe and cos θ is + νe, z lies in the fourth quadrant.
138
θ=α θ=−α
θ=α θ=π−α θ=−π+α θ=−α
Example 3.9 : Find the modulus and argument of the following complex numbers : (i) − 2 + i 2 (ii) 1 + i 3 (iii) − 1 − i 3 Solution : (i) Let − 2 + i 2 = r(cos θ + i sin θ) Equating the real and imaginary parts separately r cos θ = − 2 r sin θ = 2 r2 cos2 θ = 2
r2 sin2θ = 2
r2 (cos2θ + sin2θ) = 4 r = 4=2 − 2 −1 cos θ = 2 = 2 1 2 sin θ = 2 = 2 π 3π θ = π−4= 4 3π modulus r = 2, argument θ = 4
⇒ θ in the 2nd quadrant
3π 3π Hence − 2 + i 2 = 2 cos 4 + i sin 4 (ii) Let 1 + i 3 = r(cos θ + i sin θ) Equating the real and imaginary parts separately r cos θ = 1 2
2
r cos θ = 1
r sin θ = 3 r2 sin2θ = 3
r2 (cos2θ + sin2θ) = 4 ⇒ r = 2 1 cos θ = 2 ⇒ θ in the 1st quadrant 3 sinθ = 2
π θ=3 modulus r = 2,
‡ θ = tan−1 y x
π argument θ = 3
139
π π Hence 1 + i 3 = 2 cos 3 + i sin 3
(iii) Let − 1 − i 3 = r(cos θ + i sin θ) Equating the real and imaginary parts separately r cos θ = − 1 r sin θ = − 3 r2 cos2 θ = 1
r2 sin2θ = 3
r2 (cos2θ + sin2θ) = 4 ⇒ r=2 −1 cos θ = 2 − 3 sinθ = 2
⇒ θ in the 3rd quadrant
π − 2π θ = −π+3= 3 modulus r = 2,
− 2π argument θ = 3
2π 2π − 2π − 2π Hence − 1 − i 3 = 2 cos 3 + i sin 3 = 2 cos 3 − i sin 3 Example 3.10 : If (a1 + ib1) (a2 + ib2) … (an + ibn) = A + iB, prove that (i) (a12 + b12) (a22 + b22) … (an2 + bn2) = A2 + B2 B b1 b2 bn (ii) tan−1 a + tan−1 a + … + tan−1 a = kπ + tan−1 A, k ∈ Z
1
2
n
Solution : Given (a1 + ib1) (a2 + ib2) … (an + ibn) = A + iB | (a1 + ib1) (a2 + ib2) … (an + ibn) | = | A + iB | | (a1 + ib1) | | (a2 + ib2) | … | (an + ibn) | = | A + iB | a12 + b12
a22 + b22 …
an2 + bn2 =
A2 + B2
Squaring on both sides (a12 + b12) (a22 + b22) … (an2 + bn2) = A2 + B2 Also arg [(a1 + ib1) (a2 + ib2) … (an + ibn)] = arg (A + iB)
140
arg (a1 + ib1) + arg (a2 + ib2) … + arg (an + ibn) = arg (A + iB)
… (1)
bi Now arg (ai + ibi) = tan−1 a i
Hence (1) becomes B b1 b2 bn tan−1 a + tan−1 a + … + tan−1 a = tan−1 A 1 2 n By taking the general value, B b1 b2 bn tan−1 a + tan−1 a + … + tan−1 a = kπ + tan−1 A 1 2 n Example 3.11 : P represents the variable complex number z, find the locus of P if π z+1 z−1 (i) Re z + i = 1 (ii) arg z + 1 = 3 Solution : Let z = x + iy then (x + 1) + iy z + 1 x + iy + 1 (i) z + i = x + iy + i = x + i(y + 1) [(x + 1) + iy] [x − i(y + 1)] = x + i(y + 1) × [x − i(y + 1)] x(x + 1) + y(y + 1) + i(yx − xy − x − y − 1) = x2 + (y + 1)2 x(x + 1) + y(y + 1) + i(− x − y − 1) = x2 + (y + 1)2 z+1 Given that Re z + i = 1 x(x + 1) + y(y + 1) =1 ∴ x2 + (y + 1)2 x2 + y2 + x + y = x2 + y2 + 2y + 1 ⇒ x − y = 1 which is a straight line. ∴ The locus of P is a straight line. π z−1 (ii) arg z + 1 = 3 ⇒
141
π ∴ arg (z − 1) − arg (z + 1) = 3 π arg(x + iy − 1) − arg (x + iy + 1) = 3 π arg [(x − 1) + iy] − arg [(x + 1) + iy] = 3 y y π tan−1 − tan−1 x + 1 = 3 x−1 y y −x+1 x−1 π tan−1 y y = 3 1+ x − 1 x + 1
π 2y 2 = tan 3 x −1+y 2y = 3 2 x + y2 − 1
⇒
2
2y =
3x2 + 3y2 − 3
∴ 3x2 + 3y2 − 2y − 3 = 0 is the required locus. Result : (without proof) : If | z − z1 | = | z − z2 | then the locus of z is the perpendicular bisector of the line joining the two points z1 and z2.
3.6.4 Geometrical meaning of conjugate of a complex number : y
Let z = x + iy be a complex number represented by P in the Argand diagram. Then we know that its conjugate z is given by z = x − iy.
i.e., z = (x, y) ⇒ z = (x, − y)
∴ If Q represents the conjugate z , then conjugate of z is obviously the mirror image of the complex point z on the real axis (Fig. 3.3). This clearly indicates that = z = z ⇔ z is purely a real number. Also z = z. In polar coordinates let 142
P( z )
r O
)θ )-θ
x
r Q( z )
Fig. 3.3
z = r(cos θ + i sin θ) then z = r(cos (− θ) + i sin (− θ)) so if z = (r, θ) then z = (r, − θ)
Thus the moduli of both z and z are same i.e., r = x2 + y2 But the amplitude of z is θ and that of z is − θ. Hence | z | = | z | and amp z = − amp z. y
Fig. 3.4 gives the simple geometric relations among the complex number z, its negation − z and its conjugate z − z = (− x, − y). It is the point symmetrical to z about the origin.
z
O
-z
x
z
Fig. 3.4
3.6.5 Geometrical representation of sum of two complex numbers Let A and B represent the two complex numbers z1 = x1 + iy1 and z2 = x2 + iy2 in the Argand diagram. Complete the parallelogram OACB. Then C represents the complex number z1 + z2. Proof : y Since OACB is a parallelogram, C (z1+z2) diagonals OC and AB bisect at M. ∴ From Fig. 3.5, the midpoint M of the B( z2 ) line joining A(x1, y1) and B(x2, y2) is x1 + x2 y1 + y2 M 2 , 2 … (1) A ( z1 ) If C is (h, k) then midpoint M of OC is x 0+h 0+k O also given by 2 , 2 Fig. 3.5 k h … (2) i.e., M is 2 , 2 ∴ From (1) and (2) y1 + y2 h x1 + x2 k = ; = 2 2 2 2 ⇒ h = x1 + x2 ; k = y1 + y2
143
∴ C is (x1 + x2, y1 + y2) Hence C represents the complex number, z1 + z2 Note : OA = | z1 |, OB = | z2 | and OC = | z1 + z2 | In any triangle the sum of two sides is greater than the third side. ∴ from ∆OAC we have OA + AC > OC or OC < OA + AC … (1) | z 1 + z 2 | < | z1 | + | z 2 | Further, if the points are collinear … (2) | z 1 + z 2 | = | z1 | + | z 2 | Combining (1) and (2) we get | z 1 + z 2 | ≤ | z1 | + | z 2 | This is the reason why this inequality is called the triangle inequality.
3.6.6 Vector interpretation of complex numbers : Let A and B represent the two complex numbers z1 = x1 + iy1 and z2 = x2 + iy2 in the Argand diagram. Complete the parallelogram OACB. Then C represents the complex number z1 + z2. A complex number z = x + iy can y B be considered as a vector OP whose initial point is the origin O and whose terminal point is A P = P(x, y). We sometimes call OP = x + iy the position vector of P (x,y) P. Two vectors having the same length or magnitude and the same x direction but different initial points O such as OP and AB are considered Fig. 3.6 equal. (Fig. 3.6) ∴ OP=AB = x + iy Based on the above interpretation y B of complex numbers as vectors, z2 addition of complex numbers A corresponds to the “parallelogram z1 z2 law” for addition of vectors. Thus z1 + z1 to add the complex numbers z1 and C z2, we complete the parallelogram z2 O x OABC whose sides OA and OC correspond to z1 and z2. The Fig. 3.7 diagonal OB of this parallelogram corresponds to z1 + z2. (Fig. 3.7)
144
3.6.7. Geometrical representation of difference of two complex numbers Let A and B represent the two complex numbers z1 = x1 + iy1 and z2 = x2 + iy2 respectively in the Argand diagram. Produce BO to B′ such that OB′ = OB. Then B′ represents the complex number − z2. Complete the parallelogram OADB′. Then D represents the sum of the complex numbers z1 and − z2 or z1 − z2 i.e., D represents the difference of complex numbers z1 and z2. (Fig. 3.8)
y
C(z1+z2) B(z2) A(z1)
O
x D(z1 - z2)
B′(-z2)
Fig. 3.8
Result : From the diagram OD = AB. But OD = | z1 − z2 |. AB is the distance between z1 and z2. Thus, distance between two complex numbers z1 and z2 is | z1 − z 2 | . Note : Complete the parallelogram OACB. Then C represents the complex number z1 + z2.
3.6.8 Geometrical representation of product of two complex numbers: Let A and B represent the two complex numbers z1 and z2 respectively in the Argand diagram. Let z1 = r1 (cos θ1 + i sin θ1) and z2 = r2 (cos θ2 + i sin θ2)
y C (z1, z2)
ϕ(
Then OA = r1, XOA = θ1 )θ
) θ1
r1
A (z1)
(
O
θ2
ϕ
145
1
r2
OB = r2, XOB = θ2 . Take a point L on OX such that OL = 1 unit. Draw the triangle OBC directly similar to ∆ OLA. (Fig. 3.9)
B (z2)
L (1,0)
Fig. 3.9
x
then
OB OC OL = OA
r2 OC i.e., 1 = r 1
∴ OC = r1r2 Also
XOC = XOB + BOC = XOB + XOA = θ2 + θ1 or θ1 + θ2
(‡
XOA = BOC
)
∴ The point C represents the complex number z1z2 with polar coordinates (r1 r2, θ1 + θ2) y
Note : If P represents the complex number
z
Q (-y, x) iz x
z = r (cos θ + i sin θ) = reiθ
y
-y
then the effect of multiplication by (cos α+ i sin α) = eiα is the rotation
P (x,y)
x
O
x
Fig. 3.10
of P(z) counter clockwise about O through an angle α. π π iπ In particular, since i = cos 2 + i sin 2= e 2 , the effect of multiplication of any complex numbers P(z) by i is the rotation of P counter clockwise about the origin through an angle 90°. (Fig. 3.10)
3.6.9 Geometrical representation of the quotient of two complex numbers Let A and B represent two complex numbers z1 and z2 in the Argand diagram. Let z1 = r1(cos θ1 + i sin θ1) and z2 = r2 (cos θ2 + i sin θ2) ; (z2 ≠ 0)
y
ϕ(
A (z1)
r1
C (z1/z2)
146
)θ
OB = r2, XOB = θ2 . Take a point L on OX such that OL = 1 unit. Draw the triangle OAC directly similar to ∆ OBL. (Fig. 3.11)
2
Then OA = r1, XOA = θ1 O
θ1
) θ2 1
r2
ϕ(
B (z2)
L (1,0)
Fig. 3.11
x
OA OC OB = OL
then
r1 OC i.e., r = 1 2
r1 ∴ OC = r 2 XOC = XOA − COA = θ1 − θ2
r1 ∴ C is the point whose polar coordinates are r , θ1 − θ2 . Hence C 2 z1 represents the complex number z 2 Example 3.12 : Graphically prove that | z1 + z2 + z3 | ≤ | z1 | + | z2 | + | z3 | Solution : y By triangle inequality in ∆OAB, B | z 1 + z 2 | ≤ | z 1 | + | z2 | …(1) |z 2| A
|z | 1
By triangle inequality in ∆OBP, | z1 + z2 + z3 | ≤ | z1+ z2 | + | z3| ≤ | z1 | + | z2 | + | z3 | from (1)
| +z 2 |z 1 +z | |z 1+z 2 3
O
∴| z1 + z2 + z3 | ≤ | z1 | + | z2 | + | z3 |
|z3|
P x
Fig. 3.12 Example 3.13 : Prove that the complex numbers 3 + 3i, − 3 − 3i, − 3 3 + 3 3 i are the vertices of an equilateral triangle in the complex plane. Solution : y Let A, B and C represent the C(-3 √3, 3 √3 ) complex numbers (3 + 3i), A (3,3) (− 3 − 3i) and (− 3 3 + i3 3) in the Argand diagram. AB = | (3 + 3i) − (− 3 − 3i) | x
= | 6 + 6i | = 72 B (-3,-3 )
Fig. 3.13 BC = |(− 3 − 3i) − (− 3 3 + 3 3i)| =
|(− 3 + 3 3) + i (− 3 − 3 3)| =
147
72
|(− 3 = |(− 3
CA =
3 + 3 3i) − (3 + 3i)| 3 − 3) + i (3 3 − 3)| = 72
A (0, 2)
AB = BC = CA ∴ ∆ ABC is an equilateral triangle. Example 3.14 : Prove that the points representing the complex numbers 2i, 1 + i, 4 + 4i and 3 + 5i on the Argand plane are the vertices of a rectangle. Solution : y Let A, B, C and D represent the D (3,5) complex numbers 2i, (1 + i), (4 + 4i) and (3 + 5i) in the Argand diagram C (4,4) respectively. AB = | 2i − (1 + i) | = | − 1 + i | = (− 1)2 + (1)2 = 2 BC = | (1 + i) − (4 + 4i) |
B (1,1 )
x
= | − 3 − 3i | Fig. 3.14 2
2
= (− 3) + (− 3) = 9 + 9 = 18 CD = | (4 + 4i) − (3 + 5i) | = |1−i| =
12 + (− 1)2 = 2
DA = | (3 + 5i) − 2i | = | 3 + 3i | = ∴ AB = CD and BC = DA AC = | (0 + 2i) − (4 − 4i) | = | − 4 − 2i | = 2
(− 4)2 + (− 2)2 =
32 + 32 = 18
16 + 4 = 20
2
AB + BC = 2 + 18 = 20 AC2 = 20 Hence
AB2 + BC2 = AC2
As pairs of opposite sides are equal and B = 90°, ∴ ABCD is a rectangle. Example 3.15 : Show that the points representing the complex numbers 7 + 9i, − 3 + 7i, 3 + 3i form a right angled triangle on the Argand diagram.
148
Solution : Let A, B and C represent the complex numbers 7 + 9i, − 3 + 7i and 3 + 3i in the Argand diagram respectively. AB = | (7 + 9i) − (− 3 + 7i) | = | 10 + 2i | =
A(7,9)
) B (-3, 7
C(3,3)
102 + 22 = 104
x
O
BC = | (− 3 + 7i) − (3 + 3i) |
Fig. 3.15
= | − 6 + 4i | = (− 6)2 + 42 = 36 + 16 = 52 CA = | (3 + 3i) − (7 + 9i) | = | − 4 − 6i | =
(− 4)2 + (− 6)2 = 16 + 36 = 52
⇒ AB2 = BC2 + CA2 ⇒ BCA = 90° Hence ∆ABC is a right angled isosceles triangle. Example 3.16 : Find the square root of (− 7 + 24i) Solution : Let − 7 + 24i = x + iy On squaring, − 7 + 24i = (x2 − y2) + 2ixy Equating the real and imaginary parts x2 − y2 = − 7 and 2xy = 24 x2 + y2 =
2
(x2 − y2) + 4x2y2
= (− 7)2 + (24)2 = 25 Solving, x2 − y2 = − 7 and x2 + y2 = 25 we get x2 = 9 and y2 = 16 ∴ x = ± 3 and y = ± 4 Since xy is positive, x and y have the same sign. ∴ (x = 3, y = 4) or (x = − 3, y = − 4) ∴
− 7 + 24i = (3 + 4i) or (− 3 − 4i)
149
EXERCISE 3.2 (1) If (1 + i) (1 + 2i) (1 + 3i) … (1 + ni) = x + iy (2) (3) (4)
(5)
(6) (7) (8)
show that 2.5.10 … (1 + n2) = x2 + y2 Find the square root of (− 8 − 6i) If z2 = (0, 1) find z. Prove that the triangle formed by the points representing the complex numbers (10 + 8i), (− 2 + 4i) and (− 11 + 31i) on the Argand plane is right angled. Prove that the points representing the complex numbers (7 + 5i), (5 + 2i), (4 + 7i) and (2 + 4i) form a parallelogram. (Plot the points and use midpoint formula). Express the following complex numbers in polar form. (ii) − 1 + i 3 (iii) − 1 − i (iv) 1 − i (i) 2 + 2 3 i π π If arg (z − 1) = 6 and arg (z + 1) = 2 3 then prove that | z | = 1 P represents the variable complex number z. Find the locus of P, if 2z + 1 (ii) | z − 5i | = | z + 5i | (i) Im iz + 1 = − 2 z−1 z−1 π (iv) | 2z − 3 | = 2 (v) arg z + 3 = 2 (iii) Re z + i = 1
3.7 Solutions of polynomial equations : Consider the quadratic equation x2 − 4x + 7 = 0 b2 − 4ac = (− 4)2 − (4) (7) (1) = 16 − 28 = − 12 which is negative. ∴ The roots of this quadratic equation are not real. The roots are given by Its discriminant is
− (− 4) ± 2
− 12
=
4±
− 12 2
=2±i 3
Thus we see that the roots 2 + i 3 and 2 − i 3 are conjugate to each other. We shall now consider the cube roots of unity. Let x be the cube root of unity then 1
x = (1)3 ⇒ x3 = 1 ⇒ (x − 1) (x2 + x + 1) = 0
150
∴ x − 1 = 0 ; x2 + x + 1 = 0 Hence x = 1 and
x=
∴ Cube roots of unity are 1,
−1±
1 − (4) (1) (11) 2
− 1 + 3 i − 1 − 3i , 2 2
Here again the two complex roots
−1+ 3i − 1 − 3i and are conjugate 2 2
to each other. From the above two examples one can infer that in an equation with real coefficients, imaginary roots occur in pairs (i.e., one root is the conjugate of the other). This paved the way for the following theorem. Theorem : For any polynomial equation P(x) = 0 with real coefficients, imaginary (complex) roots occur in conjugate pairs. Proof : Let P(x) = anxn + an−1xn− 1 + … + a1x + a0 = 0 be a polynomial equation of degree n with real coefficients.
Let z be a root of P(x) = 0. We show that z is also a root of P(x) = 0. Since z is a root of P(x) = 0 P(z) = anzn + an − 1zn − 1 + … + a1z + a0 = 0 Taking the conjugate on both sides
… (1)
P(z) = anzn + an − 1zn − 1 + … + a1z + a0 = 0
Using the idea that the conjugate of the sum of the complex numbers is equal to the sum of their conjugates, anzn
+ an− 1zn− 1 + … + a1z + a0 = 0
i.e. an zn + an−1 zn−1 + … + a1 z + a0 = 0
n
Since zn = ( z ) and a0, a1, a2 … an are real numbers, each of them is its own conjugate and hence we get
an zn + an − 1zn−1 + … + a1 z + a0 = 0
151
which is same as P( z ) = 0
This means z is also a root of P(x) = 0. Hence the result. Example 3.17 : Solve the equation x4 − 4x2 + 8x + 35 = 0, if one of its roots is 2+ 3i Solution : Since 2 + i 3 is a root, 2 − i 3 is also a root. Sum of the roots = 4 Product of the roots (2 + i 3) (2 − i 3) = 4 + 3 = 7 ∴The corresponding factor is x2 − 4x + 7 ∴ x4 − 4x2+ 8x + 35 = (x2 − 4x + 7) (x2 + px + 5) Equating x term, we get 8 = 7p − 20 ⇒ p = 4 ∴ Other factor is (x2 + 4x + 5) ∴ x2 + 4x + 5 = 0 ⇒ x = − 2 ± i Thus the roots are 2 ± i 3 and − 2 ± i EXERCISE 3.3 4
3
(1) Solve the equation x − 8x + 24x2 − 32x + 20 = 0 if 3 + i is a root. (2) Solve the equation x4 − 4x3 + 11x2 − 14x + 10 = 0 if one root is 1 + 2i (3) Solve : 6x4 − 25x3 + 32x2 + 3x − 10 = 0 given that one of the roots is 2 − i
3.8 De Moivre’s Theorem and its applications : Theorem : For any rational number n, cos nθ + i sin nθ is the value or one of the values of (cos θ + i sin θ)n Proof : Case I : Let n be a positive integer. By simple multiplication we have (cosθ1 + isinθ1) (cosθ2 + isinθ2) = cos(θ1 + θ2) + isin(θ1 + θ2) Similarly (cosθ1 + isinθ1) (cosθ2 + isinθ2) (cosθ3 + i sin θ3) = cos(θ1 + θ2 + θ3) + i sin(θ1 + θ2 + θ3) By extending it to the product of n complex number we have (cosθ1 + i sinθ1) (cosθ2 + i sinθ2) … (cosθn + i sinθn) = cos(θ1 + θ2 + … + θn) + i sin(θ1 + θ2 + … + θn)
152
In this expression put θ1 = θ2 … = θn = θ, then we have (cos θ + isin θ)n = cosnθ + isin nθ Case II : Let n be a negative integer and equal to − m ; (m is a +ve integer) ∴ (cos θ + isinθ)n = (cosθ + isinθ)− m 1 = (cosθ + i sinθ)m 1 by case I = cosmθ + i sin mθ cos mθ − i sin mθ = (cos mθ + i sinmθ) (cosmθ − i sin mθ) cosmθ − i sinmθ = cos2mθ + sin2mθ = cosmθ − i sinmθ = cos(− m)θ + i sin(− m)θ = cosnθ + i sinnθ p Case III : Let n be a fraction and equal to q , where q is a positive integer and p is any integer. θ θ q Consider cos q + i sin q = cosθ + i sinθ
θ θ Therefore cos q + i sin q is such that its qth power is cosθ + i sinθ. 1
θ θ Hence cos q + i sin q is one of the values of (cos θ + i sin θ)q. Raise each of these quantities to the pth power. 1 θ θ P ∴ cos q + i sin q is one of the values of (cos θ + i sinθ)q
p
p
p p i.e., cos q θ + i sin q θ is one of the values of (cosθ + i sinθ)q ie., cos nθ + i sin nθ is one of the values of (cosθ + i sinθ)n. Note : De Moivre’s theorem holds good for irrational values also but the proof is beyond the scope of this book.
153
Properties : (i)
(cosθ + i sinθ)− n = cos(− nθ) + i sin (− nθ) = cos nθ − i sin nθ
(ii)
(cosθ − i sinθ)n = {cos(−θ) + i sin(− θ)}n = cos (− nθ) + i sin(− nθ) = cosnθ − i sin nθ n
(iii)
π π (sinθ + i cosθ) = cos 2 − θ + i sin 2 − θ n
π π = cos n 2 − θ + i sin n2 − θ
Example 3.18 : Simplify :
(cos 2θ + i sin 2θ)3 (cos 3θ − i sin 3θ)− 3 (cos 4θ + i sin 4θ )− 6 (cos θ + i sin θ)8
Solution : The given expression =
(cos 2θ + i sin 2θ)3 (cos 3θ − i sin 3θ)− 3 (cos 4θ + i sin 4θ )− 6 (cos θ + i sin θ)8 3
=
(ei2θ) . (e−i3θ) i4θ −6
(e
)
−3
iθ 8
(e )
ei6θ ei9θ = −i24θ i8θ e .e
= ei15θ . ei16θ = ei31θ = cos31θ + i sin 31θ Alternative method : The given expression =
(cos θ + i sin θ)6 (cos θ + i sin θ)9 (cos θ + i sin θ )− 24 (cos θ + i sin θ)8
= (cos θ + i sin θ)6 + 9 + 24 − 8 = (cos θ + i sin θ)31 = cos 31θ + i sin 31θ Example 3.19 : Simplify :
(cos θ + i sinθ)4 (sinθ + i cosθ)5
154
Solution : (cos θ + i sinθ)4 (cos θ + i sinθ)4 = 5 (sinθ + i cosθ)5 cosπ − θ + i sin π − θ 2 2 π π = cos 4θ − 5 2 − θ + i sin 4θ − 5 2 − θ
5π 5π = cos 9θ − 2 + i sin 9θ − 2 5π 5π = cos 2 − 9θ − i sin 2 − 9θ π π = cos 2 − 9θ − i sin 2 − 9θ
= sin 9θ − i cos 9θ Alternative method : (cos θ + i sinθ)4 1 (cos θ + i sinθ)4 = (sinθ + i cosθ)5 i5 (cosθ − i sinθ)5 = − i (cos 4θ + i sin 4θ ) (cos 5θ + i sin 5θ) = − i [cos 9θ + i sin 9θ] = sin 9θ − i cos 9θ 1 Result : | z | = 1 ⇔ z = z Example 3.20 : If n is a positive integer, prove that n
1 + sinθ + i cosθ = cos n π − θ + i sin n π − θ 1 + sinθ − i cosθ 2 2 Solution : Let z = sinθ + i cosθ 1 ∴ z = sinθ − i cosθ n
n
1 + sinθ + i cosθ 1 + z n n = ∴ 1 = z = (sin θ + i cos θ) 1 + sinθ − i cosθ 1 + z π π = cos2 − θ + i sin 2 − θ
155
n
π π = cos n2 − θ + i sin n2 − θ
Example 3.21 : If n is a positive integer, prove that nπ ( 3 + i)n + ( 3 − i)n = 2n + 1 cos 6 Solution : Let ( 3 + i)= r(cosθ + i sin θ) Equating real and imaginary parts separately, we have r cos θ = 3 and r sin θ = 1 ∴r=
( 3 )2 + 1 2
=
4=2
1 π 3 cos θ = 2 , sinθ = 2 ⇒ θ = 6 Hence
(
π π 3 + i) = 2 cos 6 + i sin 6
n
(
3 + i)
n
π π π π = 2 cos 6 + i sin 6 = 2ncos 6 + i sin 6 nπ nπ = 2ncos 6 + i sin 6 … (1)
n
To determine 3 − i, we replace i in the above result by − i we get π π ( 3 − i) = 2cos 6 − i sin 6 nπ nπ n ∴ ( 3 − i) = 2n cos 6 − i sin 6 Adding (1) and (2) we have nπ n ( 3 + i)n + ( 3 − i) = 2n 2 cos 6 nπ = 2n + 1 . cos 6
… (2)
Example 3.22 : If α and β are the roots of x2 − 2x + 2 = 0 and cot θ = y + 1, (y + α)n − (y + β)n sin nθ show that = α−β sinnθ Solution : The roots of the equation x2 − 2x + 2 = 0 are 1 ± i.
156
Let α = 1 + i and β = 1 − i (y + α)n = [(cotθ − 1) + (1 + i)]n
Then
= (cotθ + i)n 1 = [cosθ + i sinθ]n n sin θ 1 [cosnθ + i sin nθ] = sinnθ 1 [cosnθ − i sin nθ] Similarly (y + β)n = sinnθ 2i sin nθ (y + α)n − (y + β)n= sinnθ Further α − β = (1 + i) − (1 − i) = 2i sin nθ (y + α)n − (y + β)n 2i sin nθ = = n α−β 2i sin θ sinnθ EXERCISE 3.4 (1) Simplify :
(cos 2θ − i sin 2θ)7 (cos 3θ + i sin 3θ)− 5 (cos 4θ + i sin 4θ)12 (cos 5θ − i sin 5θ)− 6
(cos α + i sin α)3 (sin β + i cos β)4 (3) If cos α + cos β + cos γ = 0 = sin α + sin β + sin γ, prove that (i) cos 3α + cos 3β + cos 3γ = 3 cos (α + β + γ) (ii) sin 3α + sin 3β + sin 3γ = 3 sin (α + β + γ) (iii) cos 2α + cos 2β + cos 2γ = 0 (iv) sin 2α + sin 2β + sin 2γ = 0 (Hints : Take a = cis α, b = cis β, c = cis γ a + b + c = 0 ⇒ a3+ b3 + b3 = 3abc
(2) Simplify :
1/a + 1/b + 1/c = 0 ⇒ a2 + b2 + c2 = 0) 3 (v) cos2α +cos 2β + cos2γ = sin2α + sin2β + sin2γ = 2 For problems 4 to 9, m, n ∈ N (4) Prove that n+2
(i) (1 + i)n + (1 − i)n = 2 2
nπ cos 4
157
nπ n n (ii) (1 + i 3) + (1 − i 3) = 2n + 1 cos 3 n nθ (iii) (1 + cos θ + i sin θ)n + (1+cos θ − i sin θ)n = 2n + 1 cos (θ / 2) cos 2
(iv) (1 + i)4n and (1 + i)4n + 2 are real and purely imaginary respectively (5) If α and β are the roots of the equation x2 − 2px + (p2 + q2) = 0 and q (y + α)n − (y +β)n sin nθ tan θ = y + p show that = qn − 1 α−β sinnθ (6) If α and β are the roots of x2 − 2x + 4 = 0 nπ Prove that αn − βn = i2n + 1 sin 3 and deduct α 9 − β 9 1 (7) If x + x = 2 cos θ prove that 1 1 (i) xn + n = 2 cos nθ (ii) xn − n = 2 i sin nθ x x 1 1 (8) If x + x = 2 cos θ and y + y = 2 cos φ show that xm yn (i) n + m = 2 cos (mθ − nφ) y x
xm yn (ii) n − m = 2 i sin (mθ − nφ) y x
(9) If x = cos α + i sin α ; y = cos β + i sin β 1 prove that xmyn + m n = 2 cos (mα + nβ) x y (10) If a = cos2α + i sin 2α, b = cos2β + i sin 2β and c = cos 2γ + i sin 2γ prove that 1 (i) abc + = 2 cos (α + β + γ) abc (ii)
a2 b2 + c2 = 2 cos 2(α + β − γ) abc
3.9. Roots of a complex number Definition : A number ω is called an nth root of a complex number z, if ωn = z and 1
we write ω = zn
158
Working rule to find the nth roots of a complex number : Step 1 Step 2 Step 3 Step 4 Illustration : Let
: : : :
Write the given number in polar form. Add 2kπ to the argument apply De Moivre’s theorem (bring the power to inside) Put k = 0, 1 … upto n − 1
z = r(cosθ + i sinθ) = r{cos (2kπ + θ) + i sin (2kπ + θ)}, k is an integer. 1
1
∴ zn = [r{cos (2kπ + θ) + i sin (2kπ + θ)]n 1
2kπ + θ 2kπ + θ = rn cos n + i sin n
where k = 0, 1, 2 … (n − 1) 1
Only these values of k will give n different values of zn provided z ≠ 0 Note : (1) The number of nth roots of a non-zero complex number is n. (2) The moduli of these roots is the same non negative real number. (3) The argument of these n roots are equally spaced. That is if θ is the principal value of arg z i.e., − π ≤ θ ≤ π then the arguments of other 2π 4π θ roots of z are obtained by adding respectively n , n , … to n (4) If k be given integral values greater than or equal to n, these n values are repeated and no fresh root is obtained.
3.9.1 The nth roots of unity 1= (cos 0 + i sin 0) = cos 2kπ + i sin 2kπ 1
1
nth roots of unity = 1n = (cos 2kπ + i sin 2kπ)n 2kπ 2kπ = cos n + i sin n where k = 0, 1, 2, … n − 1 2π 2π ∴ The nth roots of unity are cos0 + i sin 0 , cos n + i sin n , 4π 4π 6π 6π 2π 2π cos n + i sin n , cos n + i sin n , ……, cos (n − 1) n + i sin (n − 1) n 2π
2π 2π i Let ω = cos n + i sin n = e n
159
Then the nth roots of unity are 2π
4π
2(n − 1)
6π
i i i i n π become 1, ω, ω2 … ωn − 1 . e0, e n , e n , e n , … e It is clear that the roots are in geometric progression. Results : (1) ωn = 1 n
2π 2π ω = cos n + i sin n = cos 2π + i sin 2π = 1 n
(2) Sum of the roots is 0 i.e., 1 + ω + ω2 + … + ωn – 1 = 0 ‡ LHS = 1 + ω + ω2 + … + ωn – 1 is a G.P. with n terms. 1 − rn 1.(1 − ωn) ‡ 1 + r + r2 + … + rn − 1 = 1−r 1−ω = 0 = R.H.S. (3) The roots are in G.P with common ratio ω 2π (4) The arguments are in A.P with common difference n =
(5) Product of the roots = (− 1)n + 1 1
3.9.2. Cube roots of unity : (1)3 1
Let x = (1)3 1
∴ x = (cos 0 + i sin 0)3 1
= (cos 2kπ + i sin 2kπ)3 , where k is an integer. 2kπ 2kπ x = cos 3 + i sin 3 , where k = 0, 1, 2 The three roots are 2π 2π 4π 4π cos 0 + i sin0, cos 3 + i sin 3 , cos 3 + i sin 3 1 1 3 3 i.e., 1, − 2 + i 2 , − 2 − i 2 1 3 ∴ The roots are 1, − 2 ± i 2
160
Result : 1
y
12 0°
120°
The modulus of each root of (1)3 is 1 B ∴ All these roots lie on the circumference of the unit circle. Let A, B and C be points represented by the three 1 3 x roots 1, − 2 ± i 2 in ordered pair form. A The angles between OA and OB, OB and 2π OC, OC and OA are each 3 radians or C 120°. Hence when these points are joined by straight lines they will form the Fig. 3.16 vertices of an equilateral triangle. 2π 2π If we denote the second root cos 3 + i sin 3 by ω then the other root 0° 12
2
4π 4π 2π 2π cos 3 + i sin 3 = cos 3 + i sin 3 becomes ω2. Hence we observe that the cube roots of unity, namely 1, ω, ω2 are in G.P. Note : −1+i 3 −1−i 3 is taken as ω it can be proved that = ω2 (i) Even if 2 2 (ii) 1 + ω + ω2 = 0 (by actual addition) i.e., the sum of the cube roots of unity is zero. (iii) Since ω is a root of the equation x3 = 1, we see that ω3 = 1.
Fourth roots of unity : Let x be a fourth root of unity. Then x =
1 (1)4
∴ x4 = 1 = (cos 2kπ + i sin 2kπ) where k is an integer. 1
x = (cos 2kπ + i sin 2kπ)4 2kπ 2kπ = cos 4 + i sin 4 kπ kπ = cos 2 + i sin 2 where k = 0, 1, 2, 3
161
The four roots are π π 3π 3π cos 0 + i sin 0, cos 2 + i sin 2, cos π + i sin π, cos 2 + i sin 2 π π i.e., 1, i, − 1(= i2), − i (= i3) . Let us denote cos 2 + i sin 2 by ω. Then the four roots of unity are 1, ω, ω2, ω3. y ω
The fourth roots of unity form the vertices of a square all lying on the unit circle. We observe that the sum of the fourth roots of unity is zero. i.e., 1 + ω + ω2 + ω3 = 0 and ω4 = 1
π/2
ω2 O
1
x
ω3
Fig. 3.17 Note : The values of ω used in cube roots of unity and in fourth roots of unity are different. Sixth roots of unity : Let x be a sixth root of unity. Then x = (1)1/6 ∴ 1 = cos 0 + i sin 0 (1)1/6 = (cos 2kπ + i sin 2kπ)1/6 where k is an integer. By De Moivre’s theorem 1 2kπ 2kπ x = (1)6 = cos 6 + i sin 6 , where k = 0, 1, 2, 3, 4, 5 The six roots are y cos 0 + i sin 0 = 1 ω ω2 π π cos 3 + i sin 3 2π 2π cos 3 + i sin 3 1 ω3 3π 3π x cos 3 + i sin 3 4π 4π cos 3 + i sin 3 5π 5π ω5 ω4 cos 3 + i sin 3 Fig. 3.18
162
Then the six, sixth roots of unity are 1, ω, ω2, ω3, ω4, ω5 y
From the above figure it can be noted that the six roots of unity form the vertices of a hexagon all lying on the unit circle (Fig. 3.18). Thus it can be seen that the n, nth roots of unity form the vertices of n sided regular polygon all lying on the unit circle (Fig. 3.19).
ω3 ω2 ω 1 ωn-1 ωn-2 ωn-3
Fig. 3.19 9
5
4
Example 3.23 : Solve the equation x + x − x − 1 = 0 Solution : x9 + x5 − x4 − 1 = 0 ⇒ x5 (x4 + 1) − 1 (x4 + 1) = 0 ⇒ (x5 − 1) (x4 + 1) = 0 ⇒ x5 − 1 = 0 ; x4 + 1 = 0 1
1
x = (1)5 ; (− 1)4 1
(i)
1
x = (1)5 = (cos 0 + i sin 0)5 1
= (cos 2kπ + i sin 2kπ)5 2kπ 2kπ = cos 5 + i sin 5 , k = 0, 1, 2, 3, 4 1
(ii)
1
(− 1)4 = (cos π + i sin π)4 1
= {cos (2k + 1)π + i sin (2k + 1)π}4 = cos
(2k + 1)π (2k + 1)π + i sin k = 0, 1, 2, 3 4 4
Thus we have 9 roots. Example 3.24 : Solve the equation x7 + x4 + x3 + 1 = 0 Solution : x7 + x4 + x3 + 1 = 0 ⇒ x4 (x3 + 1) + 1 (x3 + 1) = 0 ⇒ (x4 + 1) (x3 + 1) = 0
163
x
x4 = − 1 ; x3 = − 1 1
(i)
x = (− 1)4 1
= (cosπ + i sin π)4 1 π)]4
= [cos (2kπ + π) + i sin (2kπ + (2k + 1)π (2k + 1)π = cos + i sin 4 4 ; k = 0, 1, 2, 3
i.e.,
1
(ii)
x3 = − 1
⇒ x = (− 1)3 1
= (cos π + i sin π)3 1
= [cos (2kπ + π) + i sin (2kπ + π)] 3 π π = cos (2k + 1) 3 + i sin (2k + 1) 3 , k = 0, 1, 2 Note : π π 3π 3π 5π 5π The roots are cos 4 + i sin 4 ; cos 4 + i sin 4 ; cos 4 + i sin 4 7π 7π π π and cos 4 + i sin 4 , cos 3 + i sin 3 , (cos π + i sin π) and cos 5π + i sin 5π 3 3 1 1 1 1 1 1 1 1 i.e., +i , − +i , − −i , −i 2 2 2 2 2 2 2 2 1 1 3 3 2 + i 2 , − 1, 2 − i 2 2
Example 3.25 : Find all the values of ( 3 + i)3 Solution : Let 3 + i = r (cos θ + i sin θ) ⇒ r cosθ = 3 , r sin θ = 1 2
⇒r = ( 3) + 1 = 2 1 π 3 Cos θ = 2 , sin θ = 2 ⇒ θ = 6 2
2
2
π π ∴ ( 3 + i)3 = 23 cos 6 + i sin 6 3
164
1
2 23
=
2 3 π cos + i sin π 6 6
2
1
π π = 23 cos 3 + i sin 3 3
2 23
=
1
cos 2kπ + π + i sin 2kπ + π 3 3 3
2
π π = 23 cos (6k + 1) 9 + i sin (6k + 1) 9 where k = 0, 1, 2
Note: The values are 2
2
2
π π 7π 7π 13π 13π 23 cos 9 + i sin 9 , 23 cos 9 + i sin 9 , 23 cos 9 + i sin 9
Aliter : 2
(
2
2
π π 3 + i)3 = 23 cos 6 + i sin 63
=
2 23
2
cos 2kπ + π + i sin 2kπ + π 3 6 6
2
2
π π = 23 cos (12k + 1) 6 + i sin (12k + 1) 6 3
=
2 23
cos (12k + 1) π + i sin (12k + 1) π where k = 0, 1, 2 9 9 2
2
π π 13π 13π The different values are 23 cos 9 + i sin 9 , 23 cos 9 + i sin 9 , 2 25π 25π 23 cos 9 + i sin 9 2 23
2
cos π + i sin π, 23 cos 7π + i sin 7π, and i.e., 9 9 9 9 13π 13π 25π 25π 7π 7π cos 9 + i sin 9 since cos 9 + i sin 9 = cos 9 + i sin 9
2 23
Thus we have obtained the same values in this case also. Note : If we add 2kπ before taking the power 2 inside, we will get the same answer.
165
EXERCISE 3.5 (1) Find all the values of the following : 1
1
2
(i) (i)3
(ii) (8i))3
(iii) (− 3 − i)3
(2) If x = a + b, y = aω + bω2, z = aw2 + bω show that (i) xyz = a3 + b3 (ii) x3 + y3 + z3 = 3 (a3 + b3) where ω is the complex cube root of unity. (3) Prove that if ω3 = 1, then (i) (a + b + c) (a + bω + cω2) (a + bω2 + cω) = a3 + b3 + c3 − 3abc 5
(ii)
5
− 1 + i 3 − 1 − i 3 2 2 + = −1
1 1 1 − + = 0 1 + 2ω 1+ω 2+ω (4) Solve : (i) x4 + 4 = 0 (ii) x4 − x3 + x2 − x + 1 = 0 (iii)
3
3 4 1 (5) Find all the values of 2 − i 2 and hence prove that the product of the values is 1.
166
4. ANALYTICAL GEOMETRY 4.1 Introduction : Tracing the history of Mathematics, around 430 B.C., study of conic sections or conics, i.e., study of plane sections of a right circular cone began. The study included degenerate or singular conics (comprising single point, pair of distinct lines, two coincident lines etc., which were already dealt with in detail in lower classes) and non-degenerate or non-singular conics (comprising of circles, parabolas, ellipses and hyperbolas). The study of conic sections from Greek Geometry, developed by Apollonius, is described today as graphs of quadratic equations in the co-ordinate plane. The Greek mathematicians of Plato’s time (429 − 347 B.C.) described these curves as the curves formed by slicing a double cone (right circular cone of two nappes) with a plane and hence the name conic sections. Analytic Geometry grew out of need for establishing uniform techniques for solving geometrical problems, the aim being to apply them to the study of curves, which are of particular importance in practical problems. The aim was achieved in the co-ordinate method viz., cartesian, polar, bi-polar (where calculations are fundamental and constructions play a subordinate role). Thus solving problems by the method of Analytical Geometry requires less inventiveness. This method of the ancient Greek origin (≈ 1 − 2 B.C.) was systematically developed in the first half of the 17th century by great mathematicians Fermat, Descartes, Kepler, Newton, Euler, Leibnitz, L’Hôpital, Clairaut, Cramer and the Jacobis. A major breakthrough in the study occurred with the development of the hypothesis of Planetary Phenomena by the German mathematician cum physicist Johannes Kepler. He stated that all the planets in the solar system including the earth are moving in elliptical orbits with sun at one of a foci, governed by inverse square law. This led to the development of Newton’s gravitation theory. Euler applied the co-ordinate method in a systematic study of space curves and surfaces, which was further developed by Albert Einstein in his theory of relativity. Needless to say that today the development in this area has conquered industry, medicine and scientific research. And we shall cite a few of them before getting into the depth of actual study of conics.
167
4.1.1 Geometry and Practical applications of a parabola : A parabola is a conic section obtained on slicing a right circular cone by a plane parallel to the line joining vertex and any other point of the cone (Fig.4.1) Fig. 4. 1 If P is any point on the parabola with focus F and vertex V, the angles subtended by FP and PX with the tangent at P are equal where PX is parallel to the axis VFA of the parabola. (Fig. 4.2) This property is made use of in parabolic reflectors (surface obtained by revolving the parabola about its axis and coated with silver paint) of sound, light and radio waves when the respective source is placed at the focus S as given in (Fig. 4.3). Light (or sound or radio waves)
P
X V F
A
Fig. 4. 2
V
S
x
Fig. 4. 3 from S falls on the reflecting surface gets reflected parallel to the axis of parabola. For example, Flash light, head light of motor vehicles, parabolic mirrors, spot light reflectors, selective microphone sounding boards etc. The same reflectors can be employed in intensifying signals. Electromagnetic waves arriving x parallel to the axis of the R V parabolic reflector will be focussed at the focus where a suitable receiver ‘R’ could be Fig. 4. 4 placed. (Fig. 4.4) For example, Radio telescope, television satellite dishes, solar heaters, radar antenna’s etc. The strongest simple arch is parabolic in shape.
168
The supporting cable of a uniformly loaded bridge is parabolic in shape (weight of cable neglected in comparison with weight of the bridge). The path of an object thrown or projected obliquely upwards is a parabola. Also bombs dropped from a moving war plane or food packets dropped from helicopters during cyclone time to people in need (not moving vertically upwards or downwards) traces a parabola. Some comets have parabolic path with sun at the focus.
4.1.2 Geometry and Practical Applications of an ellipse : An ellipse is a conic obtained on slicing across obliquely one nappe of a cone. (Fig. 4.5) If P is any point on the ellipse and F1and F2 its foci, the angle subtended by F1P and F2P with the tangent at P are Fig. 4. 5 equal and if a source of light or sound is placed at one focus of an ellipsoidal P reflector (surface generated by revolving an ellipse about its major axis) all the F2 F1 waves will be reflected so as to pass Fig. 4. 6 through the other focus (Fig. 4.6) This property is also used in “Wispering Gallery”, the roof or walls of which are shaped like an ellipsoidal reflector. The ellipsoidal reflectors are designed for Nd : YAG (ND3+ Neodymium ions; YAG – Yttrium Aluminium Garnetz) laser that is widely used in medicine, industry and scientific research. A light reflector in the form of a tube whose cross section is an ellipse has Nd : YAG rod and a linear flash lamp placed at the foci of the ellipse. (Fig. 4.7). Here light emitted from the lamp is effectively coupled to the Nd : YAG rod to produce laser beam. In Bohr-Sommerfeld theory of the atom electron orbit can be circular or elliptical.
M1
Laser rod
Output
Flashtube Capacitor bank
Power Supply
Ballast resistor
Fig. 4. 7
169
M2
Ellipsoidal reflector Trigger pulse
The orbits of our planet earth and all other planets and planetoids in our solar system are elliptical with sun situated at one of the foci. Also all the satellites, either natural or artificial to all the planets in the solar system have elliptic orbits (with the force binding them following inverse square law). [Fig.4.8(a)] Uranus Mercury
Jupiter Venus
Pluto Saturn
Halley’s Comet
Saturn JupiterSun Earth
Sun Neptune
Mars
Earth
Pluto
Asteroids
Fig. 4.8(b) Fig. 4.8(a) Path of Halley’s Comet (which returns after every 75 years) is an ellipse with e ≈ 0.97 and the sun at a focus [Fig. 4.8 (b)], e being the eccentricity. Elliptical arches are often used for their beauty. Steam boilers are believed to have greatest strength when heads are made elliptical with major and minor axes in the ratio 2 : 1.
Fig. 4. 9 Gears are sometimes (for particular need) made elliptical in shape (Fig. 4.9)
Fig. 4. 10 The orbit of Comet Kohoutek is an ellipse with e ≈ 0.9999 (Fig. 4.10). The shape of our mother earth is an oblate spheroid i.e., the solid of revolution of an ellipse about its minor axis, bulged along equatorial region and flat along the polar region. The area of action of an airplane which leaves a moving carrier and returns in a given time (with no wind) is an ellipse with the take off and landing positions of the carrier as foci. The track of a plane making an On-pylon turn in a wind of constant velocity is an ellipse with one focus directly over the pylon (Fig. 4.11). Fig. 4. 11
170
4.1.3 Geometry and Practical Applications of a Hyperbola : A hyperbola is a conic obtained on slicing a double napped cone by a plane parallel to the axis of the cone (Fig.4.12)
Fig. 4. 12 The lines from the foci to any point on a hyperbola make equal angles with the tangent at that point. Hence if the surface of a reflector is generated by revolving a hyperbola about its transverse axis, all rays of light converging on one focus are reflected to the other (Fig.4.13)
P F2
F1
Fig. 4. 13
This property is made use of in some telescopes in conjunction with a parabolic reflector. American space research foundation NASA’s Hubble space telescope uses hyperbolic reflectors in conjunction with parabolic reflectors (Fig. 4.14). Hyperbolas are useful in rangefinding. (The difference in the times at which a sound is heard at two listening posts is proportional to the difference of the distances from the posts to the point of emanation of sound. A third listening post serves to give another hyperbola and the point of emanation is at the point of intersection of the two curves).
Fig. 4. 14
Boyle’s law pv = constant is hyperbolic in relationship. The same is true of relationship of any two quantities, which are inversely proportional. Hyperbolic paths arise in Einstein’s theory of relativity and form a basis for LORAN (Long Range Navigation) radio navigation system. 171
4.2 Definition of a Conic :
Fixed Line (directrix)
Consider a circle C. Let A be the A line through the centre of C and perpendicular to the plane of C and α let V be a point on A not in the plane V of C. Let P be a point on C and draw the infinite straight lines through P that also passes through V. As P C moves around C, what sweeps out is P called a right circular cone with the Fig. 4. 15 axis A and vertex V. Each of the lines PV is called a generator of the cone, and the angle α between the axis and the generator is called a vertex angle (semi-vertical angle). The upper and lower portions of the cone that meet at the vertex are called nappes of the cone (Fig. 4.15). The curves obtained by slicing the cone with a plane not passing through the vertex are called conic sections or simply conics. A conic is the locus of a point which l moves in a plane, so that its distance from P (Moving point) M a fixed point bears a constant ratio to its distance from a fixed straight line. (Fig.4.16) . ) F( Fixed point The fixed point is called focus, the fixed straight line is called directrix, and Fig. 4. 16 the constant ratio is called eccentricity, which is denoted by ‘e’. FP From the figure we have PM = constant = e
4.2.1 General equation of a Conic : Let F(x1, y1) be the focus, lx + my + n = 0, the equation of the directrix ‘l’ and ‘e’ the eccentricity of the conic. Let P(x, y) be any point on the conic. l Drop a perpendicular from P to ‘l’. P(x,y) FP = (x − x1)2 + (y − y1)2 PM = Perpendicular distance from P(x, y) to the line lx + my + n = 0 lx + my + n = ± l2 + m 2
172
M
F(x1,y1)
Fig. 4. 17
FP By the definition of a conic, PM = e ∴ FP2 = e2 PM2 2
2
∴ (x − x1) + (y − y1) = e
2
± lx + my + n l2 + m2
2
Simplifying this we get an equation of second degree in x and y of the form Ax2 + Bxy + Cy2 + Dx + Ey + F = 0.
4.2.2 Classification with respect to the general equation of a conic : The equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 represents either a (non–degenerate) conic or a degenerate conic. If it is a conic, then it is (ii) an ellipse if B2 − 4AC < 0 (i) a parabola if B2 − 4AC = 0 (iii) a hyperbola if B2 − 4AC > 0
4.2.3. Classification of conics with respect to eccentricity : 1. If e < 1, then the conic is an ellipse. From the figure 4.18 we observe that F2Pi is always less than PiMi. F2Pi i.e., P M = e < 1, (i = 1, 2, 3…) i i
y M3 M2 M1
P3 P2 P1
F2
x
F1
Fig. 4. 18
2. If e = 1, then the conic is a parabola. From the figure 4.19 we observe that FPi is always equal to PiMi. FPi i.e., P M = e = 1, (i = 1, 2, 3…) i i
y
M3
P2
M2 M1
P3
P1
x
F
Fig. 4. 19 y
3. If e > 1, then the conic is a hyperbola. From the figure 4.20 we observe that F1Pi is always greater than PiMi. F1Pi i.e., P M = e > 1, (i = 1, 2, 3…) i i
M3 M2 M1 F2
P3 P2 P1
O
Fig. 4. 20
173
F1
x
4.3 Parabola : The locus of a point whose distance from a fixed point is equal to its distance from a fixed line is called a parabola. That is a parabola is a conic whose eccentricity is 1. Note : Eventhough the syllabus does not require the derivation of standard equation and the tracing of parabola (4.3.1, 4.3.2) and it needs only the standard equation and the diagram, the equation is derived and the curve is traced for better understanding. Now we derive and trace the standard equation of a parabola.
4.3.1. Standard equation of a parabola : y
Given : Fixed point (F) Fixed line (l) Eccentricity (e = 1) Moving point P(x, y) Construction : Plot the fixed point F and draw the fixed line ‘l’.
M (-a, y)
Z (-a,0)
l
P (x,y)
O
F (a,0)
Fig. 4. 21
Drop a perpendicular (FZ) from F to l. Take FZ = 2a and treat it as x-axis. Draw a perpendicular bisector to FZ and treat it as y-axis. Let V(0, 0) be the origin. Drop a perpendicular (PM) from P to l. The known points are F(a, 0), Z(− a, 0) and hence M is (− a, y). FP 2 2 By the definition of a conic, PM = e = 1 ⇒ FP = PM (x − a)2 + (y − 0)2 = (x + a)2 + (y − y)2 x2 − 2ax + a2 + y2 = x2 + 2ax + a2 which simplifies to y2 = 4ax. This is the standard equation of the parabola. To trace a curve, we shall use the tools dealt in detail in chapter 6.
4.3.2.Tracing of the parabola y2 = 4ax : (i) Symmetry property : It is symmetrical about x-axis. i.e., x-axis divides the curve into two symmetrical parts.
174
x
(ii) Special points : The parabola passes through the origin since (0, 0) satisfies the equation y2 = 4ax. To find the points on x-axis, put y = 0. We get x = 0 only. ∴ the parabola cuts the x-axis only at the origin (0, 0). To find the points on y-axis, put x = 0. We get y = 0 only. ∴ the parabola cuts the y-axis only at the origin (0, 0). (iii) Existence of the curve : For x < 0, y2 becomes negative. i.e., y is imaginary. Therefore the curve does not exist for negative values of x. i.e., the curve exists only for non-negative values of x. y (iv) The curve at infinity : As x increases, y2 also increases.
P (x, y)
M (-a, y)
i.e., as x → ∞, y2 → ∞ i.e., as x → ∞, y → ± ∞ ∴ the curve is open rightward. [Fig. 4.22]
Z (-a, o)
V
F(a, o)
x
Fig. 4. 22
4.3.3. Important definitions regarding a parabola : Focus : The fixed point used to draw the parabola is called the focus (F). Here, the focus is F(a, 0). Directrix : The fixed line used to draw a parabola is called the directrix of the parabola. Here, the equation of the directrix is x = − a. Axis : The axis of the parabola is the axis of symmetry. The curve 2 y = 4ax is symmetrical about x-axis and hence x-axis or y = 0 is the axis of the parabola y2 = 4ax. Note that the axis of the parabola passes through the focus and perpendicular to the directrix. Vertex : The point of intersection of the parabola and its axis is called its vertex. Here, the vertex is V(0, 0). Focal distance : The focal distance is the distance between a point on the parabola and its focus. Focal chord : A chord which passes through the focus of the parabola is called the focal chord of the parabola. Latus Rectum : It is a focal chord perpendicular to the axis of the parabola. Here, the equation of the latus rectum is x = a.
175
End points of latus rectum and length of latus rectum : y
To find the end points, solve the equation of latus rectum x = a and y2 = 4ax. x = a in y2 = 4ax Using we get y2 = (4a)a = 4a2 ∴ y = ± 2a
L
V
x
F L′
Fig. 4. 23 ′ If L and L are the end points of latus rectum then L is (a, 2a) and L′ is (a, − 2a). The length of latus rectum = LL′ = 4a. Length of semi-latus rectum = FL = FL′ = 2a. So far we have discussed standard equation of a parabola which is open rightward. But we have parabolas which are open leftward, open upward and open downward.
4.3.4. Other standard parabolas : 1. Open leftward : y2 = − 4ax [a > 0] If x > 0, then y becomes imaginary. i.e., the curve does not exist for x > 0 i.e., the curve exist for x ≤ 0. 2. Open upward : x2 = 4ay [a > 0] If y < 0, then y becomes imaginary. i.e., the curve does not exist for y < 0 i.e., the curve exist for y ≥ 0.
y
y2 = -4ax
x =a
V
x
F(-a,0)
Fig. 4. 24 x 2=
4ay
y
F(0, a) x
V
y =- a
Fig. 4. 25 3. Open downward : x2 = − 4ay [a > 0] If y > 0, then y becomes imaginary. i.e., the curve does not exist for y > 0 i.e., the curve exist for y ≤ 0.
y y =a
x 2=
- 4a
y
V
F(0,- a)
Fig. 4. 26
176
x
Now, we summarise the results of the four standard types of parabolas. Type
Equation
Diagram
Focus
Equation of Directrix
Axis
x =−a
y=0
(0, 0)
x=a
4a
x=a
y=0
(0, 0)
x=−a
4a
y =−a
x=0
(0, 0)
y=a
4a
y=a
x=0
(0, 0)
y=−a
4a
Vertex
Equation of Latus Rectum
Length of Latus Rectum
y
Open rightwards
2
y = 4ax
(a, 0)
x
y 2
Open leftwards
y = − 4ax
Open upwards
x2 = 4ay
x
(− a, 0)
y
(0, a) x y
Open downwards
2
x = − 4ay
x
(0, − a)
177
Remark : So far we have discussed four standard types of parabolas. There are plenty of parabolas which cannot be classified under these standard types. For example, consider the following parabolas. y
y
y
y x
x x
x
Fig. 4. 27 For the above parabolas, the axes are neither parallel to x-axis nor parallel to y-axis. In such cases the equation of the parabolas include xy term, which is beyond the scope of this book, eventhough we will find the equation of the parabolas which are not in standard form. Note that for the standard types the axis is either parallel to x-axis or parallel to y-axis. We will study only these four types. All the parabolas discussed so far have vertex at the origin. In general the vertex need not be at the origin for any parabola. Hence we need the concept of shifting the origin or translation of the axes.
4.3.5 The process of shifting the origin or translation of axes : Consider the xoy system. Draw a line parallel to x-axis (say X-axis) and draw a line parallel to y-axis (say Y-axis). Let P(x, y) be a point with respect to xoy system and P(X, Y) be the same point with respect to XOY system. Let the co-ordinates of O′ with respect to xoy system be (h, k) The co-ordinate of P with Y y (X, Y) respect to xoy system : P(x, y) OL = OM + ML = h + X Y O′ X i.e., x = X + h k X Similarly y = Y + k h x O (0,0) M L ∴ The new co-ordinates of P with respect to XOY system X=x−h Fig. 4. 28 Y=y−k
178
4.3.6 General form of the standard equation of a parabola, which is open rightward (i.e., the vertex other than origin) : Consider a parabola with vertex V whose co-ordinates with respect to XOY system is (0, 0) and with respect to xoy system is (h, k). Since it is open rightward, the equation of the parabola w.r.t. XOY system is Y2 = 4aX. By shifting the origin X = x − h and Y = y − k, the equation of the parabola with respect to old xoy system is (y − k)2 = 4a(x − h). This is the general form of the standard equation of the parabola, which is open rightward. Similarly the other general forms are (y − k)2 = − 4a (x − h) (open leftwards) (x − h)2 = 4a (y − k) (open upwards) (x − h)2 = − 4a (y − k) (open downwards) Note : To find the general form, replace x by x − h and y by y − k if the vertex is (h, k) Remark : The above form of equations do not have xy term. Example 4.1: Find the equation of the following parabola with indicated focus and directrix. (i) (a, 0)
;
x=−a
(ii) (− 1, − 2)
;
x − 2y + 3 = 0
(iii) (2, − 3)
;
y−2=0
a>0
Solution: (i) Let P(x, y) be any point on the parabola. If PM is drawn perpendicular to the directrix, x =- a FP PM = e = 1 P(x,y) ⇒
M
FP2 = PM2
x + a (x − a)2 + (y − 0)2 = ± 12
2
F(a,0)
⇒ (x − a)2 + y2 = (x + a)2 ⇒ y2 = 4ax This is the required equation.
Fig. 4. 29
179
Alternative method : From the given data, the parabola is open rightward. ∴ The equation of the parabola is of the form (y − k)2 = 4a (x − h) We know that the vertex is the midpoint of Z(− a, 0) and focus F(a, 0), where Z is the point of intersection of the directrix and the x-axis.
y x =- a F(a,0) Z (-a,0)
V (0,0)
x
Fig. 4. 30
−a+a ∴ Vertex is 2 ,
0 + 0 2 = (0, 0) = (h, k) Again the distance between the vertex and the focus VF = a ∴ The required equation is (y − 0)2 = 4a (x − 0) i.e., y2 = 4ax (ii) Let P(x, y) be any point on the parabola. If PM is drawn perpendicular to the directrix, FP P(x, y) M PM = e = 1 FP2 = PM2
⇒
(x + 1)2 + (y + 2)2 = ±
x − 2y + 3
12+22
2
x - 2y + 3 = 0
F( -1, -2 )
Fig. 4. 31 ⇒ 4x2 + 4xy + y2 + 4x + 32y + 16 = 0 Note : Here the directrix is parallel to neither x-axis nor y-axis. This type is not standard type. Therefore we can’t do this problem as in the alternative method of previous problem. (iii) Let P(x, y) be any point on the parabola. If PM is drawn perpendicular to the directrix FP M P(x,y) PM = e = 1 2
y − 2 i.e., (x − 2) + (y + 3) = ± 1 2
2
2 2
(x − 2) + (y + 3) = (y − 2)
2
y - 2 =0
FP2 = PM2
⇒
F(2,-3)
Fig. 4. 32
2
⇒ x − 4x + 10y + 9 = 0
180
Note : Since the directrix y = 2 is parallel to x-axis, the type is standard and hence this problem can be solved by alternative method of 4.1(i). Example 4.2 : Find the equation of the parabola if (i) the vertex is (0, 0) and the focus is (− a, 0), a > 0 (ii) the vertex is (4, 1) and the focus is (4, − 3) Solution: (i) From the given data the parabola is open leftward y The equation of the parabola is of the form (y − k)2 = − 4a(x − h) Here, the vertex (h, k) is (0, 0) and VF = a V (o,o) x ∴ The required equation is F 2 (y − 0) = − 4a (x − 0) (-a,o) 2 y = − 4ax Fig. 4. 33 (ii) From the given data the parabola is open downward. ∴ The equation is of the form y 2 (x − h) = − 4a (y − k) x =4 Here, the vertex (h, k) is (4, 1) and the distance between the vertex and V(4,1) the focus x VF = a ⇒ (4 − 4)2 + (1 + 3)2 = 4 = a ∴ the required equation is (x − 4)2 = − 4(4) (y − 1)
F(4,-3)
Fig. 4. 34 (x − 4)2 = − 16(y − 1) Example 4.3: Find the equation of the parabola whose vertex is (1, 2) and the equation of the latus rectum is x = 3. Solution: From the given data the parabola is open rightward. y ∴ The equation is of the form (y − k)2 = 4a(x − h) Here, the vertex V (h, k) is (1, 2) Draw a perpendicular from V to the latus rectum. It passes through the focus. ∴ F is (3, 2) Again VF = a = 2
181
x =3 V
F(3,2)
(1,2)
Fig. 4. 35
y =2 x
∴ The required equation is (y − 2)2 = 4(2) (x − 1) (y − 2)2 = 8(x − 1) Example 4.4: Find the equation of the parabola if the curve is open rightward, vertex is (2, 1) and passing through point (6, 5). Solution: Since it is open rightward, the equation of the parabola is of the form (y − k)2 = 4a(x − h) The vertex V(h, k) is (2, 1) ∴ (y − 1)2 = 4a (x − 2) But it passes through (6, 5) ∴ 42 = 4a (6 − 2) ⇒ a = 1 ∴ The required equation is (y − 1)2 = 4(x − 2) Example 4.5 : Find the equation of the parabola if the curve is open upward, vertex is (− 1, − 2) and the length of the latus rectum is 4. Solution: Since it is open upward, the equation is of the form (x − h)2 = 4a(y − k) Length of the latus rectum = 4a = 4 and this gives a = 1 The vertex V (h, k) is (− 1, − 2) Thus the required equation becomes (x + 1)2 = 4 (y + 2) Example 4.6 : Find the equation of the parabola if the curve is open leftward, vertex is (2, 0) and the distance between the latus rectum and directrix is 2. Solution: Since it is open leftward, the equation is of the form (y − k)2 = − 4a(x − h) The vertex V(h, k) is (2, 0) The distance between latus rectum and directrix = 2a = 2 giving a = 1 and the equation of the parabola is (y − 0)2 = − 4(1) (x − 2) or y2 = − 4(x − 2) Example 4.7 : Find the axis, vertex, focus, directrix, equation of the latus rectum, length of the latus rectum for the following parabolas and hence draw their graphs. (i) y2 = 4x (ii) x2 = − 4y (iii) (y + 2)2 = − 8(x + 1) (iv) y2 − 8x + 6y + 9 = 0 (v) x2 − 2x + 8y + 17 = 0
182
Solution: (i)
y2 = 4x
(y − 0)2 = 4(1) (x − 0) Here (h, k) is (0, 0) and a = 1 y Axis : The axis of symmetry is x-axis. x = -1 Vertex : The vertex V (h, k) is (0, 0) x =1 Focus : The focus F (a, 0) is (1, 0) F(1,0) x Directrix : The equation of the directrix is V (0,0) x = − a i.e. x = − 1 Latus Rectum : The equation of the latus Fig. 4. 36 rectum is x = a i.e. x = 1 and its length is 4a = 4(1) = 4 ∴ the graph of the parabola looks as in Fig. 4.36. (ii) x2 = − 4y (x − 0)2 = − 4(1) (y − 0) y Here (h, k) is (0, 0) and a = 1 y =1 Axis : y-axis or x = 0 V (0,0) x Vertex : V (0, 0) y = -1 Focus : F (0, − a) i.e. F (0, − 1) F Directrix : y = a i.e. y = 1 (0,-1) Latus rectum : y = − a i.e. y = − 1 : length = 4 Fig. 4. 37 ∴ the graph looks as in Fig. 4.37 (iii)
(y + 2)2 = − 8 (x + 1) Y2 = − 8X
where X = x + 1
2
Y = − 4(2) X Y=y+2 a=2 The type is open leftward. Referred to X, Y Referred to x, y X = x + 1, Y = y + 2 Axis Y=0 Y=0 ⇒ y+2=0 Vertex (0, 0) X=0 ; Y=0 ⇒ x+1=0 ; y+2=0 x = − 1, y = − 2 ∴ V (− 1, − 2)
183
Focus
(− a, 0) i.e. (− 2, 0)
Directrix
X=a
Latus rectum
X=−a
Length of Latus rectum
4a = 8
i.e. X = 2 i.e. X = − 2
Y
X=−2 ; Y=0 ⇒ x + 1 = − 2, y + 2 = 0 x = − 3, y = − 2 F (− 3, − 2) X=2 ⇒ x+1=2 ⇒ x=1 X=−2 ⇒ x+1=−2 ⇒ x=−3 8
y x =1
(0,0) F (-3,-2 )
V (-1,-2)
x X
Fig. 4. 38 (iv)
2
y − 8x + 6y + 9 = 0 y2 + 6y = + 8x − 9 (y + 3)2 − 9 = + 8x − 9 (y + 3)2 = 8x Y2 = 8X 2
Y = 4(2)X a=2 The type is open rightward Referred to X, Y Axis Vertex
Y=0 (0, 0)
184
where X = x Y = y+3
Referred to x, y X=x,Y=y+3 Y=0 ⇒ y+3=0 X=0 ; Y=0 ⇒ x=0 ; y+3=0 ∴ V (0, − 3)
Focus
(a, 0) i.e. (2, 0)
Directrix Latus rectum Length
X = − a i.e. X = − 2 X = a i.e. X = 2 4a = 8 x = -2
Y
X=+2 ; Y=0 ⇒ x = 2, y + 3 = 0 F (2, − 3) X=−2 ⇒ x=−2 X=2 ⇒ x=2 8
y x x =2
V ( 0,-3)
x2 − 2x + 8y + 17 x2 − 2x (x − 1)2 − 1 (x − 1)2 (x − 1)2 X2 X2 a The type is open downward (v)
F(2,-3)
y = -3 X
Fig. 4. 39 =0 = − 8y − 17 = − 8y − 17 = − 8y − 16 = − 8(y + 2) = − 8Y where X = x − 1 = − 4(2)Y Y = y+2 =2 Y y x V (1,-2) y = -2
y = -4
F (1,-4)
Fig. 4. 40
185
X
Referred to X, Y Axis
X=0
Vertex
(0, 0)
Referred to x, y X=x−1,Y=y+2 X=0 ⇒ x−1=0 ⇒ x=1 X=0 ; Y=0 ⇒ x −1 = 0, y + 2 = 0 ∴ V (1, − 2)
Focus
(0, − a) i.e. (0, − 2)
X=0 ; Y=−2 ⇒ x − 1 = 0, y + 2 = − 2 F (1, − 4)
Directrix
Y=a
Y=2 ⇒ y+2=2 ⇒ y=0
Latus rectum
Y=−a
Length
4a = 8
i.e. Y = 2 i.e. Y = − 2
Y=−2 ⇒ y+2= −2 y=−4 8
4.3.7 Some practical problems : Example 4.8 : The girder of a railway bridge is in the parabolic form with span 100 ft. and the highest point on the arch is 10 ft. above the bridge. Find the height of the bridge at 10 ft. to the left or right from the midpoint of the bridge. y
Solution: Girder
C 10 (-50,-10)
Fig. 4. 41 Consider the parabolic girder as open downwards i.e.,
x2 = − 4ay
It passes through (50, − 10) ∴ 50 × 50 = − 4a (− 10)
186
x
B 10 A
(50,-10)
⇒ ∴
250 a= 4 250 x2 = − 4 4 y
x2 = − 250y Let B(10, y1) be a point on the parabola. ∴ 100 = − 250y1 100 2 y1 = − 250 = − 5 Let AB be the height of the bridge at 10 ft to the right from the mid point 2 AC = 10 and BC = 5 2 3 AB = 10 − 5 = 9 5 ft 3 i.e. the height of the bridge at the required place = 9 5 ft. Example 4.9 : The headlight of a motor vehicle is a parabolic reflector of diameter 12cm and depth 4cm. Find the position of bulb on the axis of the reflector for effective functioning of the headlight. Solution: By the property of parabolic reflector the position of the bulb should be placed at the focus. By taking the vertex at the origin, the equation of the reflector is y2 = 4ax Let PQ be the diameter of the reflector
y
P
12 cm
-- - 4 - -V
x
F
Q
Fig. 4. 42 ∴ P is (4, 6) and since P(4, 6) lies on the parabola, 36 = 4a × 4 ⇒ a = 2.25 The focus is at a distance of 2.25cm from the vertex on the x-axis.
187
∴ The bulb has to be placed at a distance of 2.25 cms from the centre of the mirror. Example 4.10 : On lighting a rocket cracker it gets projected in a parabolic path and reaches a maximum height of 4mts when it is 6 mts away from the point of projection. Finally it reaches the ground 12 mts away from the starting point. Find the angle of projection. Solution:
V (0,0) 4
6 mts
12
4
(- 6, - 4)
Fig. 4. 43
6 mts (6, - 4)
Fig. 4. 44 2
The equation of the parabola is of the form x = − 4ay (by taking the vertex at the origin). It passes through (6, − 4) 9 ∴ 36 = 16a ⇒ a = 4 …(1) The equation is x2 = − 9y Find the slope at (− 6, − 4) Differentiating (1) with respect to x, we get dy dy 2 2x = − 9 dx ⇒ dx = − 9 x 2 4 4 dy At (− 6, − 4), dx = − 9 × − 6 = 3 i.e. tan θ = 3 4 θ = tan−1 3 4 ∴ The angle of projection is tan−1 3 Example 4.11 : A reflecting telescope has a parabolic mirror for which the distance from the vertex to the focus is 9mts. If the distance across (diameter) the top of the mirror is 160cm, how deep is the mirror at the middle?
188
Solution: Let the vertex be at the origin. VF = a = 900 The equation of the parabola is y2 = 4 × 900 × x Let x1 be the depth of the mirror at the middle Since (x1, 80) lies on the parabola 16 802 = 4 × 900 × x1 ⇒ x1 = 9 16 ∴ depth of the mirror = 9 cm Example 4.12 :
y
(x1, 80)
x1
V(0,0)
F
x
----------900------
(x1, - 80)
Fig. 4. 45
Assume that water issuing from the end of a horizontal pipe, 7.5m above the ground, describes a parabolic path. The vertex of the parabolic path is at the end of the pipe. At a position 2.5m below the line of the pipe, the flow of water has curved outward 3m beyond the vertical line through the end of the pipe. How far beyond this vertical line will the water strike the ground? Solution: y
Pipe line 2.5 mts 7.5 mts
V (0,0)
3 mts
2.5
?
x P(3, -2.5)
3
7.5
Ground A
Fig. 4. 46
x1
Q (x1, -7.5)
Fig. 4. 47 As per the given information, we can take the parabola as open downwards i.e. x2 = − 4ay Let P be the point on the flow path, 2.5m below the line of the pipe and 3m beyond the vertical line through the end of the pipe. ∴ P is (3, − 2.5) Thus 9 = − 4a (− 2.5) 9 ⇒ a = 10
189
9 ∴ The equation of the parabola is x2 = − 4 × 10 y Let x1 be the distance between the bottom of the vertical line on the ground from the pipe end and the point on which the water touches the ground. But the height of the pipe from the ground is 7.5 m The point (x1, − 7.5) lies on the parabola 9 x12 = − 4 × 10 × (− 7.5) = 27 x1 = 3 3 ∴ The water strikes the ground 3 3 m beyond the vertical line. Example 4.13 : A comet is moving in a parabolic orbit around the sun which is at the focus of a parabola. When the comet is 80 million kms from the sun, the line segment π from the sun to the comet makes an angle of 3 radians with the axis of the orbit. find (i) the equation of the comet’s orbit (ii) how close does the comet come nearer to the sun? (Take the orbit as open rightward). Solution: Take the parabolic orbit as open rightward and the vertex at the origin. Let P be the position of the comet in which FP = 80 million kms. Draw a perpendicular PQ from P to the axis of the parabola. Let FQ = x1 From the triangle FQP, π PQ = FP . sin 3 3 = 80 × 2 = 40 3
y Comet P 80 V
/3 F ) π Sun x1 Q
x
Fig. 4. 48
π 1 Thus FQ = x1 = FP . cos 3 = 80 × 2 = 40 ∴ VQ = a + 40 if VF = a ; P is (VQ, PQ) = (a + 40, 40 3) Since P lies on the parabola y2 = 4ax
190
(40 3)
2
= 4a(a + 40)
⇒ a = − 60 or 20 a = − 60 is not acceptable. ∴ The equation of the orbit is y2 = 4 × 20 × x y2 = 80x The shortest distance between the Sun and the Comet is VF i.e. a ∴ The shortest distance is 20 million kms. Example 4.14 : A cable of a suspension bridge hangs in the form of a parabola when the load is uniformly distributed horizontally. The distance between two towers is 1500 ft, the points of support of the cable on the towers are 200ft above the road way and the lowest point on the cable is 70ft above the roadway. Find the vertical distance to the cable (parallel to the roadway) from a pole whose height is 122 ft. Solution : B
D Q
P C′ C
750
70
V
130
52
R′
70
R
70
A′ A
Fig. 4. 49 Take the lowest point on the cable as the vertex and take it as origin. Let AB and CD be the towers. Since the distance between the two towers is 1500 ft. VA′ = 750 ft ; AB = 200 ft ∴A′B = 200 − 70 = 130 ft Thus the point B is (750, 130) The equation of the parabola is x2 = 4ay Since B is a point on x2 = 4ay (750)2 = 4a(130) 75 × 750 ⇒ 4a = 13
191
∴ The equation is x2 =
75 × 750 y 13
Let PQ be the vertical distance to the cable from the pole RQ. RQ = 122, RR′ = 70 ⇒ R′Q = 52 Let VR′ be x1 ∴ Q is (x1, 52) Q is a point on parabola x12 =
75 × 750 × 52 13
x1 = 150 10 PQ = 2x1 = 300 10 ft. EXERCISE 4.1 (1) Find the equation of the parabola if (i) Focus : (2, − 3) ; directrix : 2y − 3 = 0 (ii) Focus : (− 1, 3) ; directrix : 2x + 3y = 3 (iii) Vertex : (0, 0) ; focus : (0, − 4) (iv) Vertex : (1, 4) ; focus : (− 2, 4) (v) Vertex : (1, 2) ; latus rectum : y = 5 (vi) Vertex : (1, 4) ; open leftward and passing through the point (− 2, 10) (vii) Vertex : (3, − 2) ; open downward and the length of the latus rectum is 8. (viii) Vertex : (3, − 1) ; open rightward ; the distance between the latus rectum and the directrix is 4. (ix) Vertex : (2, 3) ; open upward ; and passing through the point (6, 4). (2) Find the axis, vertex, focus, equation of directrix, latus rectum, length of the latus rectum for the following parabolas and hence sketch their graphs. (i) y2 = − 8x
(ii) x2 = 20y
(iii) (x − 4)2 = 4(y + 2)
(iv) y2 + 8x − 6y + 1 = 0
(v) x2 − 6x − 12y − 3 = 0 (3) If a parabolic reflector is 20cm in diameter and 5cm deep, find the distance of the focus from the centre of the reflector.
192
(4) The focus of a parabolic mirror is at a distance of 8cm from its centre (vertex). If the mirror is 25cm deep, find the diameter of the mirror. (5) A cable of a suspension bridge is in the form of a parabola whose span is 40 mts. The road way is 5 mts below the lowest point of the cable. If an extra support is provided across the cable 30 mts above the ground level, find the length of the support if the height of the pillars are 55 mts.
4.4 Ellipse : Definition : The locus of a point in a plane whose distance from a fixed point bears a constant ratio, less than one to its distance from a fixed line is called ellipse. Note : Eventhough the syllabus does not require the derivation of standard equation and the tracing of ellipse (4.4.1, 4.4.2) and it needs only the standard equation and the diagram, the equation is derived and the curve is traced for better understanding. Now, we will derive the standard equation of an ellipse.
4.4.1 Standard equation of the ellipse : Given : y x =a / e Fixed point F Fixed line l P(x,y) M Eccentricity e (e < 1) Moving point P (x, y) (a,0) C x Construction : Z F(ae,0) A A′ (-a,0) (a/e, 0) Plot the fixed point F and draw the fixed Fig. 4. 50 line l Drop a perpendicular (FZ) from F to l Drop a perpendicular (PM) Plot the points A, A′ which divides FZ internally and externally in the ratio e : 1 Take AA′ = 2a and treat it as x-axis. Draw a perpendicular bisector to AA′ and treat it as y-axis. Let C be the origin. The known points are the origin C(0, 0), A(a, 0), A′(− a, 0) To find the co-ordinates of F and M, do the following : Since A, A′ divides FZ internally and externally in the ratio e : 1 respectively, 193
FA e AZ = 1 ∴ FA = e AZ i.e., CA − CF = e (CZ − CA) ∴ a − CF = e (CZ − a) …(1) (2) + (1) (2) − (1)
∴ FA′ = e A′Z i.e. A′C + CF = e (A′C + CZ)
∴ a + CF = e(a + CZ) a 2a = e [2CZ] ⇒ CZ = e
⇒ ⇒
FA′ e =1 A′Z
2CF = e(2a)
⇒
… (2)
CF = ae
a ∴ M is e, y and F is (ae, 0)
To obtain the equation of the ellipse, do the following : Since P is any point on the ellipse FP 2 2 2 PM = e ⇒ FP = e PM
a i.e. (x − ae)2 + (y − 0)2 = e2x − e + (y − y)2 2
⇒ x2 − e2x2 + y2 = a2 − a2e2 (1 − e2) x2 + y2 = a2 (1 − e2) Dividing by a2 (1 − e2), we get y2 x2 + =1 a2 a2(1 − e2) x2 y2 + = 1, where b2 = a2 (1 − e2) a2 b2 This is known as the standard equation of an ellipse. i.e.
x2 y2 4.4.2 Tracing of the ellipse 2 + 2 = 1, a > b a b (i) Symmetry property : It is symmetrical about x-axis and y-axis simultaneously and hence about the origin. (ii) Special points : It does not pass through the origin.
194
To find the points on x-axis, put y = 0, we get x = ± a. Therefore the curve meets the x-axis at A(a, 0) and A′(− a, 0). To find the points on y-axis, put x = 0, we get y = ± b. Therefore the curve meets the y-axis at B(0, b) and B′(0, − b) (iii) Existence of the curve : b a2 − x2 . y is real only if Write the equation of the ellipse as y = ± a a2 − x2 ≥ 0. i.e., the curve does not exist for a2 − x2 < 0 or x2 − a2 > 0 Equivalently the curve does not exist for x > a and x < − a. Thus the curve exists only when − a ≤ x ≤ a. a b2 − y2 . x is real only if Write the equation of the ellipse as x = ± b b2 − y2 ≥ 0. The curve does not exist for b2 − y2 < 0 i.e., y2 − b2 > 0 i.e., the curve does not exist for y > b and y < − b. The curve exist only when − b ≤ y ≤ b. ∴ Ellipse is a closed curve bounded by the lines x = ± a and y = ± b. Thus the curve is y x = - a/e M′
Z′ (- a/e, 0)
A′ (- a,0)
x = a/e
B
C (0,0)
P(x,y)
F1 (ae,0)
M
A Z (a,0) (a/e, 0)
x
B′
Fig. 4. 51 4.4.3 Important definitions regarding ellipse : Focus : The fixed point is called focus, denoted as F1 (ae, 0). Directrix : The fixed line is called directrix l of the ellipse and its equation a is x = e . Major axis : The line segment AA′ is called the major axis and the length of the major axis is 2a. The equation of the major axis is y = 0.
195
Minor axis : The line segment BB′ is called the minor axis and the length of minor axis is 2b. Equation of the minor axis is x = 0. Note that the length of major axis is always greater than minor axis. Centre : The point of intersection of the major axis and minor axis of the ellipse is called the centre of the ellipse. Here C(0, 0) is the centre of the ellipse. Note that the centre need not be the origin of the ellipse always. End points of latus rectum and length of latus rectum : To find the end points, solve x = ae … (1)
and
x2 y2 + = 1 … (2) a2 b2
Using (2) in (1) we get y2 a2 e2 + =1 a2 b2 ∴
y2 = 1 − e2 b2
∴ y2 = b2 (1 − e2)
‡ b2 = a2 (1 − e2) b = b2 2 b2 2 a or 2 = 1 − e a 2
b2 ∴ y=± a
b2 If L1 and L1′ are the end points of the latus rectum then L1 is ae, a and b2 L1′ is ae, − a . b2 The end points of the other latus rectum are − ae, ± a . 2b2 The length of the latus rectum is a . For the above discussed ellipse, the major axis is along x-axis. There is another standard ellipse in which the major axis is along the y-axis. Vertices : The points of intersection of the ellipse and its major axis are called its vertices. Here the vertices of the ellipse are A(a, 0) and A′(− a, 0).
196
Focal distance : The focal distance with respect to any point P on the ellipse is the distance of P from the referred focus. Focal chord : A chord which passes through the focus of the ellipse is called the focal chord of the ellipse. A special property : Thanks to the symmetry about the origin, it permits (i) the second focus F2 (− ae, 0) a (ii) the second directrix x = − e Latus rectum : It is a focal chord perpendicular to the major axis of the ellipse. The equations of latus rectum are x = ae, x = − ae. Eccentricity e=
b2 1− 2 a
Remark : b In the case of an ellipse 0 < e < 1. As e → 0, a → 1 i.e., b → a or the length of the minor and major axes are close in size. i.e., the ellipse is close to being a circle. b As e → 1, a → 0 and the ellipse degenerates into a line segment (degenerate conic) i.e., the ellipse is flat.
4.4.4 The other standard form of the ellipse : If the major axis of the ellipse is along the y-axis, then the equation of the x2 y2 ellipse takes the form 2 + 2 = 1, for a > b. b a For this type of ellipse, we have the following as explained in the earlier ellipse. Centre : C (0, 0) Vertices : A (0, a), A′ (0, − a) Foci
:
F1 (0, ae), F2 (0, − ae)
Equation of major axis is x = 0 Equation of minor axis is y = 0 End points of minor axis : B (b, 0), B′(− b, 0) Equation of directrices : y = ± a/e
197
b2 b2 End points of latus rectums : ± a , ae , ± a , − ae y y =a/e
A(0,a)
(-b2/a ,ae) L11
L1 (-b2/a,ae)
B1
x
B
C
(-b2/a -,ae)L21
x
L2 (b2/a,-ae)
A′(0,-a)
y =- a/e
Fig. 4. 52 4.4.5 General forms of standard ellipses : To obtain the general forms of standard ellipses, replace x by x − h and y by (y − k) if the centre is C(h, k). Thus the general forms of standard ellipses are
(y − k)2 (x − h)2 + = 1, a2 b2
(x − h)2 (y − k)2 + = 1, a > b b2 a2 Focal property of an ellipse : The sum of the focal distances of any point on an ellipse is constant and is equal to the length of the major axis. y x = - a/e
x = a/e P
M′ Z′
F2 C
F1
Fig. 4. 53
198
M Z
x
Proof : To prove : F1P + F2P = 2a Let P be a point on the ellipse. Drop the perpendiculars PM and PM′ to the a a directrices x = e and x = − e respectively . F2P F1P =e We know that PM = e, PM ′ ∴ F1P = ePM, F2P = e PM′ ∴ F1P + F2P = e(PM + PM′) = e(MM′) 2a = e. e = 2a = length of the major axis Example 4.15 : Find the equation of the ellipse whose foci are (1, 0) and 1 (− 1, 0) and eccentricity is 2 . Solution: The centre of the ellipse is the midpoint of FF′ where F is (1, 0) and F′ is (− 1, 0). y 1−1 0+0 ∴ Centre C is 2 , 2 = (0, 0) 1 C x But F1F2 = 2ae = 2 and e = 2 F1 (1,0) (-1,0) F2 1 2a × 2 = 2 Fig. 4. 54 a=2 1 b2 = a2 (1 − e2) = 4 1 − 4 = 3
From the given data the major axis is along x-axis. ∴ the equation of the ellipse is of the form (y − k)2 x2 y2 (x − h)2 + = 1 ⇒ 4 + 3 =1 2 2 a b
199
Example 4.16 : Find the equation of the ellipse whose one of the foci is (2, 0) 1 and the corresponding directrix is x = 8 and eccentricity is 2 Solution: Let P(x, y) be a moving point. By definition FP PM = e
x =8
y P(x,y)
∴ FP2 = e2 PM2
M (8,y)
2
1 x − 8 (x − 2)2 + (y − 0)2 = 4 ± 1 1 (x − 2)2 + y2 = 4 (x − 8)2 2
2
4 [(x − 2) + y ] = (x − 8)
x
F(2,0)
Fig. 4. 55
2
3x2 + 4y2 = 48 x2 y2 16 + 12 = 1 Aliter : From the given data, the major axis is along the x-axis and the equation of the ellipse may be taken as
y x =8
x2 y2 + =1 a2 b2 a FZ = e − ae = 6 1 1 But e = 2 ⇒ 2a − 2 a = 6 3 ⇒ 2a = 6 ⇒ a=4
C
F(2,0)
Z(8,0)
Fig. 4. 56
1 3 b2 = a2 (1 − e2) = 16 1 − 4 = 16 × 4 = 12
2
2
x y ∴ The required equation is 16 + 12 = 1
200
x
Example 4.17 : Find the equation of the ellipse with focus (− 1, − 3), directrix 4 x − 2y = 0 and eccentricity 5
∴ FP2 = e2 PM2 16 x − 2y (x + 1) + (y + 3) = 25 ± 1 + 4 2
2
2
x - 2y =0
Solution: Let P(x, y) be a moving point. By definition FP PM = e
F(-1,-3)
125 [(x + 1)2 + (y + 3)2 ] = 16 (x − 2y)2 2
P(x,y)
Fig. 4. 57
2
⇒ 109x + 64xy + 61y + 250x + 750y + 1250 = 0 Example 4.18 : Find the equation of the ellipse with foci (± 4, 0) and vertices (± 5, 0) Solution: y Let the foci be F1(4, 0) and F2 (− 4, 0), vertices be A(5, 0) and A′(− 5, 0). The centre is C (0,0) F1 A A′ F2 the midpoint of AA′ x −5+5 0+0 i.e., C is 2 , 2 = (0, 0)
(-5,0) (-4,0)
(4,0) (5,0)
Fig. 4. 58 From the given data, the major axis is along the x-axis and the equation of the ellipse is of the form x2 y2 + =1 a2 b2 Here CA = a = 5 4 since e = 5
CF = ae = 4
b2 = a2 (1 − e2) = 25 − 16 = 9 and the x2 y2 required equation of the ellipse is 25 + 9 = 1
201
Example 4.19 : The centre of the ellipse is (2, 3). One of the foci is (3, 3). Find the other focus. Solution: From the given data the major axis is parallel to the x axis. Let F1 be (3, 3) F C F 2
Let F2 be the point (x, y). Since C (2, 3) is the midpoint of F1 and F2 on the major axis y = 3 x+3 y+3 2 = 2 and 2 = 3
y =3
1
Fig. 4. 59
This gives x = 1 and y = 3. Thus the other focus is (1, 3). Example 4.20 : Find the equation of the ellipse whose centre is (1, 2), one of 1 the foci is (1, 3) and eccentricity is 2 Solution: The major axis is parallel to y-axis. ∴ The equation is of the form
y x =1 F1(1,3) C(1,2)
(x − h)2 (y − k)2 + =1 b2 a2 CF1 = ae = 1
x
1 But e = 2 ⇒ a = 2, a2 = 4
Fig. 4. 60
1 b2 = a2(1 − e2) = 41 − 4 = 3 ; C(h, k) = (1, 2) ∴ The required equation is
(x − 1)2 (y − 2)2 + =1 3 4
Example 4.21 : Find the equation of the ellipse whose major axis is along 1 x-axis, centre at the origin, passes through the point (2, 1) and eccentricity 2 Solution: Since the major axis is along the x-axis and the centre is at the origin, the x2 y2 equation of the ellipse is of the form 2 + 2 = 1 a b
202
4
1
… (1)
It passes through the point (2, 1). ∴ 2 + 2 = 1 a b 1 e=2
1 b2 = a2 (1 − e2) ⇒ b2 = a2 1 − 4
2
2
∴ 4b = 3a … (2) 16 2 2 Solving (1) and (2) we get a = 3 , b =4 2 x y2 ∴ The required equation is 16/3 + 4 = 1 Example 4.22 : Find the equation of the ellipse if the major axis is parallel to y-axis, semi-major axis is 12, length of the latus rectum is 6 and the centre is (1, 12) Solution: Since the major axis is parallel to y-axis the equation of the ellipse is of the form (x − h)2 (y − k)2 + =1 b2 a2 The centre C (h, k) is (1, 12) a = 12 ⇒ a2 = 144 Semi major axis 2b2 2b2 Length of the latus rectum a = 6 ⇒ 12 = 6 (x − 1)2 (y − 12)2 ∴ b2 = 36 and the required equation is + 144 =1 36 Example 4.23 : Find the equation of the ellipse given that the centre is (4, − 1), focus is (1, − 1) and passing through (8, 0). Solution : y From the given data since the major axis is parallel to the x axis, x C(4,-1) the equation is of the form y =-1 F2(1,-1)
(x − h)2 (y − k)2 + =1 a2 b2 The centre C(h, k) is (4, − 1) 2
Fig. 4. 61
2
(x − 4) (y + 1) + =1 a2 b2
203
16 1 It passes through (8, 0) ∴ 2 + 2 = 1 b a But CF1 = ae = 3
(1)
⇒
b2 = a2 (1 − e2) = a2 − a2 e2 = a2 − 9 16 1 =1 2+ 2 a a −9
⇒
16a2 − 144 + a2 = a4 − 9a2
⇒
a4 − 26a2 + 144 = 0 a2 = 18 or 8
⇒ Case (i)
:
… (1)
a2 = 18 b2 = a2 − 9 = 18 − 9 = 9
Case (ii)
:
a2 = 8 b2 = 8 − 9 = − 1 which is not possible
∴ Thus the equation is
a2 = 18,
b2 = 9
(x − 4)2 (y + 1)2 + =1 18 9
Example 4.24 : Find the equation of the ellipse whose foci are (2, 1), (− 2, 1) and length of the latus rectum is 6. Solution : From the given data the major axis is parallel to the x axis. ∴ The equation is of the form (x − h)2 (y − k)2 + =1 a2 b2 Since the centre is the midpoint of F1F2 − 2 + 2 1 + 1 , 2 C is 2 = (0, 1) and the equation becomes x2 (y − 1)2 + =1 a2 b2
204
y
F2(-2,1)
F1(2,1)
y =1 x
Fig. 4. 62
F1 F2 = 2ae = 4 ⇒ a2 e2 = 4 a2 e2 = a2 − b2 ∴ a2 − b2 = 4 2b a
The length of the latus rectum
2
a2 − 3a − 4 a a ∴a
(1) ⇒ ⇒
=6 = = = =
… (1) b2 = 3a
0 4 or − 1 − 1 is absurd 4
… (2) (by (2))
b2 = 3a = 12 x2 (y − 1)2 =1 16 + 12 Example 4.25 : Find the equation of the ellipse whose vertices are (− 1, 4) and 1 (− 7, 4) and eccentricity is 3 . y Solution : y From the given data the major axis is parallel to x axis. A′ A y =4 ∴ The equation is of the form (-7,4) (-1,4) Thus the equation is
(x − h)2 (y − k)2 + =1 a2 b2
x
The centre is the midpoint of AA′ − 1 − 7 4 + 4 ∴ C is 2 , 2 = (− 4, 4) Thus the equation becomes (y − 4)2 (x + 4)2 + =1 a2 b2 We know that AA′ = 2a = 6
Fig. 4. 63
⇒ a=3 1 b2 = a2 (1 − e2) = 9 1 − 9 = 8
The required equation is
(x + 4)2 (y − 4)2 + =1 9 8
205
Example 4.26 : Find the equation of the ellipse whose foci are (1, 3) and (1, 9) 1 and eccentricity is 2 y Solution : x =1 From the given data the major axis is parallel to y axis. F2(1,9) ∴ The equation is of the form (x − h)2 (y − k)2 + =1 b2 a2 The centre of the ellipse is the midpoint of F1 F2 1+1 3+9 ∴ C is 2 , 2 = (1, 6)
F1(1,3) x
Fig. 4. 64
F1 F2 = 2ae = 6 ae = 3 1 But e = 2
∴ a=6
1 b2 = a2 (1 − e2) = 36 1 − 4 = 27
Thus the required equation is (y − 6)2 (x − 1)2 + 27 36 = 1 Property (without proof) : A point moves such that the sum of its distances from two fixed points in a plane is a constant. The locus of this point is an ellipse. Example 4.27 : Find the equation of a point which moves so that the sum of its distances from (− 4, 0) and (4, 0) is 10. y P(x,y)
F1(4,0)
F2(-4,0)
Fig. 4. 65
206
x
Solution : Let F1 and F2 be the fixed points (4, 0) and (− 4, 0) respectively and P(x1, y1) be the moving point. F1P + F2P = 10
It is given that i.e.,
(x1 − 4)2 + (y1 − 0)2 +
(x1 + 4)2 + (y1 − 0)2 = 10
Simplifying we get 9x12 + 25y12 = 225. ∴ The locus of (x1, y1) is x2 y2 + 25 9 =1 Example 4. 28 : Find the equations and lengths of major and minor axes of (x − 1)2 (y + 1)2 x2 y2 (iii) + 16 =1 (i) 9 + 4 = 1 (ii) 4x2 + 3y2 = 12 9 Solution : (i) The major axis is along x-axis and the minor axis is along y-axis. This gives the equation of major axis as y = 0 and the equation of minor axis as x = 0. We have a2 = 9 ; b2 = 4 ⇒ a = 3, b = 2 ∴ The length of major axis is 2a = 6 and the length of minor axis is 2b = 4 (ii)
x2 y2 3 + 4 =1
The major axis is along y-axis and the minor axis is along x-axis. ∴ The equation of major axis is x = 0 and the equation of minor axis is y = 0. Here a2 = 4 ; b2 = 3 ⇒ a = 2, b = 3 ∴ The length of major axis (2a) = 4 The length of minor axis (2b) = 2 3 (iii) Let x − 1 = X and y + 1 = Y X2 Y 2 ∴ The given equation becomes 9 + 16 = 1 Clearly the major axis is along Y-axis and the minor axis is along X-axis. ∴ The equation of major axis is X = 0 and the equation of minor axis is Y=0 i.e., the equation of major axis is x − 1 = 0 and the equation of minor axis is y + 1 = 0
207
Here a2 = 16, b2 = 9 ⇒ a = 4, b = 3 ∴ Length of major axis (2a) = 8 ∴ Length of minor axis (2b) = 6 Example 4. 29 : Find the equations of axes and length of axes of the ellipse 6x2 + 9y2 + 12x − 36y − 12 = 0 Solution : 6x2 + 9y2 + 12x − 36y − 12 = 0 (6x2 + 12x) + (9y2 − 36y) = 12 6(x2 + 2x) + 9(y2 − 4y) = 12 6{(x + 1)2 − 1} + 9 {(y − 2)2 − 4} = 12 6(x + 1)2 + 9 (y − 2)2 = 12 + 6 + 36
Let X = x + 1 ;
6(x + 1)2 + 9(y − 2)2 = 54 (y − 2)2 (x + 1)2 + =1 9 6 Y=y−2
X 2 Y2 9 + 6 =1 Clearly the major axis is along X-axis and the minor axis is along Y-axis. ∴ The equation of the major axis is Y = 0 and the equation of the minor axis is X = 0. The equation of the major axis is y − 2 = 0 and of minor axis is x + 1 = 0 i.e., the equation of the major axis is y − 2 = 0 ∴ The equation becomes
Here a2 = 9, b2 = 6 ⇒ a = 3, b = 6 ∴ The length of major axis (2a) = 6 The length of minor axis (2b) = 2 6 Example 4.30 : Find the equations of directrices, latus rectum and length of latus rectums of the following ellipses. x2 y2 (i) 16 + 9 = 1 (ii) 25x2 + 9y2 = 225 Solution : (i) The major axis is along x-axis Here a2 = 16, b2 = 9
208
(iii) 4x2 + 3y2 + 8x + 12y + 4 = 0
b2 1− 2 = a a x = ±e
e= Equations of directrices are
x= Equations of the latus rectums are
9 7 1 − 16 = 4
± 16 7
x = ± ae x=± 7
Length of the latus rectum (ii) 25x2 + 9y2 = 225 Here a2 = 25,
2b a
2
=
2×9 9 4 = 2
x2 y2 ∴ 9 + 25 = 1
b2 = 9 e=
b2 1− 2 = a
9 4 1 − 25 = 5
a y =±e
The equations of the directrices are
y=
± 25 4
Equations of the latus rectum are y = ± ae y = ±4 Length of the latus rectum is (iii)
2×9 18 2b2 a = 5 = 5
4x2 + 3y2 + 8x + 12y + 4 = 0 (4x2 + 8x) + (3y2 + 12y) + 4 = 0 4(x2 + 2x) + 3(y2 + 4y) = − 4 4{(x + 1)2 − 1} + 3 {(y + 2)2 − 4} = − 4 4(x + 1)2 + 3(y + 2)2 = 12 (y + 2)2 (x + 1)2 + =1 3 4
209
X 2 Y2 3 + 4 = 1 where X = x + 1 , Y = y + 2 The major axis is along Y axis. Here a2 = 4,
1 b2 = 3 and e = 2
a Y =±e
Equations of the directrices are
2 i.e. Y = ± (1/2)
Y = ±4 (i) Y = 4 ⇒ y+2=4 ⇒ y=2 (ii) Y = − 4 ⇒ y + 2 = − 4 ⇒ y = − 6 The directrices are y = 2 and y = − 6 Y = ± ae
Equations of the latus rectum are
1 i.e. Y = ± 2 2
Y = ±1 ⇒ y+2=1 ⇒ y=−1 (ii) Y = − 1 ⇒ y + 2 = 1 ⇒ y=−3 ∴ Equation of the latus rectum are y = − 1 and y = − 3
(i) Y = 1
2×3 2b2 Length of the latus rectum is a = 2 = 3 Example 4.31 : Find the eccentricity, centre, foci, vertices of the following x2 y2 x2 y2 ellipses : (i) 25 + 9 = 1 (ii) 4 + 9 = 1 (iii)
(x + 3)2 (y − 5)2 + =1 6 4
(iv) 36x2 + 4y2 − 72x + 32y − 44 = 0
x2 y2 Solution : (i)25 + 9 = 1 The major axis is along x-axis a2 = 25, b2 = 9 4 e = 5 and ae = 4 Clearly centre C is (0, 0), A′ F2 Foci are (± ae, 0) = (± 4, 0) (-5,0) (-4,0) Vertices are (± a, 0) = (± 5, 0)
y
C (0,0)
A
(4,0) (5,0)
Fig. 4. 66
210
F1
x
(ii) The major axis is along y-axis a2 = 9, b2 = 4 5 e = 3 and ae = 5 Clearly centre C is (0, 0) Foci are (0, ± ae) = (0, ± 5 ) Vertices are (0, ± a) = (0, ± 3)
y A(0,3) F1 (0,√5)
C (0,0)
x
F2 (0,-√5)
A′(0,-3)
Fig. 4. 67 (iii) Let x + 3 = X, y − 5 = Y Y2 X2 ∴ The equation becomes 6 + 4 = 1 The major axis is along X-axis a2 = 6, b2 = 4 1 e= and ae = 2 3 Referred to X, Y Centre
(0, 0)
(± a, 0) i.e. (± 6, 0) (i) ( 6, 0) Vertices
(ii) (− 6, 0)
Referred to x, y X=x+3; Y=y−5 X = 0 ; Y=0 ⇒ x + 3 = 0, y − 5 = 0 x = − 3, y = 5 Centre C(− 3, 5) (i) X = 6, Y=0 x+3 = 6 , y−5=0 x = 6 − 3, y = 5 A (− 3 + 6, 5) (ii)
211
X = − 6,Y=0 x+3 = − 6,y−5=0 x = −3− 6 ,y=5 A′ (− 3 − 6, 5)
(± ae, 0) i.e. (± 2 , 0)
(i)
X = 2, Y=0 x+3 = 2, y−5=0 x = −3+ 2,y=5 F1 (− 3 + 2 , 5)
(ii)
X = − 2 , Y=0 x+3 = − 2,y−5=0 x = −3− 2,y=5 F2 (− 3 − 2 , 5)
(i) ( 2 , 0) foci
(ii) (− 2, 0)
Y A1 (-3 - √6, 5)
F2
y
C (-3,5) F1
AA (-3 +√6, 5) (-3+√2,5)
(-3- √2, 5)
y =5 X
x
Fig. 4. 68 (iv) 36x2 + 4y2 − 72x + 32y − 44 = 0 36 (x2 − 2x) + 4 (y2 + 8y) = 44 36 {(x − 1)2 − 1} + 4 {(y + 4)2 − 16} = 44 36(x − 1)2 + 4 (y + 4)2 = 144 (y + 4)2 (x − 1)2 + =1 4 36 i.e.,
Y2 X2 + 4 36 = 1 where X = x − 1 , Y = y + 4
The major axis is along Y-axis. a2 = 36, b2= 4 2 2 e = 3 and ae = 4 2
212
Referred to x, y X=x−1; Y=y+4 X = 0 ; Y=0 ⇒ x − 1 = 0, y + 4 = 0 x = 1, y = − 4 Centre C(1, − 4) (i) X = 0, Y = 6 x−1 = 0, y+4=6 x = 1, y = 2 A (1, 2) X = 0, Y = − 6 (ii) x − 1 = 0, y + 4 = − 6 x−1=0,y+4=−6 x = 1, y = − 10 A′ (1, − 10) (i) X = 0; Y=4 2 x − 1 = 0, y + 4 = 4 2 x = 1, y = 4 2 − 4 F1 (1, 4 2 − 4)
Referred to X, Y Centre
(0, 0)
(0, ± a) i.e. (0, ± 6) (i) (0, 6) Vertices (ii) (0, − 6)
(0, ± ae) i.e. (0, ± 4 2) (i) (0, 4 2) Foci (ii) (0, − 4 2)
(ii)
X = 0, Y=−4 2 x − 1= 0 ; y + 4 = − 4 2 x = 1, y = − 4 − 4 2 F2 (1, − 4 − 4 2)
y Y
x =1 A(1,2) F1
C (1,-4) F2
′
A (1,-10)
Fig. 4. 69
213
x X Where F1(1, 4 √2 – 4) F2(1, - 4 – 4 √2)
4.4.6 Some practical problems : Example 4.32 : An arch is in the form of a semi-ellipse whose span is 48 feet wide. The height of the arch is 20 feet. How wide is the arch at a height of 10 feet above the base? Solution : Take the mid point of the (x1,10) base as the centre C (0, 0) 20 Since the base wide is 48 feet, 10 x1 Q P the vertices A and A′ are (24, 0) A C (0,0) A′ and (− 24, 0) respectively. (24,0) (-24,0) Fig. 4. 70 Clearly 2a = 48 and b = 20. x2 y2 + =1 … (1) 242 202 Let x1 be the distance between the pole whose height is 10m and the centre.
The corresponding equation is
Then (x1, 10) satisfies the equation (1) ∴
x12 24
2
+
102 =1 202
⇒ x1 = 12 3
Clearly the width of the arch at a height of 10 feet is 2x1 = 24 3 Thus the required width of arch is 24 3 feet. Example 4.33 : The ceiling in a hallway 20ft wide is in the shape of a semi ellipse and 18 ft high at the centre. Find the height of the ceiling 4 feet from either wall if the height of the side walls is 12ft. Solution : R (6, y1) Let PQR be the height of the ceiling which is 4 feet from the 6 10 6 4 wall. A A′ C (0,0) Q (10,0) (-10,0) From the diagram PQ = 12 ft To find the height QR 12 Since the width is 20ft, take A, A′ as vertices with A as (10, 0) and P A′ as (− 10, 0). Take the midpoint Fig. 4. 71 of AA′ as the centre which is (0, 0)
214
From the diagram and
AA′ = 2a = 20 ⇒ a = 10 b = 18 − 12 = 6
y2 x2 + 100 36 = 1 Let QR be y1 then R is (6, y1) ∴
Since R lies on the ellipse, y12 36 100 + 36 = 1 ⇒ y1 = 4.8 ∴ PQ + QR = 12 + 4.8 ∴ The required height of the ceiling is 16.8 feet. Example 4.34 : The orbit of the earth around the sun is elliptical in shape with sun at a focus. The semi major axis is of length 92.9 million miles and eccentricity is 0.017. Find how close the earth gets to sun and the greatest possible distance between the earth and the sun. Solution : y Semi-major axis CA is H a = 92.9 million miles RT EA Given e = 0.017 SUN A A′ x The closest distance of the C (0,0) F earth from the sun = FA and farthest distance of the earth from the sun = FA′ Fig. 4. 72 CF = ae = 92.9 × 0.017 FA = CA − CF = 92.9 − 92.9 × 0.017 = 92.9 [1 − 0.017] = 92.9 × 0.983 = 91.3207 million miles FA′ = CA′ + CF = 92.9 + 92.9 × 0.017 = 92.9 (1 + 0.017) = 92.9 × 1.017 = 94.4793 million miles Example 4.35 : A ladder of length 15m moves with its ends always touching the vertical wall and the horizontal floor. Determine the equation of the locus of a point P on the ladder, which is 6m from the end of the ladder in contact with the floor.
215
Solution : Let AB be the ladder and P(x1, y1) be a point on the ladder such that AP = 6m. Draw PD perpendicular to x-axis and PC perpendicular to y-axis. Clearly the triangles ADP and PCB are similar. PC PB BC ∴ DA = AP = PD i.e.,
B 9 C
x1
y1 O
x1 9 BC DA = 6 = y1 ⇒
y
P(x1y1) 6
y1 x1
D
A
x
Fig. 4. 73
6x1 2x1 DA = 9 = 3
9y1 3 ; BC = 6 = 2 y1
2x1 3y1 5 5 OA = OD + DA = x1 + 3 = 3 x1 ; OB = OC + BC = y1 + 2 = 2 y1 But
OA2 + OB2 = AB2
⇒
25 2 25 2 9 x1 + 4 y1 = 225
x12 y12 9 + 4 =9 x2 y2 ∴ The locus of (x1, y1) is 81 + 36 = 1, which is an ellipse. EXERCISE 4.2 (1) Find the equation of the ellipse if (i)
one of the foci is (0, − 1), the corresponding directrix is 3 3x + 16 = 0 and e = 5
(ii)
1 the foci are (2, − 1), (0, −1) and e = 2
(iii)
the foci are (± 3, 0) and the vertices are (± 5, 0)
(iv)
3 the centre is (3, − 4), one of the foci is (3 + 3, − 4) and e = 2
216
2 the centre at the origin, the major axis is along x-axis, e = 3 and −5 passes through the point 2, 3 80 (vi) the length of the semi major axis, and the latus rectum are 7 and 7 respectively, the centre is (2, 5) and the major axis is parallel to y-axis. (vii) the centre is (3, − 1), one of the foci is (6, − 1) and passing through the point (8, − 1). 32 (viii) the foci are (± 3, 0), and the length of the latus rectum is 5 . (v)
3 the vertices are (± 4, 0) and e = 2 (2) If the centre of the ellipse is (4, − 2) and one of the focus is (4, 2), find the other focus? (3) Find the locus of a point which moves so that the sum of its distances from (3, 0) and (− 3, 0) is 9 (4) Find the equations and length of major and minor axes of (ix)
(i) 9x2 + 25y2 = 225
(iii) 9x2 + 4y2 = 20
(ii) 5x2 + 9y2 + 10x − 36y − 4 = 0 (iv) 16x2 + 9y2 + 32x − 36y − 92 = 0 (5) Find the equations of directrices, latus rectum and lengths of latus rectums of the following ellipses : (i) 25x2 + 169y2 = 4225
(ii) 9x2 + 16y2 = 144
(iv) 3x2 + 2y2 − 30x − 4y + 23 = 0 (iii) x2 + 4y2 − 8x − 16y − 68 = 0 (6) Find the eccentricity, centre, foci, vertices of the following ellipses and draw the diagram : (i) 16x2 + 25y2 = 400 (ii) x2 + 4y2 − 8x − 16y − 68 = 0 (iv) 16x2 + 9y2 + 32x − 36y = 92 (iii) 9x2 + 4y2 = 36 (7) A kho-kho player in a practice session while running realises that the sum of the distances from the two kho-kho poles from him is always 8m. Find the equation of the path traced by him if the distance between the poles is 6m. (8) A satellite is travelling around the earth in an elliptical orbit having the earth at a focus and of eccentricity 1/2 . The shortest distance that the satellite gets to the earth is 400 kms. Find the longest distance that the satellite gets from the earth.
217
(9) The orbit of the planet mercury around the sun is in elliptical shape with sun at a focus. The semi-major axis is of length 36 million miles and the eccentricity of the orbit is 0.206. Find (i) how close the mercury gets to sun? (ii) the greatest possible distance between mercury and sun. (10) The arch of a bridge is in the shape of a semi-ellipse having a horizontal span of 40ft and 16ft high at the centre. How high is the arch, 9ft from the right or left of the centre.
4.5 Hyperbola : Definition: The locus of a point whose distance from a fixed point bears a constant ratio, greater than one to its distance from a fixed line is called a hyperbola. Note : Eventhough the syllabus does not require the derivation of standard equation and the tracing of hyperbola (4.5.1, 4.5.2) and it needs only the standard equation and the diagram, the equation is derived and the curve is traced for better understanding. We shall now derive the standard equation of the hyperbola.
4.5.1. Standard equation of the hyperbola : y Given : l Fixed point (F) Fixed line (l) P(x,y) M Eccentricity e, (e > 1) Moving point P(x, y) x Construction A′ C Z A F(ae,0) Plot the fixed point F and draw the fixed line ‘l’. Drop a perpendicular (FZ) Fig. 4. 74 from F to l. Drop a perpendicular (PM) from P to l. Plot the points A, A′ which divides FZ internally and externally in the ratio e : 1 respectively. Take AA′ = 2a and treat it as x-axis. Draw a perpendicular bisector of AA′ and treat it as y-axis. Let C be the origin. The known points are C(0, 0), A(a, 0), A′(− a, 0). To find the co-ordinates of F and M do the following : Since A, A′ divides FZ internally and externally in the ratio e : 1 respectively,
218
FA e AZ = 1 ∴ FA = e AZ i.e. CF − CA = e (CA − CZ) ∴ CF − a = e (a − CZ) …(1) (2) − (1)
⇒
FA′ e =1 A′Z ∴ FA′ = e A′Z i.e. A′C + CF = e (A′C + CZ)
∴ a + CF = e(a + CZ) a 2a = e [2CZ] ⇒ CZ = e
… (2)
(2) + (1) ⇒ 2CF = e(2a) ⇒ CF = ae a ∴ M is e, y and F is (ae, 0) To obtain the equation of the hyperbola we do the following: Since P is a point on the hyperbola, FP 2 2 2 We have PM = e ⇒ FP = e PM a 2 ∴ (x − ae)2 + (y − 0)2 = e2 x − e + (y − y)2
2 2
2
[e x − 2aex + a ] x2 − 2aex + a2e2 + y2 = e2 e2 x2 − e2x2 + y2 = a2 − a2e2 (e2 − 1)x2 − y2 = a2(e2 − 1) 2 x y2 − =1 a2 a2(e2 − 1) y2 x2 2 2 2 ∴ 2 − 2 = 1 where b = a (e − 1) is a positive quantity. a b This is the required standard equation of the hyperbola.
4.5.2 Tracing of the hyperbola
x2 y2 − =1 a2 b2
(i) Symmetry : The hyperbola is symmetric about x-axis, y-axis and hence the hyperbola is symmetric about the origin. (ii) Special points : The hyperbola does not pass through the origin. To find the points on x-axis, put y = 0, we get x = ± a. Therefore the curve meets the x-axis at A(a, 0) and A′(− a, 0).
219
To find the points on y-axis, put x = 0, we get y2 = −b2. i.e., y is imaginary. Therefore the curve does not meet the y-axis. (iii) Existence of the curve : b Write the equation of the hyperbola as y = ± a x2 − a2 . If x2 − a2 < 0 i.e., − a < x < a, y is imaginary. i.e., the curve does not exist for − a < x < a. Therefore the curve exists for x ≤ − a and x ≥ a. Note that for all y, the curve exists. y (iv) The curve at infinity : As x increases y2 also increases i.e., as x → ∞, y2 → ∞. as x → ∞, y → ± ∞. Thus the curve branches out to infinity on either side.
x = a/e
x = -a/e
F2 A′ (-ae,0)
B A
C
Z
Z′ B′
F1 (ae,0)
x
Fig. 4. 75
4.5.3 Important definitions regarding hyperbola : Focus : The fixed point is called a focus F1 (ae, 0) of the hyperbola. Directrix : The fixed line is called the directrix of the hyperbola and its a equation is x = e . Transverse axis : The line segment AA′ joining the vertices is called the transverse axis and the length of the transverse axis is 2a. The equation of transverse axis is y = 0. Note that the transverse axes cut both the branches of the curve. Conjugate axis : The line segment joining the points B(0, b) and ′ B (0, − b) is called the conjugate axis. The length of the conjugate axis is 2b. The equation of the conjugate axis is x = 0 Centre : The point of intersection of the transverse and conjugate axes of the hyperbola is called the centre of the hyperbola. Here C(0, 0) is called the centre of the hyperbola. Vertices : The points of intersection of the hyperbola and its transverse axis is called its vertices. The vertices of the hyperbola are A(a, 0) and A′(− a, 0). As in the case of ellipse, hyperbola also has the special property of the a second focus F2(− ae, 0) and the second directrix x = − e .
220
Eccentricity : e =
b2 1+ 2 a
Remark : b In the case of a hyperbola e > 1. As e → 1, a → 0 i.e., as e → 1, b is very small related to a and the hyperbola becomes a pointed nose. As e → ∞, b is very large related to a and the hyperbola becomes flat. Latus rectum : It is a focal chord perpendicular to the transverse axis of the hyperbola. The equations of the latus rectum are x = ± ae. End points of latus rectum and length of latus rectum : To find the end points, solve x = ae … (1)
and
x2 y2 − = 1 … (2) a2 b2
Using (1) in (2) we get a2 e2 y2 − =1 a2 b2 ∴
y2 = e2 − 1 b2
∴ y2 = b2 (e2 − 1)
b = b 2 . 2 a 2
(‡ b2 = a2 (e2 − 1))
b2 ∴ y=± a
b2 If L1 and L1′ are the end points of one latus rectum then L1 is ae, a and b2 L1′ is ae, − a . b2 Similarly the end points of the other latus rectum are − ae, ± a and the 2b2 length of the latus rectum is a . For the above discussed hyperbola, the transverse axis is along x-axis. There is another standard hyperbola in which the transverse axis is along y-axis.
221
4.5.4 The other form of the hyperbola: If the transverse axis is along y-axis and the conjugate axis is along x-axis, x2 y2 then the equation of the hyperbola is of the form 2 − 2 = 1 a b For this type of hyperbola, we have the following as explained in the earlier hyperbola Centre : C (0, 0) Vertices
:
A (0, a), A′ (0, − a)
Foci
:
F1 (0, ae), F2 (0, − ae)
Equation of transverse axis is :
x=0
Equation of conjugate axis is
:
y=0
End points of conjugate axis
:
(b, 0), (− b, 0)
Equations of latus rectum
:
Equations of directrices
:
y = ± ae a y=±e
End points of latus rectum
:
b2 b2 ± a , ae , ± a , − ae y F1 (0, ae) A(0,a)
y = a/e
Z C B
B′ y = -a/e
Z′ A′(0, -a) F2 (0, - ae)
Fig. 4. 76
222
x
Example 4. 36 : Find the equation of hyperbola whose directrix is 2x + y = 1, focus (1, 2) and eccentricity 3 . Solution: Let P(x, y) be any point on the hyperbola. Draw PM perpendicular to the directrix. FP By definition, PM = e ⇒ ∴ FP2 = e2 . PM2 2x + y − 1 4+1
i.e., (x − 1)2 + (y − 2)2= 3
2
P(x,y)
M 2x +y - 1=0
3 (x − 1)2 + (y − 2)2 = 5 (2x + y − 1)2 i.e., 7x2 + 12xy − 2y2 − 2x + 14y − 22 = 0 This is the required equation of the hyperbola.
F(1,2)
Fig. 4. 77 Example 4.37 : Find the equation of the hyperbola whose transverse axis is along x-axis. The centre is (0, 0) length of semi-transverse axis is 6 and eccentricity is 3. Solution: Since the transverse axis is along x-axis and the centre is (0, 0), the x2 y2 equation of the hyperbola is of the form 2 − 2 = 1 a b y Given that semi-transverse axis a = 6, eccentricity e = 3 We know that b2 = a2 (e2 − 1) ∴ b2 = 36(8) = 288 ∴ The equation of the hyperbola is x2 y2 − 36 288 = 1
A′ C F2
A 6
F1
x
Fig. 4. 78
Example 4.38 : Find the equation of the hyperbola whose transverse axis is parallel to x-axis, centre is (1, 2), length of the conjugate axis is 4 and eccentricity e = 2.
223
Solution: Since the transverse axis is parallel to x-axis, the equation is of the form (x − h)2 (y − k)2 − =1 a2 b2 Here centre C(h, k) is (1, 2). The length of conjugate axis 2b = 4 and e = 2 b2 = a2 (e2 − 1) 4 = a2 (4 − 1) 4 ⇒ a2 = 3 (x − 1)2 (y − 2)2 ∴ The required equation is 4/3 − =1 4 Example 4.39 : Find the equation of the hyperbola whose centre is (1, 2). The 20 distance between the directrices is 3 , the distance between the foci is 30 and the transverse axis is parallel to y-axis. Solution: Since the transverse axis is parallel to y-axis, the equation is of the form (y − k)2 (x − h)2 − =1 a2 b2 Here centre C(h, k) is (1, 2) 2a 20 a 10 The distance between the directrices e = 3 ⇒ e = 3 The distance between the foci, 2ae = 30 ⇒ ae = 15 10 a 2 e (ae) = 3 × 15 ⇒ a = 50 9 ae Also a/e ⇒ e2 = 2 9 b2 = a2 (e2 − 1) ⇒ b2 = 50 2 − 1 = 175 (y − 2)2 (x − 1)2 − 50 175 = 1 Example 4.40 : Find the equation of the hyperbola whose transverse axis is parallel to y-axis, centre (0, 0), length of semi-conjugate axis is 4 and eccentricity is 2. The required equation is
224
Solution: From the given data the hyperbola is of the form
y2 x2 = 1 − a2 b2
Given that semi-conjugate axis b = 4 and e = 2, b2 = a2 (e2 − 1) 42 = a2 (22 − 1) 16 ∴ a2 = 3 y2 x2 Hence the equation of the hyperbola is 16/3 − 16 = 1 or 3y2 − x2 = 16 Example 4.41 : Find the equation of the hyperbola whose foci are (± 6, 0) and length of the transverse axis is 8. Solution: y From the given data the transverse axis is along x-axis. ∴ The equation is of the form C (0,0) (x − h)2 (y − k)2 x − =1 2 2 F2(-6,0) F1(6,0) a b The centre is the midpoint of Fig. 4. 79 F1 and F2 − 6 + 6 0 + 0 , 2 i.e., C is 2 = (0, 0) The length of the transverse axis 2a = 8, ⇒ a = 4 F1F2 = 2ae = 12 ae = 6 ∴ 4e = 6 6 3 e=4 = 2 16 × 5 9 b2 = a2 (e2 − 1) = 16 4 − 1 = 4 = 20
2
2
x y ∴ The required equation is 16 − 20 = 1
225
Example 4.42 : Find the equation of the hyperbola whose foci are (5, ± 4) and 3 eccentricity is 2 . Solution: From the given data the transverse axis is parallel to y-axis and hence the equation of the hyperbola is of the form (x − h)2 (y − k)2 − =1 a2 b2 The centre C (h, k) is the midpoint of F1 and F2
y F1(5,4) x
C (0,0) F2(5,-4)
Fig. 4. 80
5+5 4−4 i.e., C is 2 , 2 = (5, 0) F1F2 = 2ae = (5 − 5)2 + (4 + 4)2 = 8 ae = 4 3 8 But e = 2 ∴ a = 3 64 9 b2 = a2 (e2 − 1) = 9 4 − 1 80 = 9 ∴ The required equation is (y − 0)2 (x − 5)2 9y2 9(x − 5)2 − = 1 or − =1 64/9 80/9 64 80 Example 4.43 : Find the equation of the hyperbola whose centre is (2, 1), one of the foci is (8, 1) and the corresponding directrix is x = 4. Solution: y From the given data the equation x =4 is of the form C(2,1)
(x − h)2 (y − k)2 − =1 a2 b2 Centre C (h, k) is (2, 1) CF1 = ae = 6
F1(8,1)
Fig. 4. 81
(Draw CZ perpendicular to x = 4)
226
y =1 x
The distance between the centre and directrix a CZ = e = 2 a ∴ ae . e = 6 × 2 ⇒ a2 = 12 6 ae 2 a/e = 2 ⇒ e = 3 b2 = a2 (e2 − 1) ∴ b2 = 12(3 − 1) = 24 ∴ The required equation is (y − 1)2 (x − 2)2 − 12 24 = 1 Example 4.44 : Find the equation of the hyperbola whose foci are (0, ± 5) and the length of the transverse axis is 6. Solution: y From the given data the transverse axis is along y-axis and hence the equation is of F1(0,5) the form (y − k)2 (x − h)2 =1 − a2 b2 The centre C (h, k) is the midpoint of F1 and F2
(0,0) C F2(0,-5)
Fig. 4. 82 0+0 5−5 i.e. C is 2 , 2 = (0, 0) F1F2 = 2ae = 10 The length of the transverse axis = 2a = 6 5 ⇒ a = 3 and e = 3 b2 = a2 (e2 − 1) 25 = 9 9 − 1 = 16 y2 x2 ∴ The required equation is 9 − 16=1
227
x
Example 4.45 : Find the equation of the hyperbola whose foci are (0, ± 10) and passing through (2, 3). Solution: From the data, the transverse axis is along the y-axis. ∴ it is of the form y2 x2 − =1 a 2 b2 Given that the foci are (0, ± ae) = (0, ± 10) ⇒ ae = 2
F1(0, √10) (0,0) C
x
F2(0, -√10)
10 2
y
2
2 2
Also b = a (e − 1) = a e − a
2
b2 = 10 − a2 ∴Equation of the hyperbola is
Fig. 4. 83
x2 y2 − = 1 10 − a2 a2
It passes through (2, 3), 9 4 =1 2 − a 10 − a2 9(10 − a2) − 4a2 =1 a2 (10 − a2) 90 − 9a2 − 4a2 = 10a2 − a4 or a4 − 23a2 + 90 = 0 (a2 − 18) (a2 − 5) = 0 a2 = 18 or 5 If a2 = 18, b2 = 10 − 18 = − 8 which is impossible. If a2 = 5, b2 = 10 − 5 = 5 x2 y2 ∴ Equation of the hyperbola is 5 − 5 = 1 or y2 − x2 = 5 Example 4.46 : Find the equations and length of transverse and conjugate axes x2 y2 of the hyperbola 9 − 4 = 1
228
Solution: The centre is at the origin, the transverse axis is along x-axis and the conjugate axis is along the y-axis. i.e., transverse axis is x-axis i.e., y = 0 and the conjugate axis y-axis i.e., x = 0. Hence a2 = 9, b2 = 4 ⇒ a = 3, b = 2 ∴ Length of transverse axis = 2a = 6 Length of conjugate axis = 2b = 4 Example 4.47 : Find the equations and length of transverse and conjugate axes of the hyperbola 16y2 − 9x2 = 144 y2 x2 Solution: 9 − 16 = 1 The centre is at the origin, the transverse axis is along y-axis, and the conjugate axis is along x-axis. ∴ The transverse axis is y-axis, i.e. x = 0 The conjugate axis is x-axis i.e. y = 0. Here a2 = 9, b2 = 16 ⇒ a = 3, b = 4 ∴ The length of transverse axis = 2a = 6 The length of conjugate axis = 2b = 8 Example 4.48 : Find the equations and length of transverse and conjugate axes of the hyperbola 9x2 − 36x − 4y2 − 16y + 56 = 0 Solution: 9(x2 − 4x) − 4(y2 + 4y) = − 56 9{(x − 2)2 − 4} − 4 {(y + 2)2 − 4} = − 56 9(x − 2)2 − 4(y + 2)2 = 36 − 16 − 56 9(x − 2)2 − 4(y + 2)2 = − 36 4(y + 2)2 − 9(x − 2)2 = 36 (x − 2)2 (y + 2)2 =1 − 9 4 X = x − 2 Y2 X2 − 9 4 = 1 where Y = y + 2 Clearly the transverse axis is along y-axis and the conjugate axis is along xaxis. i.e. transverse axis is y-axis or X = 0 i.e., x − 2 = 0 The conjugate axis is X-axis or Y = 0 i.e., y + 2 = 0
Here a2 = 9,
b2 = 4
⇒ a = 3, 229
b =2
∴ The length of transverse axis = 2a = 6 The length of conjugate axis = 2b = 4 Example 4.49 : Find the equations of directrices, latus rectum and length of x2 y2 latus rectum of the hyperbola 9 − 4 = 1 Solution: The centre is at the origin and the transverse axis is along x-axis. a The equations of the directrices are x = ± e The equations of the latus rectum are x = ± ae 2b2 Length of the latus rectum = a Here a2 = 9,
b2 = 4 b2 1+ 2 = a
e=
4 1+9 =
13 3
∴ The equations of the directrices are x=±
9 3 i.e. x = ± 13/3 13
The equation of the latus rectum are x = ± 13 2b2 8 Length of the latus rectum is a = 3 Example 4. 50 : Find the equations of directrices, latus rectum and length of latus rectum of the hyperbola 16y2 − 9x2 = 144 x2 y2 Solution: 9 − 16 = 1 5 e = 3 The transverse axis is along the y-axis. Here a2 = 9,
b2 = 16
a ∴ The equations of the directrices are y = ± e
9 i.e., y = ± 5
The equation of the latus rectum are y = ± ae i.e., y = ± 5 2b2 32 Length of the latus rectum is a = 3
230
Example 4.51 : Find the equations of directrices, latus rectum and length of latus rectum of the hyperbola 9x2 − 36x − 4y2 − 16y + 56 = 0 Y = y + 2 Y2 X2 Solution: By simplifying we get 9 − 4 = 1 where X = x − 2
Here a2 = 9,
b2 = 4 b2 13 1+ 2 = 3 a a 9 13 , e = 13
e= ae = The transverse axis is along Y-axis.
a ∴ The equations of the directrices are Y = ± e
i.e. Y = ±
9 13
9 9 9 ⇒ y+2 = ⇒ y= −2 13 13 13 9 −9 −9 ⇒ y+2 = ⇒ y= −2 (ii) Y = − 13 13 13 The equations of the latus rectum are Y = ± ae i.e. Y = ± 13 ⇒ y+2 = 13 ⇒ y = 13 − 2 (i) Y = 13 (ii) Y = − 13 ⇒ y + 2 = − 13 ⇒ y = − 13 − 2 (i) Y =
8 2b2 Length of the latus rectum is a = 3 Example 4.52 : The foci of a hyperbola coincide with the foci of the ellipse x2 y2 25 + 9 = 1. Determine the equation of the hyperbola if its eccentricity is 2. Solution : x2 y2 The equation of the ellipse is 25 + 9 = 1 ⇒ a2 = 25, b2 = 9,
e =
b2 1− 2 = a
9 4 1 − 25 = 5
∴ ae = 4 The foci of the ellipse are (± ae, 0) = (± 4, 0) Given that the foci of the hyperbola coincide with the foci of the ellipse, foci of the hyperbola are (± ae, 0) = (± 4, 0)
231
∴ ae = 4 Given that the eccentricity of the hyperbola is 2 a(2) = 4 ⇒ a = 2 For a hyperbola
b2 = a2 (e2 − 1) = a2 e2 − a2 = 16 − 4 = 12
x2 y2 ∴ The equation of the hyperbola is 4 − 12 = 1 Property (without proof) : A point moves such that the difference of its distances from two fixed points in a plane is a constant. The locus of this point is a hyperbola and this difference is equal to the length of the transverse axis. Example 4.53 : Find the equation of the locus of all points such that the differences of their distances from (4, 0) and (− 4, 0) is always equal to 2. Solution : By the property, the locus is a hyperbola. Take the fixed points as foci. ∴ F1 is (4, 0) and F2 is (− 4, 0) Let P(x, y) be a point on the hyperbola. F1P F2P = length of transverse axis = 2a = 2 ∴a=1 Centre is the midpoint of F1F2 = (0, 0)
y
Hence from
the given data the x2 y2 hyperbola is of the form 2 − 2 = 1 a b F1F2 = 2ae = 8 ae = 4 ⇒ e = 4
P(x,y)
F2
b2 = a2 (e2 − 1) = 1(16 − 1) = 15
C (0,0)
F1
x
Fig. 4. 84
y2 x2 ∴ The equation is 1 − 15 = 1 Alternate method: Let P(x, y) be a point on the hyperbola and let F1and F2 be the fixed points (4, 0) and (− 4, 0).
232
It is given that F1P (x − 4)2 + (y − 0)2
F2P = 2 (x + 4)2 + (y − 0)2 = 2
x2 y2 Simplifying, we get 1 − 15 = 1 Example 4.54 : Find the eccentricity, centre, foci and vertices of the hyperbola y2 x2 − 4 5 = 1 and also trace the curve Solution : y a2 = 4, b2 = 5 b2 3 1+ 2 = 2 F2 A′ C (0,0) A F1 x a (2,0) (3,0) (-3,0) (-2,0) 3 ∴ ae = 2 × 2 = 3. The transverse axis is along the Fig. 4. 85 x-axis Centre : (0, 0) Foci : (± ae, 0) = (± 3, 0) vertices : (± a, 0) = (± 2, 0) Example 4.55 : Find the eccentricity, centre, foci and vertices of the hyperbola x2 y2 − 6 18 = 1 and also trace the curve. y Solution : ⇒ e=
F1 (0,2√6)
a2 = 6 b2 = 18 ⇒
e=
b2 1+ 2= a
24 6 =2
∴ ae = 2 6 The transverse axis is along the y-axis
A(0, √6) C (0,0) A′(0, -√6) F2 (0,-2√6)
Fig. 4. 86 Centre : (0, 0) Foci are : (0, ± ae) = (0, ± 2 6) vertices are : (0, ± a) = (0, ± 6)
233
x
Example 4.56 : Find the eccentricity, centre, foci and vertices of the hyperbola 9x2 − 16y2 − 18x − 64y − 199 = 0 and also trace the curve. 9(x2 − 2x) − 16 (y2 + 4y) = 199
Solution:
9{(x − 1)2 − 1} − 16 {(y + 2)2 − 4} = 199 9(x − 1)2 − 16(y + 2)2 = 199 + 9 − 64 9(x − 1)2 − 16(y + 2)2 = 144 (x − 1)2 (y + 2)2 =1 − 16 9 i.e., a2 = 16,
b2 = 9
X = x − 1 Y2 X2 − 16 9 = 1 where Y = y + 2
⇒ e=
b2 5 1+ 2= 4 a 5 ae = 4 × 4 = 5
The transverse axis is parallel to X-axis. Referred to X, Y Centre
Referred to x, y X = x − 1, Y = y + 2 X=0 ; Y=0 x−1=0 ; y+2=0 x=1 ; y=−2 ∴ C (1, − 2) (i) X = 5 ; Y=0 x−1=5 ; y+2=0 x=6 ; y=−2 ∴ F1 (6, − 2)
(0, 0)
(± ae, 0) is (± 5, 0) (i) (5, 0) Foci
(ii) X = − 5 ; Y=0 x − 1 = − 5; y + 2 = 0 ∴ F2 (− 4, − 2)
(ii) (− 5, 0)
234
Vertices
(± a, 0) i.e. (± 4, 0) (i) (4, 0)
(i) X = 4 x−1=4 ∴ A (5, − 2)
(ii) (− 4, 0)
(ii) X = − 4 ;Y=0 x−1=−4 ; y+2=0 ∴ A′ (− 3, − 2)
y
;Y=0 ; y+2=0
Y
x F2 (-4, -2)
A′ (-3,-2)
C (1,-2) A (5,-2)
F1 (6,-2)
X
Fig. 4. 87 Example 4.57 : Find the eccentricity, centre, foci and vertices of the following hyperbola and draw the diagram : 9x2 − 16y2 + 36x + 32y + 164 = 0 Solution: 9(x2 + 4x) − 16(y2 − 2y) = − 164 9{(x + 2)2 − 4} − 16 {(y − 1)2 − 1} = − 164 9(x + 2)2 − 16(y − 1)2 = − 164 + 36 − 16 16(y − 1)2 − 9(x + 2)2 = 144 (x + 2)2 (y − 1)2 − =1 9 16 X = x + 2 X2 Y2 − 9 16 = 1 where Y = y − 1
a2 = 9,
b2 = 16
⇒ e=
b2 5 1+ 2= 3 a ae = 5
235
The transverse axis is parallel to Y-axis. Referred to X, Y Centre
Referred to x, y X = x + 2, Y = y − 1 X=0 ; Y=0 x+2=0 ; y−1=0 x=−2 ; y=1 ∴ C (− 2, 1) (i) X = 0 ; Y=5 x+2=0 ; y−1=5 x=−2 ; y=6 ∴ F1 (− 2, 6)
(0, 0)
(0, ± ae) i.e., (0, ± 5) (i) (0, 5)
(ii) (0, − 5)
(ii) X = 0 ; Y=− 5 x+2=0 ; y−1=−5 x=−2 ; y=−4 ∴ F2 (− 2, − 4) (i) X = 0 ; Y=3 x+2=0 ; y−1=3 ∴ A (− 2, 4) (ii) X = 0 ; Y=−3 x+2=0 ; y−1=−3 x=−2 ; y=−2
Foci (0, ± a) (i) (0, 3) Vertices (ii) (0, − 3)
∴ A′ (− 2, − 2)
Y
x =-2 y
F1 (-2,6) A (-2,4) y =1 C (-2,1) A′ (-2,-2) F2 (-2,-4)
Fig. 4. 88 236
X x
Example 4.58 : Points A and B are 10 km apart and it is determined from the sound of an explosion heard at those points at different times that the location of the explosion is 6 km closer to A than B. Show that the location of the explosion is restricted to a particular curve and find an equation of it. y
P(x,y)
A(-5, 0)
B(5, 0)
x
Fig. 4. 89 Given : i.e.,
PB − PA = 6 (x − 5)2 + y2 −
Simplifying we get
(x + 5)2 + y2 = 6 − 9y2 + 16x2 = 144
− y2 x2 x2 y2 + = 1 i.e., − 16 9 9 16 = 1 which is a hyperbola. EXERCISE 4.3 (1) Find the equation of the hyperbola if (i) (ii) (iii)
(iv) (v)
focus : (2, 3) ; corresponding directrix : x + 2y = 5, e = 2 7 centre : (0, 0) ; length of the semi-transverse axis is 5 ; e = 5 and the conjugate axis is along x-axis. centre : (0, 0) ; length of semi-transverse axis is 6 ; e = 3, and the transverse axis is parallel to y-axis. 5 centre : (1, − 2) ; length of the transverse axis is 8 ; e = 4 and the transverse axis is parallel to x-axis. centre : (2, 5) ; the distance between the directrices is 15, the distance between the foci is 20 and the transverse axis is parallel to y-axis.
237
(2)
(3)
(4) (5)
(vi) foci : (0, ± 8) ; length of transverse axis is 12 (vii) foci : (± 3, 5) ; e = 3 (viii) centre : (1, 4) ; one of the foci (6, 4) and the corresponding 9 directrix is x = 4 . (ix) foci : (6, − 1) and (− 4, − 1) and passing through the point (4, − 1) Find the equations and length of transverse and conjugate axes of the following hyperbolas : (i) 144x2 − 25y2 = 3600 (ii) 8y2 − 2x2 = 16 (iii) 16x2 − 9y2 +96x + 36y − 36 = 0 Find the equations of directrices, latus rectums and length of latus rectum of the following hyperbolas : (ii) 9x2 − 4y2 − 36x + 32y + 8 = 0 (i) 4x2 − 9y2 = 576 Show that the locus of a point which moves so that the difference of its distances from the points (5, 0) and (− 5, 0) is 8 is 9x2 − 16y2 = 144. Find the eccentricity, centre, foci and vertices of the following hyperbolas and draw their diagrams. y2 x2 (i) 25x2 − 16y2 = 400 (ii) 9 − 25 = 1 (iii) x2 − 4y2 + 6x + 16y − 11 = 0 (iv) x2 − 3y2 + 6x + 6y + 18 = 0
4.6 Parametric form of Conics: Conic Parabola Ellipse Hyperbola
Parametric equations x = at2 y = 2at x = a cos θ y = b sin θ x = a sec θ y = b tan θ
Parameter t
Range of parameter −∞
θ
0 ≤ θ ≤ 2π
θ
0≤ θ ≤ 2π
Any point on the conic ‘t’ or (at2, 2at) ‘θ’ or (a cosθ, b sinθ) ‘θ’ or (a sec θ, b tan θ)
Note: For ellipse, we have another parametric form of equations x = y=
a(1 − t2) , 1 + t2
θ b.2t 2 , − ∞ < t < ∞. This result will be obtained by putting tan 2 = t in the 1+t
parametric equations x = a cosθ and y = b sin θ.
238
Thus we have two forms of representations of conics i.e., cartesian form and parametric form. Now we will derive the equations of chord, tangent and normal to the conics.
4.7 Chords, tangents and normals We derive these equations using both forms of conics.
4.7.1 Cartesian form (i) Parabola Equation of the chord joining A(x1, y1) and B(x2, y2) on the parabola y2 = 4ax y A(x1,y1) Since (x1, y1) and (x2, y2) lie on the parabola, y12 = 4ax1, y22 = 4ax2 x
y12 − y22 = 4a (x1 − x2) ⇒
y1 − y2 4a = x1 − x2 y1 + y2
B(x2,y2) y2 =4ax
Fig. 4. 90 4a i.e., the slope (m) of the chord AB = y + y 1 2 The equation of the chord, using slope (m) and point (x1, y1) is 4a (y − y1) = y + y (x − x1) 1 2
y
⇒
Ta
4a (y − y1) = y + y (x − x1) 1 1 yy1 = 2a(x + x1) (use
y12
ng en t
If the point (x2, y2) coincides with (x1, y1) then the chord becomes the tangent at (x1, y1). Therefore, to obtain tangent at (x1, y1), put x2 = x1 and y2 = y1 in the equation of the chord. ∴ the equation of the tangent is
(x1,y1) x
= 4ax1)
Thus the equation of the tangent at (x1, y1) to the parabola y2 = 4ax is yy1 = 2a(x + x1)
y2 =4ax
Fig. 4. 91
239
No rm al
2ax − y1y + 2ax1 = 0
y
Ta
Equation of the tangent is
ng en t
To find the equation of the normal using perpendicularity.
(x1,y1)
∴ the normal is of the form
x
y1x + 2ay = k
y2 =4ax
But it passes through (x1, y1) ∴ k = x1y1 + 2ay1
Fig. 4. 92
Thus the equation of the normal at (x1, y1) to the parabola is y1x + 2ay = x1y1 + 2ay1 (ii) Ellipse Equation of the chord joining A(x1, y1) and B(x2, y2) on the ellipse y2 x + =1 a2 b2 2
Since (x1, y1) and (x2, y2) lie on the ellipse, 2
2
x22 2
y1 + 2 = 1, a b a
x1
2
A(x1,y1)
2
y2 + 2 = 1 b
x
B(x2,y2)
By simplification, the slope m=
y
y1 − y2 − b2 (x1 + x2) = 2 x1 − x2 a (y1 + y2)
Fig. 4. 93
∴ the equation of the chord is (y − y1) =
− b2 (x1 + x2) a2(y1 + y2)
(x − x1)
To get the equation of the tangent at (x1, y1) put x2 = x1 and y2 = y1 in the equation of the chord.
240
∴The equation of the tangent at (x1, y1) is
y (x1,y1)
Tan g
2
(y − y1) = ⇒
− b (x1 + x1) 2
a (y1 + y1)
(x − x1)
en t
x
yy1 + 2 =1 a b
xx1 2
No rm a
l
Fig. 4. 94 To get the equation of the normal, use the perpendicularity property to a straight line. ∴The equation of the tangent is y x1b2x + y1a2y − a2b2 = 0 ∴The equation of the normal is of the form y1a2x − x1b2y = k
(x1,y1) x
But it passes through (x1, y1) ∴ k = (a2 − b2) x1 y1 Fig. 4. 95 ∴ The required equation is a2x b2y y1a2x − x1b2y = (a2 − b2) x1y1 or x − y = a2 − b2 1 1 (iii) Hyperbola Following the same procedure as in the case of ellipse we get the equation of the chord as y − y1 =
b2(x1 + x2) a2(y1 + y2)
The equation of the tangent at (x1, y1) as
(x − x1) xx1 a
2
−
yy1 b2
= 1
a2 x b 2 y and the normal at (x1, y1) as x + y = a2 + b2 1 1 Note : To get the results for the hyperbola replace b2 as − b2 in the results of ellipse. 241
4.7.2 Parametric form : To get the parametric forms of equations of chord, tangent and normal to conics, replace (x1, y1), by the corresponding ‘any point’ in the parametric form. (i) Parabola : The equation of the chord joining (x1, y1) and (x2, y2) on the parabola is 4a y − y1 = y + y (x − x1) 1 2 ∴ The equation of the chord joining (at12, 2at1) and (at22, 2at2) or ‘t1’ and ‘t2’ on the parabola is 4a y − 2at1 = 2at + 2at (x − at12) 1 2 i.e.
y(t1 + t2) = 2x + 2a t1t2
To find the equation of the tangent at ‘t’ put t1 = t2 = t in the equation of the chord. We get y(2t) = 2x + 2at2 i.e. yt = x + at2 Another method: The tangent at (x1, y1) to y2 = 4ax is yy1 = 2a(x + x1) ∴ The tangent at (at2, 2at) is y(2at) = 2a (x + at2) i.e.,
yt = x + at2
Applying the perpendicularity, we get the equation of the normal at ‘t’ as y + tx = 2at + at3 Similarly we can derive the equation of chord, tangent and normal for ellipse and hyperbola.
242
Thus we get the following : Cartesian form : Equation of chord joining (x1, y1) and (x2, y2)
Parabola 4a y − y1 = y + y (x − x1) 1 2
y − y1 = −
Equation of tangent at (x1, y1)
yy1 = 2a(x + x1)
xx1 / a2 + yy1/b2 = 1
Equation of normal at (x1, y1)
xy1 + 2ay = x1y1 + 2ay1
Parametric form : Equation of chord
Parabola Chord joining the points ‘t1’ and ‘t2’ is y(t1 + t2) = 2x + 2at1t2
Equation of tangent
at ‘t’ is yt = x + at2
Equation of normal
at ‘t’ is tx + y = 2at + at3
Ellipse
Hyperbola
2
b (x1+x2) 2
a (y1+y2)
2
(x − x1)
b (x1+x2) (x − x1) y − y1 = 2 a (y1+y2) xx1 / a2 − yy1/b2 = 1
a2x b2y 2 2 x1 − y1 = a − b Ellipse Chord joning the points ‘θ’1 and ‘θ’2 is
a2x b2y 2 2 x1 + y1 = a + b Hyperbola Chord joning the points ‘θ’1 and ‘θ’2 is
(θ1 − θ2) x (θ1 + θ2) y (θ1 + θ2) cos + sin = cos a b 2 2 2 x y at ‘θ’ is a cos θ + b sin θ = 1 by ax − = a2 − b2 cosθ sinθ
(θ1 + θ2) x (θ1 − θ2) y (θ1 + θ2) − cos sin = cos a b 2 2 2 x y at ‘θ’ is a sec θ − b tan θ = 1 by ax + = a2 + b2 secθ tanθ
243
Note : The equation of tangent at (x1, y1) is obtained from the equation of the 1 1 curve by replacing x2 by xx1, y2 by yy1, xy by 2 (xy1 + x1y), x by 2 (x + x1) and 1 y by 2 (y + y1) To find the condition that y = mx + c may be a tangent to the conics (1) Parabola : Let y = mx + c be a tangent to the parabola y2 = 4ax at (x1, y1). We know that at (x1, y1), the equation of the tangent is yy1 = 2a(x + x1) ∴ The above two equations represent the same tangent and hence their corresponding coefficients are proportional ∴ 2ax − y1y + 2ax1 = 0 mx − y + c = 0 ⇒
2ax1 2a − y1 = = m c −1
⇒
c 2a x1 = m , y1 = m
Since (x1, y1) lies on the parabola, y12 = 4ax1 ,
c 4a2 = 4a . m m2
a i.e. , c = m Thus we have three results to the parabola y2 = 4ax. a (1) The condition for the tangency is c = m c 2a a 2a (2) The point of contact is m , m i.e., 2 , m . m a (3) The equation of any tangent is of the form y = mx + m Note : Instead of taking the equation of the tangent in the cartesian form, we can prove the same result by taking the tangent in the parametric form. Similarly, we can derive the results for other conics also.
244
Results connected with ellipse : x2 y2 (i) The condition that y = mx + c may be a tangent to the ellipse 2 + 2 = 1 a b is c2 = a2m2 + b2 − a2m b2 (ii) The point of contact is c , c where c2 = a2m2 + b2 a2m2 + b2
(iii) The equation of any tangent is of the form y = mx ± Note : In y = mx ±
a2m2 + b2 , either y = mx +
a2m2 + b2 holds
or y = mx − a2m2 + b2 holds Results connected with hyperbola : (i) The condition that y = mx + c may be a tangent to the hyperbola is c2 = a2m2 − b2 − a2m − b2 (ii) The point of contact is c , c where c2 = a2m2 − b2 a2m2 − b2
(iii) The equation of any tangent is of the form y = mx ± Note : In y = mx ± or y = mx −
a2m2 − b2 , either y = mx +
a2m2 − b2
a2m2 − b2 is correct but not both.
4.7.3 Equation of chord of contact of tangents from a point (x1, y1)
Q
(x
2
,y
2)
x2 y2 x2 y2 to the (i) Parabola y2 = 4ax (ii) ellipse 2 + 2 = 1 (iii) hyperbola 2 − 2 = 1 a b a b Solution : y The equation of tangent at Q(x2, y2) is yy2 = 2a(x + x2) It passes through the point P(x1, y1) y1y2 = 2a (x1 + x2) … (1) (x1,y1) P V The equation of tangent at x R(x3, y3) is yy3 = 2a(x + x3) It passes through the point P(x1, y1) ∴ y1y3 = 2a(x1 + x3) … (2) The result (1) and (2) show that Q(x2, y2) and R(x3, y3) lie on the straight line yy1 = 2a(x + x1). Fig. 4. 96 ) ,y 3 x3 R(
245
∴ Equation of QR, the chord of contact of tangents is yy1 = 2a(x + x1) Similarly we can find the required equations of the chord of contact for yy1 xx1 yy1 xx1 ellipse as 2 + 2 = 1 and for the hyperbola as 2 − 2 = 1 a b a b Example 4.59 : Find the equations of the tangents to the parabola y2 = 5x from the point (5, 13). Also find the points of contact. Solution: 5 Here 4a = 5 ⇒ a = 4 The equation of the parabola is y2 = 5x a 5 Let the equation of the tangent be y = mx + m i.e., y = mx + 4m … (1) Since it passes through (5, 13) we have 5 13 = 5m + 4m ∴ 20m2 − 52m + 5 = 0 (10m − 1) (2m − 5) = 0 5 1 ∴ m = 10 or m = 2 Using the values of m, we get the equations of tangents are 2y = 5x + 1, 10y = x + 125. 5 5 1 a 2a The points of contact are given by 2 , m , where a = 4 m = 2 , 10 m 1 ∴ the points of contact are 5 , 1 , (125, 25) Example 4.60 : Find the equation of the tangent at t = 1 to the parabola y2 = 12x Solution: Equation of the parabola is y2 = 12x. Here 4a = 12, a = 3 ‘t’ represents the point (at2, 2at). ∴ t = 1 represents the point = (3, 6) (x + x1) Equation of tangent at (x1, y1) to the parabola y2 = 12x is yy1 = 12 2 12 (x + 3) ∴ Equation of tangent at (3, 6) is y(6) = i.e., x − y + 3 = 0 2 Alternative form : The equation of the tangent at ‘t’ is yt = x + at2 Here 4a = 12 ⇒ a = 3 Also t=1
246
∴ The equation of the tangent is y = x + 3 x−y+3 = 0 Example 4.61 : Find the equation of the tangent and normal to the parabola x2 + x − 2y + 2 = 0 at (1, 2) Solution: The equation of the parabola is x2 + x − 2y + 2 = 0 Equation of the tangent at (x1, y1) to the given parabola is x + x1 (y + y1) x+1 (y + 2) − 2 + 2 = 0 i.e., x(1) + 2 − 2 2 + 2 = 0 2 2 On simplification we get 3x − 2y + 1 = 0 Equation of the normal is of the form 2x + 3y + k = 0 This normal passes through (1, 2) ∴ 2+6+k = 0 ∴ k=−8 ∴ Equation of the normal is 2x + 3y − 8 = 0 Example 4.62 : Find the equations of the two tangents that can be drawn from the point (5, 2) to the ellipse 2x2 + 7y2 = 14 Solution: Equation of the ellipse is xx1 +
2x2 + 7y2 = 14 i.e.,
x2 y2 7 + 2 =1
Here a2 = 7, b2 = 2 Let the equation of the tangent be y = mx +
a2m2 + b2
∴ y = mx + 7m2 + 2 Since this line passes through the point (5, 2) we get 2 = 5m + i.e.
2 − 5m =
7m2 + 2
7m2 + 2
∴ (2 − 5m)2 = 7m2 + 2 4 + 25m2 − 20m = 7m2 + 2 18m2 − 20m + 2 = 0 9m2 − 10m + 1 = 0
247
∴ (9m − 1) (m − 1) = 0 ∴ m=1
1 or m = 9
To find the equations of the tangents, use slope-point form (i) m = 1, The equation is y − 2 = 1(x − 5) i.e., x − y − 3 = 0 (ii) m = 1/9 1 The equation is y − 2 = 9 (x − 5), i.e., x − 9y + 13 = 0. Thus the equations of the tangents are x − y − 3 = 0, x − 9y + 13 = 0 Example 4.63 : Find the equation of chord of contact of tangents from the point (2, 4) to the ellipse 2x2 + 5y2 = 20 Solution: The equation of chord of contact of tangents from (x1, y1) to 2x2 + 5y2 − 20 = 0 is 2xx1 + 5yy1 − 20 = 0 ∴ the required equation from (2, 4) is 2x(2) + 5y(4) − 20 = 0 i.e. x + 5y − 5 = 0 EXERCISE 4.4 (1) Find the equations of the tangent and normal (i) to the parabola y2 = 12x at (3, − 6) (ii) to the parabola x2 = 9y at (− 3, 1) (iii) to the parabola x2 + 2x − 4y + 4 = 0 at (0, 1) (iv) to the ellipse 2x2 + 3y2 = 6 at ( 3 , 0) (v) to the hyperbola 9x2 − 5y2 = 31 at (2, − 1) (2) Find the equations of the tangent and normal 1 (i) to the parabola y2 = 8x at t = 2 π (ii) to the ellipse x2 + 4y2 = 32 at θ = 4 1 (iii) to the ellipse 16x2 + 25y2 = 400 at t = 3 x2 y2 π (iv) to the hyperbola 9 − 12 = 1 at θ = 6
248
(3) Find the equations of the tangents (i) to the parabola y2 = 6x, parallel to 3x − 2y + 5 = 0 (ii) to the parabola y2 = 16x, perpendicular to the line 3x − y + 8 = 0 x2 y2 (iii) to the ellipse 20 + 5 = 1, which are perpendicular to x + y + 2 = 0 (iv) to the hyperbola 4x2 − y2 = 64, which are parallel to 10x − 3y + 9 = 0 (4) Find the equation of the two tangents that can be drawn (i) from the point (2, − 3) to the parabola y2 = 4x (ii) from the point (1, 3) to the ellipse 4x2 + 9y2 = 36 (iii) from the point (1, 2) to the hyperbola 2x2 − 3y2 = 6. (5) Prove that the line 5x + 12y = 9 touches the hyperbola x2 − 9y2 = 9 and find its point of contact. (6) Show that the line x − y + 4 = 0 is a tangent to the ellipse x2 + 3y2 = 12. Find the co-ordinates of the point of contact. (7) Find the equation to the chord of contact of tangents from the point (i) (− 3, 1) to the parabola y2 = 8x (ii) (2, 4) to the ellipse 2x2 + 5y2 = 20 (iii) (5, 3) to the hyperbola 4x2 − 6y2 = 24 Results without Proof : (1) Two tangents can be drawn to (i) a parabola (ii) an ellipse and (iii) a hyperbola, from any point on the plane. (2) (a) Three normals can be drawn to a parabola (b) Four normals can be drawn to (i) an ellipse and (ii) a hyperbola from any point on the plane. (3) The equation of chord of contact of tangents from a point (x1, y1) (i) a parabola y2 = 4ax is yy1 = 2a(x + x1) xx1 yy1 x2 y2 (ii) an ellipse 2 + 2 = 1 is 2 + 2 = 1 a b a b 2 2 xx1 yy1 y x (iii) a hyperbola 2 − 2 = 1 is 2 − 2 = 1 a b a b (4) The chord of contact of tangents from any point on the directrix (i) of a parabola passes through its focus (ii) passes through the corresponding focus for ellipse and hyperbola 249
(5) The condition that lx + my + n = 0 may be a tangent to (i) the parabola y2 = 4ax is am2 = ln x2 y2 (ii) the ellipse 2 + 2 = 1 is a2l2 + b2m2 = n2 a b x2 y2 (iii) the hyperbola 2 − 2 = 1 is a2l2 − b2m2 = n2 a b (6) The condition that lx + my + n = 0 may be a normal to (i) the parabola y2 = 4ax is al3 + 2alm2 + m2n = 0 a2 b2 (a2 − b2) x2 y2 (ii) the ellipse 2 + 2 = 1 is 2 + 2 = a b l m n2
2
2
a2 b2 (a2 + b2) x2 y2 (iii) the hyperbola 2 − 2 = 1 is 2 − 2 = a b l m n2 (7) The locus of the foot of the perpendicular from a focus to a tangent to (i) the parabola y2 = 4ax is x = 0 x2 y2 (ii) the ellipse 2 + 2 = 1 is the circle x2 + y2 = a2 b a x2 y2 (iii) the hyperbola 2 − 2 = 1 is the circle x2 + y2 = a2 a b (This circle is also called auxiliary circle) (8) The locus of the point of intersection of perpendicular tangents to (i) the parabola y2 = 4ax is x = − a (the directrx) x2 (ii) the ellipse 2 + a director circle)
y2 = 1 is x2 + y2 = a2 + b2 (This circle is called b2
x2 y2 (iii) an hyperbola 2 − 2 = 1 is x2 + y2 = a2 − b2 (This circle is also a b called director circle) (9) The point of intersection of the tangents at ‘t1’ and ‘t2’ to the
parabola y2 = 4ax is [at1t2, a(t1 + t2)] (10) The normal at the point ‘t1’ on the parabola y2 = 4ax meets the parabola 2 again at the point ‘t2’, then t2 = − t1 + t 1 250
(11) If ‘t1’ and ‘t2’ are the extremities of any focal chord of the parabola y2 = 4ax, then t1t2 = − 1 Note : For the proof of above results one may refer the Solution Book.
4.8. Asymptotes Consider the graph of a function y = f(x). As a point P on the curve moves farther and farther away from the origin, it may happen that the distance between P and some fixed line tends to zero. This fixed line is called an asymptote. Note that it is possible only when the curve is open. Since hyperbola is open and y → ± ∞ as x → + ∞ and x → − ∞ hyperbola have asymptotes. Definition : An asymptote to a curve is the tangent to the curve such that the point of contact is at infinity. In particular the asymptote touches the curve at + ∞ and − ∞. y
xe Fi
A′ F2 P
P
C
P
in dL A
e(
as
pto ym
P
te)
P
P
F1
x
Fig. 4. 97 x 2 y2 The equations of the asymptotes to the hyperbola 2 − 2 = 1 a b Assume that the equation of an asymptote is of the form y = mx + c. To find the points of intersection of the hyperbola and the asymptote, solve 2 y2 x − = 1 and y = mx + c. a2 b2 ∴
(mx + c)2 x2 − =1 a2 b2
251
1 m2 2 2mc c2 2 − 2 x − 2 x − 2 + 1 = 0 b a b b The points of contact are at infinity. i.e., the roots of the equations are infinite. Since the roots are infinite, the coefficients of x2 and x must be zero. m2 1 − 2mc =0 ∴ 2 − 2 = 0 and b b2 a b i.e., m = ± a and c = 0 b Then y = ± a x ∴ there are two asymptotes to the hyperbola whose equations are −b b y = a x and y = a x x y x y i.e. a − b = 0 and a + b = 0 The combined equation of asymptotes is
x − y x + y = 0 i.e. a b a b x/a
x2 y2 2 − 2=0 a b y
+
y/b =0
F2 (-ae,0)
l
l1
A' Z'
α C
-y x/a
A Z
0 /b=
F1 (ae,0)
x
Fig. 4. 98 Results regarding asymptotes : (1) The asymptotes pass through the centre C(0, 0) of the hyperbola.
252
b b (2) The slopes of asymptotes are a and − a i.e., the transverse axis and conjugate axis bisect angles between the asymptotes. y x (3) If 2α is the angle between the asymptotes then the slope of a − b = 0 is b tan α = a . b ∴ angle between the asymptotes is 2α = 2 tan−1 a (4) We know that sec2α = 1 + tan2α a2 + b2 b2 sec2α = 1 + 2 = = e2 a a2 ⇒ sec α = e ⇒ α = sec−1e ∴ angle between the asymptotes 2α = 2 sec−1e Important Note : Eventhough the asymptotes are straight lines, if the angle between the asymptotes is obtuse, take obtuse angle as the angle between them and not the corresponding acute angle. (5) The standard equation of hyperbola and combined equation of asymptotes differs only by a constant. (6) If l1 = 0 and l2 = 0 are the separate equations of asymptotes, then the combined equation of the asymptotes is l1 l2 = 0. ∴ the equation of the corresponding hyperbola is of the form l1l2 = k, where k is a constant. To find this k, we need a point on the hyperbola. Example 4.64 : Find the separate equations of the asymptotes of the hyperbola 3x2 − 5xy − 2y2 + 17x + y + 14 = 0 Solution: The combined equation of the asymptotes differs from the hyperbola by a constant only. ∴ the combined equation of the asymptotes is 3x2 − 5xy − 2y2 + 17x + y + k = 0 Consider 3x2 − 5xy − 2y2 = 3x2 − 6xy + xy − 2y2 = 3x (x − 2y) + y(x − 2y) = (3x + y) (x − 2y) ∴ The separate equations are 3x + y + l = 0, x − 2y + m = 0
253
∴ (3x + y + l) (x − 2y + m) = 3x2 − 5xy − 2y2 + 17x + y + k Equating the coefficients of x, y terms and constant term, we get l + 3m = 17 … (1) − 2l + m = 1 …(2) lm = k Solving (1) and (2) we get l = 2, m = 5 and k = 10 Hence separate equations of asymptotes are 3x + y + 2 = 0, x − 2y + 5 = 0 The combined equation of asymptotes is 3x2 − 5xy − 2y2 + 17x + y + 10 = 0 Note : The hyperbola, discussed above is not a standard hyperbola. Example 4.65 : Find the equation of the hyperbola which passes through the point (2, 3) and has the asymptotes 4x + 3y − 7 = 0 and x − 2y = 1. Solution: The separate equations of the asymptotes are 4x + 3y − 7 = 0, x − 2y − 1 = 0 ∴ combined equation of asymptotes is (4x + 3y − 7) (x − 2y − 1) = 0 The equation of the hyperbola differs from this combined equation of asymptotes by a constant only. ∴ the equation of the hyperbola is of the form (4x + 3y − 7) (x − 2y − 1) + k = 0 But this passes through (2, 3) (8 + 9 − 7) (2 − 6 − 1) + k = 0 ∴ k = 50 ∴ The equation of the corresponding hyperbola is (4x + 3y − 7) (x − 2y − 1) + 50 = 0 i.e., 4x2 − 5xy − 6y2 − 11x + 11y + 57 = 0 Example 4.66 : Find the angle between the asymptotes of the hyperbola 3x2 − y2 − 12x − 6y − 9 = 0 3x2 − y2 − 12x − 6y − 9 = 0 Solution: 3(x2 − 4x) − (y2 + 6y) = 9 3 {(x − 2)2 − 4} − {(y + 3)2 − 9} = 9 3(x − 2)2 − (y + 3)2 = 12 (x − 2)2 (y + 3)2 − 4 12 = 1 Here a = 2, b = 12 = 2 3
254
The angle between the asymptotes is b 2 3 π 2π 2α = 2 tan−1 a = 2tan−1 2 = 2 tan−1 3 = 2 × 3 = 3 Another method : a2 = 4, b2 = 12 b2 1+ 2 = a The angle between the asymptotes is
12 1+ 4 = 2
e=
π 2π 2 α = 2 sec−12 = 2 × 3 = 3 Example 4.67 : Find the angle between the asymptotes to the hyperbola 3x2 − 5xy − 2y2 + 17x + y + 14 = 0 Solution: Combined equation of the asymptotes differs from that of the hyperbola by a constant only. ∴ Combined equation of asymptotes is 3x2 − 5xy − 2y2 + 17x + y + k = 0 3x2 − 5xy − 2y2 = 3x2 − 6xy + xy − 2y2 = 3x(x − 2y) + y(x − 2y) = (x − 2y) (3x + y) ∴ Separate equations are x − 2y + l = 0, 3x + y + m = 0 1 Let m1 and m2 be the slopes of these lines, then m1 = 2 , m2 = − 3 m1 − m2 1/2 − (− 3) ∴ angle between the lines is tanθ = 1 + m m = 1 + 1/2 (− 3) = 7 1 2 θ = tan−1 (7) Alternative method : Combined equation of asymptotes is nothing but pair of straight lines. Hence the angle between the asymptotes is
2 h2 − ab a+b
tan θ =
Comparing with ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 We have a = 3, b = − 2, 2h = − 5 25 2 4 + 6 tan θ = 3−2 2×7 = 2 =7 θ = tan−1 (7)
255
Note : Since the above hyperbola is not in the standard form, it is difficult to identify whether the angle between the asymptotes is obtuse or acute. According to the above method we will get only the acute angle as the angle between the asymptotes. b Therefore if the hyperbola in the standard form, use either 2 tan−1 a or 2 sec−1e to find the angle between the asymptotes and take the angle as it is. Example 4.68 : Prove that the product of perpendiculars from any point on the x2 y2 a2b2 hyperbola 2 − 2 = 1 to its asymptotes is constant and the value is 2 a b a + b2 Solution: x12 y12 x2 y2 Let P (x1, y1) be any point on the hyperbola 2 − 2 = 1∴ 2 − 2 = 1 … (1) a b a b The perpendicular distance from (x1, y1) to the asymptote x y a − b = 0 is
x1 y1 a − b x y and to a + b = 0 is 1 1 + a 2 b2
x1 y1 a + b 1 1 + a2 b 2
y x/a
+
y/b
-y x/a
=0
0 /b=
P(x1,y1) F2
A′
A
C
F1
x
Fig. 4. 99 ∴ Product of perpendicular distances =
256
x1 y1 a + b . 1 1 + a2 b 2
x1 y1 a − b 1 1 + a2 b2
x12
y12 − a2 b2 1 = 1 1 = 2 b + a2 2+ 2 a b a2 b2
(by (1))
a2 b2 , which is a constant. = 2 a + b2 EXERCISE 4.5 (1) Find the equation of the asymptotes to the hyperbola (ii) 8x2 + 10xy − 3y2 − 2x + 4y − 2 = 0 (i) 36x2 − 25y2 = 900 (2) Find the equation of the hyperbola if (i) the asymptotes are 2x + 3y − 8 = 0 and 3x − 2y + 1 = 0 and (5, 3) is a point on the hyperbola (ii) its asymptotes are parallel to x + 2y − 12 = 0 and x − 2y + 8 = 0, (2, 4) is the centre of the hyperbola and it passes through (2, 0). (3) Find the angle between the asymptotes of the hyperbola (ii) 9(x − 2)2 − 4(y + 3)2 = 36 (i) 24x2 − 8y2 = 27 2 2 (iii) 4x − 5y − 16x + 10y + 31 = 0
4.9 Rectangular hyperbola Definition: A hyperbola is said to be a rectangular hyperbola if its asymptotes are at right angles. b The angle between the asymptotes is given by 2tan−1 a . But angle between the asymptotes of the rectangular hyperbola is 90° . b b ∴ 2tan−1 a = 90° ∴ a = tan45° ⇒ a = b. x2 y2 2 − 2 = 1, we get the equation of a b 2 2 2 rectangular hyperbola as x − y = a . Hence the combined equation of the asymptotes is x2 − y2 = 0. The separate equations are x − y = 0 and x + y = 0. i.e., x = y and x = − y. The transverse axis is y = 0, conjugate axis is x = 0. Using a = b in the hyperbola
All the results corresponding to the rectangular hyperbola of the form x2 − y2 = a2 are obtained simply by putting a = b in the corresponding results of x2 y2 the hyperbola 2 − 2 = 1 a b
257
This type of rectangular hyperbola is not a standard one. For standard type, the asymptotes are the co-ordinate axes. The standard rectangular hyperbola xy = c2 is obtained by rotating the rectangular hyperbola x2 − y2 = a2 through an angle 45° about the origin in the anticlockwise direction. y
y x y = c2 x2 -
y 2 = a2 A
A′
C
90
C
x
° A
45°
x
A′
Fig. 4. 100 4.9.1 Standard equation of a rectangular hyperbola : For a standard rectangular hyperbola the asymptotes are co-ordinate axes. Since the axes are the asymptotes, the equations of the asymptotes are x = 0 and y = 0. The combined equation of the asymptotes is xy = 0. Therefore the equation of the standard rectangular hyperbola is of the form xy = k. To find k, we need a point on the rectangular hyperbola. y D1 D2
F1(a,a) A (a/√2, a/√2) C
45°
x
M A′ D1′
F2(-a,-a) D2′
Fig. 4. 101 258
Let the asymptotes meet at C. Let AA′ = 2a be the length of the transverse axis. Draw AM perpendicular to x-axis. Since the asymptotes bisect the angle a a between the axes, ACM = 45°. CM = a cos 45° = , AM = a sin 45° = 2 2 a a ∴ co-ordinates of A are , . This point lies on the rectangular 2 2 hyperbola xy = k. ∴ k =
a a a2 . or k = 2 and 2 2
a2 the equation of the rectangular hyperbola is xy = 2 or xy = c2
a2 where c2 = 2 .
Eccentricity of the hyperbola is given by b2 = a2(e2 − 1). Since a = b in a rectangular hyperbola, a2 = a2 (e2 − 1) Eccentricity of the rectangular hyperbola is e = 2 . a a a a Also the vertices of the rectangular hyperbola are , , − , − 2 2 2 2 and foci are (a, a), (− a, − a). The equation of transverse axis is y = x and the conjugate axis is y = − x. If the centre of the rectangular hyperbola is at (h, k) and the asymptotes are parallel to x and y-axis, the general form of standard rectangular hyperbola is (x − h) (y − k) = c2. The parametric equation of the rectangular hyperbola xy = c2 are c x = ct, y = t where ‘t’ is the parameter and ‘t’ is any non-zero real number. c Any point on the rectangular hyperbola is ct , t . This point is often referred to as the point ‘t’. Results : Equation of the tangent at (x1, y1) to the rectangular hyperbola xy = c2 is xy1 + yx1 = 2c2 Equation of the tangent at ‘t’ is x + yt2 = 2ct. Equation of normal at (x1, y1) is xx1 − yy1 = x12 − y12 .
259
c Equation of normal at ‘t’ is y − xt2 = t − ct3 Two tangents and four normals can be drawn from a point to a rectangular hyperbola. Example 4.69 : Find the equation of the standard rectangular hyperbola whose −3 −2 centre is − 2 , 2 and which passes through the point 1, 3 Solution: The equation of the standard rectangular hyperbola with centre at (h, k) is (x − h) (y − k) = c2 −3 The centre is − 2 , 2 . 3 ∴ the equation of the standard rectangular hyperbola is (x+2) y + 2 = c2 − 2 5 − 2 3 2 2 It passes through 1 , 3 ∴ (1 + 2) 3 + 2 = c ⇒ c = 2 3 5 Hence the required equation is (x + 2) y + 2 = 2 or 2xy + 3x + 4y + 1 = 0 Example 4.70 : The tangent at any point of the rectangular hyperbola xy = c2 makes intercepts a, b and the normal at the point makes intercepts p, q on the axes. Prove that ap + bq = 0 y Ta nt
N or m al
e ng
b }q } p
a
Fig. 4. 102
260
x
Solution: Equation of tangent at any point ‘t’ on xy = c2 is x + yt2 = 2ct x y or 2ct + 2c/t = 1 2c ∴ Intercept on the axes are a = 2ct, b = t . c Equation of normal at ‘t’ on xy = c2 is y − xt2 = t − ct3 x y + =1 c − ct3 c − ct3 t t
2 −t
c −1 c p = 2 t − ct3 , q = t − ct3 t 2c c −1 c ∴ ap + bq = 2ct 2 t − ct3 + t t − ct3 t 2c c 2c c = − t t − ct3 + t t − ct3 =0 Example 4.71 : Show that the tangent to a rectangular hyperbola terminated by its asymptotes is bisected at the point of contact. Solution: y The equation of tangent at c P ct, t is x + yt2 = 2ct Putting y = 0 in this equation we get the co-ordinates of A as B (2ct, 0). Putting x = 0 we get P(ct, c/t) 2c the co-ordinates of B as 0, t O x ∴ intercept on axes are
A
Fig. 4. 103 2c 2ct + 0 0 + t c The mid-point of AB is 2 , 2 = ct , t which is the point P. This shows that the tangent is bisected at the point of contact.
261
EXERCISE 4.6 (1) Find the equation of the standard rectangular hyperbola whose centre is − 1 , − 1 and which passes through the point 1 , 1 . 2 2 4 (2) Find the equation of the tangent and normal (i) at (3, 4) to the rectangular 1 hyperbolas xy = 12 (ii) at − 2 , 4 to the rectangular hyperbola 2xy − 2x − 8y − 1 = 0 (3) Find the equation of the rectangular hyperbola which has for one of its asymptotes the line x + 2y − 5 = 0 and passes through the points (6, 0) and (− 3, 0). (4) A standard rectangular hyperbola has its vertices at (5, 7) and (− 3, −1). Find its equation and asymptotes. (5) Find the equation of the rectangular hyperbola which has its centre at (2, 1), one of its asymptotes 3x − y − 5 = 0 and which passes through the point (1, − 1). (6) Find the equations of the asymptotes of the following rectangular hyperbolas. (i) xy − kx − hy = 0 (ii) 2xy + 3x + 4y +1 = 0 (iii) 6x2 + 5xy − 6y2 + 12x + 5y + 3 = 0 (7) Prove that the tangent at any point to the rectangular hyperbola forms with the asymptotes a triangle of constant area. Results without proof : (1) The foot of the perpendicular from a focus of a hyperbola on an asymptote lies on the corresponding directrix. (2) (i) Two tangents (ii) four normals can be drawn from a point to the rectangular hyperbola xy = c2. (3) The condition that the line lx + my + n = 0 may be a tangent to the rectangular hyperbola xy = c2 is 4c2lm = n2 (4) If the normal to the rectangular hyperbola xy = c2 at ‘t1’ meets the curve again at ‘t2’ prove that t13 t2 = − 1. Note : For the proof of above results one may refer the Solution Book.
262
5. DIFFERENTIAL CALCULUS APPLICATIONS - I 5.1 Introduction : In higher secondary first year we discussed the theoretical aspects of differential calculus, assimilated the process of various techniques involved and created many tools of differentiation. Geometrical and kinematical significances for first and second order derivatives were also interpreted. Now let us learn some practical aspects of differential calculus. At this level we shall consider problems concerned with the applications to (i) plane geometry, (ii) theory of real functions, (iii) optimisation problems and approximation problems.
5.2 Derivative as a rate measure : If a quantity y depends on and varies with a quantity x then the rate of dy change of y with respect to x is dx . Thus for example, the rate of change of pressure p with respect to height dp h is dh . A rate of change with respect to time is usually called as ‘the rate of change’, the ‘with respect to time’ being assumed. Thus for example, a rate of di dθ change of current ‘i’ is dt and a rate of change of temperature ‘θ’ is dt and so on. Example 5.1 : The length l metres of a certain metal rod at temperature θ°C is given by l = 1 + 0.00005θ + 0.0000004θ2. Determine the rate of change of length in mm/°C when the temperature is (i) 100°C and (ii) 400°C. dl . Solution : The rate of change of length means dθ Since length l = 1 + 0.00005θ + 0.0000004θ2, dl = 0.00005 + 0.0000008θ . dθ (i) when θ = 100°C dl = 0.00005 + (0.0000008) (100) dθ = 0.00013 m/°C = 0.13 mm/°C
1
(ii) when θ = 400°C dl = 0.00005 + (0.0000008) (400) dθ = 0.00037 m/°C = 0.37 mm/°C Example 5.2 : The luminous intensity I candelas of a lamp at varying voltage V is given by : I = 4 × 10−4V2. Determine the voltage at which the light is increasing at a rate of 0.6 candelas per volt. dI Solution : The rate of change of light with respect to voltage is given by dV . Since I = 4 × 10−4V2 dI −4 dV = 8 × 10 V. dI When the light is increasing at 0.6 candelas per volt then dV = + 0.6. Therefore we must have + 0.6 = 8 × 10-4 V, from which, 0.6 = 0.075 × 104 = 750 Volts. Voltage V = 8 × 10−4 Velocity and Acceleration : y Distance
A car describes a distance x metres in time t seconds along a straight road. If the velocity v is x constant, then v = t m/s i.e., the slope (gradient) of the distance/time graph shown in Fig.5.1 is constant.
x t x
Time
Fig. 5.1 If, however, the velocity of the car is not constant then the distance / time graph will not be a straight line. It may be as shown in Fig.5.2 The average velocity over a small time ∆t and distance ∆x is given by the gradient of the chord AB i.e., the average velocity over time ∆t ∆x is . ∆t
y Distance
B ∆x A
∆t Time
Fig. 5.2
2
x
Velocity
As ∆t → 0, the chord AB becomes a tangent, such that at point A the dx velocity is given by v = dt . Hence the velocity of the car at any instant is given by gradient of the distance / time graph. If an expression for the distance x is known in terms of time, then the velocity is obtained by differentiating the expression. The acceleration ‘a’ of the car is y defined as the rate of change of D velocity. A velocity / time graph is shown in Fig.5.3. If ∆v is the change ∆y in v and ∆t is the corresponding ∆v C change in time, then a = . As ∆t ∆t x Time ∆t → 0 the chord CD becomes a tangent such that at the point C, Fig. 5.3 dv the acceleration is given by a = dt Hence the acceleration of the car at any instant is given by the gradient of the velocity / time graph. If an expression for velocity is known in terms of time t, then the acceleration is obtained by differentiating the expression. dv dx Acceleration a = dt , where v = dt d2x d dx Hence a = dt dt = 2 dt The acceleration is given by the second differential coefficient of distance x with respect to time t. The above discussion can be summarised as follows. If a body moves a distance x meters in time t seconds then (i) distance x = f(t). dx (ii) velocity v = f ′(t) or dt , which is the gradient of the distance / time graph. d2 x dv (iii) Acceleration a = dt = f ′′(t) or 2 , which is the gradient of the dt velocity / time graph. Note : (i) Initial velocity means velocity at t = 0 (ii) Initial acceleration means acceleration at t = 0. (iii) If the motion is upward, at the maximum height, the velocity is zero. (iv) If the motion is horizontal, v = 0 when the particle comes to rest.
3
Example 5.3 : The distance x metres described by a car in time t seconds is given by: x = 3t3 − 2 t2 + 4t − 1. Determine the velocity and acceleration when (i) t = 0 and (ii) t = 1.5 s Solution : distance x = 3t3 − 2 t2 + 4t −1 dx velocity v = dt = 9t2 − 4 t + 4 m/s d2 x = 18t − 4 m/s2 dt2 When time t = 0
acceleration a = (i)
velocity v = 9(0)2 − 4(0) + 4 = 4 m/s and acceleration a = 18(0) − 4 = −4 m/s2 (ii) when time t = 1.5 sec velocity v = 9(1.5)2 − 4(1.5) + 4 = 18.25 m/sec and acceleration a = 18(1.5) − 4 = 23 m/sec2 Example 5.4 : Supplies are dropped from an helicopter and distance fallen in 1 time t seconds is given by x = 2 gt2 where g = 9.8 m/sec2. Determine the velocity and acceleration of the supplies after it has fallen for 2 seconds. 1 1 distance x = 2 gt2 = 2 (9.8) t2 = 4.9 t2 m Solution : dx velocity v = dt = 9.8t m/sec d2 x = 9.8 m/sec2 dt2 When time t = 2 seconds velocity v = (9.8)(2) = 19.6 m/sec and acceleration a = 9.8 m/sec2 which is the acceleration due to acceleration a =
gravity. Example 5.5 : The angular displacement θ radians of a fly wheel varies with time t seconds and follows the equation θ = 9t2 − 2t3. Determine (i) the angular velocity and acceleration of the fly wheel when time t = 1 second and (ii) the time when the angular acceleration is zero. Solution : (i)
angular displacement θ = 9t2 − 2t3 radians. dθ angular velocity ω = dt = 18t – 6t2 rad/s
4
When time t = 1 second, ω = 18(1) − 6(1)2 = 12 rad/s angular acceleration =
d 2θ 2 2 = 18 − 12t rad/s dt
when t = 1, angular acceleration = 6 rad/ s2 (ii) Angular acceleration is zero ⇒ 18 – 12t = 0, from which t = 1.5 s Example 5.6 : A boy, who is standing on a pole of height 14.7 m throws a stone vertically upwards. It moves in a vertical line slightly away from the pole and falls on the ground. Its equation of motion in meters and seconds is x = 9.8 t − 4.9t2 (i) Find the time taken for upward and downward motions. (ii) Also find the maximum height reached by the stone from the ground. Solution : (i) x = 9.8 t − 4.9 t2 At the maximum height v = 0 Max. Ht. dx v = dt = 9.8 − 9.8 t s =0 v = 0 ⇒ t = 1 sec ∴ The time taken for upward motion is 1 sec. For each position x, s =-147 Ground there corresponds a time ‘t’. The ground position is x = − 14.7, since Fig. 5.4 the top of the pole is taken as x = 0. To get the total time, put x = − 14.7 in the given equation. i.e., − 14.7 = 9.8 t − 4.9t2 ⇒ t = − 1, 3 ⇒ t = − 1 is not admissible and hence t = 3 The time taken for downward motion is 3 − 1 = 2 secs (ii) When t = 1, the position x = 9.8(1) − 4.9(1) = 4.9 m The maximum height reached by the stone = pole height + 4.9 = 19.6 m
5.3 Related Rates : In the related rates problem the idea is to compute the rate of change of one quantity in terms of the rate of change of another quantity. The procedure is to find an equation that relates the two quantities and then use the chain rule to differentiate both sides with respect to time. We suggest the following problem solving principles that may be followed as a strategy to solve problems considered in this section.
5
(1) Read the problem carefully. (2) Draw a diagram if possible. (3) Introduce notation. Assign symbols to all quantities that are functions of time. (4) Express the given information and the required rate in terms of derivatives. (5) Write an equation that relates the various quantities of the problem. If necessary, use the geometry of the situation to eliminate one of the variables by substitution. (6) Use the chain rule to differentiate both sides of the equation with respect to t. (7) Substitute the given information into the resulting equation and solve for the unknown rate. Illustration : Air is being pumped into a spherical balloon so that its volume increases at a rate of 100 cm3/s. How fast is the radius of the balloon increasing when the diameter is 50 cm. Solution : We start by identifying two things. (i) The given information : The rate of increase of the volume of air is 100 cm3/s. and (ii) The unknown : The rate of increase of the radius when the diameter is 50 cm. In order to express these quantities mathematically we introduce some suggestive notation. Let V be the volume of the balloon and let r be its radius. The key thing to remember is that the rates of change are derivatives. In this problem, the volume and the radius are both functions of time t. The rate of dV increase of the volume with respect to time is the derivative dt and the rate of dr increase of the radius is dt . We can therefore restate the given and the unknown as follows : dV dr Given : dt = 100 cm3/s and unknown : dt when r = 25 cm. dV dr In order to connect dt and dt we first relate V and r by the formula for 4 the volume of a sphere V = 3 πr3.
6
dy/dt =?
Wall
In order to use the given information, we differentiate both sides of this equation with respect to t. To differentiate the right side, we need to use chain rule as V is a function of r and r is a function of t. dV dV dr dr dr 4 i.e., dt = dr . dt = 3 3πr2 dt = 4πr2 dt 1 dV dr . Now we solve for the unknown quantity dt = 4πr2 dt dV If we put r = 25 and dt = 100 in this equation, 1 × 100 1 dr we obtain dt = 2 = 25π 4π(25) 1 cm/s. i.e., the radius of the balloon is increasing at the rate of 25π Example 5.7 : A ladder 10 m long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1 m/sec how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 m from the wall ? Solution : We first draw a diagram y and lable it as in Fig. 5.5 Let x metres be the distance from the bottom of the ladder to the wall and y metres be the vertical 10 y distance from the top of the ladder to the ground. Note that x and y are both dx/dt =1 x functions of time‘t’. We are given x Ground dx that dt = 1 m/sec and we are asked dy Fig. 5.5 to find dt when x = 6 m. In this question, the relationship between x and y is given by the Pythagoras theorem : x2 + y2 = 100 Differentiating each side with respect to t, using chain rule, we have dx dy 2x dt + 2y dt = 0 and solving this equation for the derived rate we obtain, dy x dx dt = − y dt
7
When x = 6, the Pythagoras theorem gives, y = 8 and so substituting these dy 6 -3 dx values and dt = 1, we get dt = − 8 (1) = 4 m/sec. 3 The ladder is moving downward at the rate of 4 m/sec. Example 5.8 : A car A is travelling from west at 50 km/hr. and car B is travelling towards north at 60 km/hr. Both are headed for the intersection of the two roads. At what rate are the cars approaching each other when car A is 0.3 kilometers and car B is 0.4 kilometers from the intersection? Solution : x A C We draw Fig. 5.6 where C is the intersection of the two roads. At a given time t, let x be the distance from car A to C, y z let y be the distance from car B to C and let z be the distance between the cars A and B B where x, y and z are measured in kilometers. Fig. 5.6 dx dy We are given that dt = − 50 km/hr and dt = − 60 km/hr. Note that x and y are decreasing and hence the negative sign. We are asked dz to find dt . The equation that relate x, y and z is given by the Pythagoras theorem z2 = x2 + y2 Differentiating each side with respect to t, dx dy dz 1 dy dz dx we have 2z dt = 2x dt + 2y dt ⇒ dt = z x dt + y dt When x = 0.3 and y = 0.4 km, we get z = 0.5 km and we get dz 1 dt = 0.5 [0.3 (− 50) + 0.4 (−60)] = −78 km/hr. i.e., the cars are approaching each other at a rate of 78 km/hr. Example 5.9 : A water tank has the shape of an inverted circular cone with base radius 2 metres and height 4 metres. If water is being pumped into the tank at a rate of 2m3/min, find the rate at which the water level is rising when the water is 3m deep.
8
Solution : 2m We first sketch the cone rm and label it as in Fig. 5.7. Let V, r 4m and h be respectively the volume of h the water, the radius of the cone and the height at time t, where Fig. 5.7 t is measured in minutes. dV dh We are given that dt = 2m3/min. and we are asked to find dt when h is 3m. 1 The quantities V and h are related by the equation V = 3 πr2h. But it is very useful to express V as function of h alone. r 2 In order to eliminate r we use similar triangles in Fig. 5.7 to write h = 4 h 1 π h 2 ⇒ r = 2 and the expression for V becomes V = 3 π 2 h = 12 h3. Now we can differentiate each side with respect to t and we have π 2 dh dh 4 dV dV dt = 4 h dt ⇒ dt = πh2 dt dV Substituting h = 3m and dt = 2m3/min. 4 8 dh we get, dt = 2 . 2= 9π m/min π(3) EXERCISE 5.1 (1) A missile fired from ground level rises x metres vertically upwards in 25 t seconds and x = 100t - 2 t2. Find (i) the initial velocity of the missile, (ii) the time when the height of the missile is a maximum (iii) the maximum height reached and (iv) the velocity with which the missile strikes the ground. (2) A particle of unit mass moves so that displacement after t secs is given by x = 3 cos (2t – 4). Find the acceleration and kinetic energy at the end of 2 K.E. = 1 mv2, m is mass secs. 2 (3) The distance x metres traveled by a vehicle in time t seconds after the brakes are applied is given by : x = 20 t − 5/3t2. Determine (i) the speed of the vehicle (in km/hr) at the instant the brakes are applied and (ii) the distance the car travelled before it stops. 9
(4) Newton’s law of cooling is given by θ = θ0° e−kt, where the excess of
temperature at zero time is θ0°C and at time t seconds is θ°C. Determine the rate of change of temperature after 40 s, given that θ0 = 16° C and [e1.2 = 3.3201)
k = − 0.03.
(5) The altitude of a triangle is increasing at a rate of 1 cm/min while the area of the triangle is increasing at a rate of 2 cm2/min. At what rate is the base of the triangle changing when the altitude is 10 cm and the area is 100 cm2. (6) At noon, ship A is 100 km west of ship B. Ship A is sailing east at 35 km/hr and ship B is sailing north at 25 km/hr. How fast is the distance between the ships changing at 4.00 p.m. (7) Two sides of a triangle are 4m and 5m in length and the angle between them is increasing at a rate of 0.06 rad/sec. Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is π/3. (8) Two sides of a triangle have length 12 m and 15 m. The angle between them is increasing at a rate of 2° /min. How fast is the length of third side increasing when the angle between the sides of fixed length is 60°? (9) Gravel is being dumped from a conveyor belt at a rate of 30 ft3/min and its coarsened such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 10 ft high ?
5.4 Tangents and Normals (Derivative as a measure of slope) y
In this section the applications of derivatives to plane geometry is discussed. For this, let us consider a curve whose equation is y = f(x). On this curve take a point P(x1,y1). Assuming that the tangent
y =f (x) P (x1,y1)
No rm al
α
at this point is not parallel to the coordinate axes, we can write the equation of the tangent line at P.
O
Time
Fig. 5.8
10
x
The equation of a straight line with slope (gradient) m passing through (x1,y1) is of the form y – y1 = m(x – x1). For the tangent line we know the slope dy m = f ′(x1) = dx at (x1,y1) and so the equation of the tangent is of the form y – y1= f ′(x1) (x – x1). If m=0, the curve has a horizontal tangent with equation lim y = y1 at P(x1,y1). If f(x) is continuous at x = x1, but x → x f ′(x) = ∞ ⇒ the 1 curve has a vertical tangent with equation x = x1. In addition to the tangent to a curve at a given point, one often has to consider the normal which is defined as follows : Definition : The normal to a curve at a given point is a straight line passing through the given point, perpendicular to the tangent at this point. From the definition of a normal it is clear that the slope of the normal m′ 1 and that of the tangent m are connected by the equation m′ = – m . 1 −1 = dy i.e., m′ = – f ′(x1) (x ,y ) dx 1 1 Hence the equation of a normal to a curve y = f(x) at a point P(x1,y1) is 1 of the form y – y1= – ( x – x1). f ′(x1) The equation of the normal at (x1,y1) is (i) x = x1 if the tangent is horizontal (ii) y = y1 if the tangent is vertical and –1 (iii) y – y1 = m (x – x1) otherwise. Example 5.10: Find the equations of the tangent and normal to the curve y = x3 at the point (1,1). Solution :
We have y = x3 ; slope y′= 3x2.
At the point (1,1), x = 1 and m = 3(1)2 = 3. Therefore equation of the tangent is y − y1 = m(x − x1) y – 1 = 3(x – 1) or y = 3x – 2 1 The equation of the normal is y − y1 = − m (x − x1) 1 4 –1 y – 1 = 3 (x – 1) or y = – 3 x + 3
11
Example 5.11 : Find the equations of the tangent and normal to the curve y = x2 – x – 2 at the point (1,− 2). dy Solution : We have y = x2 – x – 2 ; slope, m = dx = 2x – 1. At the point (1,–2), m = 1 Hence the equation of the tangent is y – y1 = m(x – x1) i.e., y – (–2) = x – 1 i.e., y = x – 3 –1 Equation of the normal is y – y1 = m (x – x1) –1 i.e., y – (–2) = 1 (x – 1) or y = – x – 1 Example 5.12 : Find the equation of the tangent at the point (a,b) to the curve xy = c2. Solution : The equation of the curve is xy = c2. Differentiating w.r.to x we get, dy y +x dx = 0 –y –b dy dy or dx = x and m = dx (a, b)= a . Hence the required equation of the tangent is –b y –b = a (x – a) i.e., ay – ab = – bx + ab x y bx + ay = 2ab or a + b = 2 π Example 5.13 : Find the equations of the tangent and normal at θ = 2 to the curve x = a (θ + sin θ), y = a (1 + cos θ). θ dx = a (1 + cosθ) = 2a cos2 2 Solution : We have dθ dy θ θ = – a sin θ = – 2a sin 2 cos 2 dθ dy dθ dy θ Then dx = dx = – tan 2 dθ
12
π dy ∴ Slope m = dx θ = π/ = – tan 4 = –1 2 π π Also for θ = 2 , the point on the curve is a 2 + a, a. π Hence the equation of the tangent at θ = 2 is π y – a = (–1) x − a 2 + 1
1 1 i.e., x + y = 2 a π + 2a or x + y – 2 a π – 2a = 0 Equation of the normal at this point is π y – a = (1) x − a 2 + 1 1 or x – y – 2 a π = 0 Example 5.14 : Find the equations of tangent and normal to the curve 16x2 + 9y2 = 144 at (x1,y1) where x1 = 2 and y1 > 0. Solution : We have 16x2 + 9y2 = 144 (x1,y1) lies on this curve, where x1 = 2 and y1 > 0 ∴ (16 × 4) + 9 y12 = 144 or 9 y12 = 144 – 64 = 80 80 y12 = 9
80 80 3 . But y1 > 0 ∴ y1 = 3 80 ∴ The point of tangency is (x1,y1) = 2 , 3 ∴ y1 = ±
We have 16x2 + 9y2 = 144 dy 32 x 16 x Differentiating w.r.to x we get dx = – 18 y = – 9 y ∴ The slope at
2 , 80 = dy 3 dx 2 , 16 =– 9 ×
13
80 3
2 8 =– 3 5 80 3
8 80 ∴ The equation of the tangent is y – 3 = – (x – 2) 3 5 On simplification we get 8x + 3 5 y = 36 Similarly the equation of the normal can be found as 9 5 x – 24 y + 14 5 = 0 Example 5.15 : Find the equations of the tangent and normal to the ellipse π x = a cosθ, y = b sin θ at the point θ = 4 . π b π π a Solution : At θ = 4 , (x1,y1) = a cos 4 , b sin 4 = 2 , 2 dx dy = – a sin θ, = b cos θ. y dθ dθ dy T dθ dy –b P = = cotθ ‘θ’ =π/4 dx a dx x dθ O N –b π –b ⇒ m = = a cot 4 = a Fig. 5.9 b –b a Thus the point of tangency is , and the slope is m = a . 2 2 b a b = − a x − or bx + ay − ab 2 = 0 The equation of the tangent is y − 2 2 a a b = b x – The equation of the normal is y – 2 2 or (ax – by) 2 – (a2 – b2) = 0. Example 5.16 : Find the equation of the tangent to the parabola, y2 = 20 x which forms an angle 45° with the x – axis. Solution : We have y2 = 20x . Let (x1,y1) be the tangential point 10 10 Now 2yy′ = 20 ∴ y′ = y ie., at (x1, y1) m = y 1 But the tangent makes an angle 45° with the x – axis. ∴ slope of the tangent m=tan 45° = 1 10 From (1) and (2) y = 1 ⇒ y1 = 10 1 But (x1,y1) lies on y2 = 20x ⇒ y12 = 20 x1
14
… (1) … (2)
100 = 20 x1 or x1 = 5 i.e., (x1,y1) = (5,10) and hence the equation of the tangent at (5, 10) is y – 10 = 1(x – 5) or y = x + 5. Note : This problem is suitable for equation of any tangent to a parabola a i.e., y = mx + m
5.5 Angle between two curves : The angle between the curves C1 and C2 at a point of intersection P is defined to be the angle between the tangent lines to C1 and C2 at P (if these tangent lines exist) Let us represent the two curves C1 and C2 by the Cartesian equation y = f(x) and y = g(x) respectively. Let them intersect at P (x1,y1) . If ψ1 and ψ2 are the angles made by the tangents PT1 and PT2 to C1 and C2 at P, with the positive direction of the x – axis, then m1 = tan ψ 1 and m2 = tan ψ2 are the slopes of PT1 and PT2 respectively. y
Let ψ be the angle between PT1 and PT2. Then ψ = ψ2 – ψ1 and
y =g (x) C2
y =f (x) C1
tan ψ = tan (ψ2 – ψ1) tan ψ2 – tan ψ1 = 1 + tan ψ1 tan ψ2
T1
P
m2 – m1 = 1+m m 1 2
ψ1
180
–ψ
O
where 0 ≤ ψ < π
T2
2
Time
ψ2
x
Fig. 5.10
We observe that if their slopes are equal namely m1 = m2 then the two curves touch each other. If the product m1 m2 = – 1 then these curves are said to cut at right angles or orthogonally. We caution that if they cut at right angles then m1 m2 need not be –1. Note that in this case ψ1 is acute and ψ2 is obtuse and ψ = ψ2 − ψ1. If ψ1 is obtuse and ψ2 is acute, then ψ = ψ1−ψ2.
15
Combining together the angle between tangents can be given as ψ1∼ψ2 or tan ψ1∼ tanψ2 m1 − m2 tan ψ = tan(ψ1∼ψ2) = = 1 + m m 1 + tan ψ1 tan ψ2 1 2 Example 5.17 : Find the angle between the curves y = x2 and y = (x – 2)2 at the point of intersection. Solution : To get the point of intersection of the curves solve the equation y =x2
y =(x-2)2
y
we get x2 = (x− 2)2 This gives x = 1. When x = 1, y = 1 ∴ The point of intersection is (1, 1) dy Now y = x2 ⇒ dx = 2x dy ⇒ m1 = dx (1,1) = 2
2 1
(1,1) 2
(0,0)
x
Tan-1(4/3)
Fig. 5.11 dy dy y = (x – 2)2 ⇒ dx = 2(x – 2) ⇒ m2 = dx (1,1) = – 2. If ψ is the angle between them, then –2–2 –4 –1 4 tan ψ = 1 – 4 = − 3 ⇒ ψ = tan 3 Example 5.18 : Find the condition for the curves ax2 + by2 = 1, a1x2 + b1y2= 1 to intersect orthogonally. Solution : If (x1,y1) is the point of intersection, then ax12 + by12 = 1 ; a1x12 + b1y12 = 1 b1 – b a – a1 then, x12 = ab – a b , y12 = ab – a b (By Cramer’s rule) 1 1 1 1 – ax dy 1 For ax2 + by2 = 1, m1 = dx (x ,y ) = by1 1 1
– a1x1 = b y (x ,y ) 1 1 1 1 For orthogonal intersection, we have m1m2 = –1. This gives
and for a1x2 + b1y2 = 1,
dy m2 = dx 2
a a1x1 – ax1 – a1x1 = –1. by b y = –1 or 1 1 1 bb1y12
16
b1 – b a – a1 aa1x12 + bb1y12 = 0 ⇒ aa1 ab – a b + bb1 ab – a b = 0
1
1
1
1
b1 – b a – a1 ⇒ aa1 (b1 – b) + bb1 (a – a1) = 0 ⇒ bb + aa = 0 1 1 1 1 1 1 1 1 1 1 or b – b + a – a = 0 or a – a = b – b which is the required condition. 1 1 1 1 Example 5.19 : Show that x2 – y2 = a2 and xy = c2 cut orthogonally. Solution : Let (x1,y1) be the point of intersection of the given curves ∴ x12 – y12 = a2 and x1 y1 = c2 dy dy x ⇒ 2x – 2y dx = 0 ⇒ dx = y x1 x1 dy = y ie., m1 = y ∴ m1 = dx (x ,y ) 1 1
x2 – y2 = a2
1 1
xy = c2
c2 ⇒ y = x
dy ⇒ dx = –
c2 x2
– c2 – c2 dy = 2 i.e., m2 = 2 ∴ m2 = dx (x ,y ) x1 x1 1 1 – c2 x1 – c2 – c2 ∴ m1m2 = y 2 = x y = 2 = –1 1 1 c 1 x
1
⇒ the curves cut orthogonally. Example 5.20 : Prove that the sum of the intercepts on the co-ordinate axes of π any tangent to the curve x = a cos4θ, y = a sin4θ, 0 ≤ θ ≤ 2 is equal to a. Solution : Take any point ‘θ’ as (a cos4θ, a sin4θ, ) dx y Now = – 4a cos3θ sin θ ; dθ θ = π/2 dy = 4a sin3θ cos θ and dθ (0,a) sin2θ dy ∴ dx = – cos2θ O
θ=0 x (a,0)
Fig. 5.12
17
i.e., slope of the tangent at ‘θ’ is = –
sin2θ cos2θ
Equation of the tangent at ‘θ’ is (y − a sin4θ) =
− sin2θ 4 2 (x − a cos θ) cos θ
or x sin2 θ + y cos2 θ = a sin2 θ cos2 θ x y ⇒ =1 2 + a cos θ a sin2θ i.e., sum of the intercepts = a cos2 θ + a sin2 θ = a EXERCISE 5.2 (1) Find the equation of the tangent and normal to the curves (i)
π (ii) y = x – sin x cos x, at x = 2
y = x2 – 4x – 5 at x = – 2
π (iii) y = 2 sin2 3x at x = 6
1 + sinx π (iv) y = cos x at x = 4
(2) Find the points on curve x2– y2=2 at which the slope of the tangent is 2. (3) Find at what points on the circle x2 + y2 = 13, the tangent is parallel to the line 2x + 3y = 7 (4) At what points on the curve x2 + y2 – 2x – 4y + 1 = 0 the tangent is parallel to (i) x – axis (ii) y – axis. (5) Find the equations of those tangents to the circle x2 + y2 = 52, which are parallel to the straight line 2x + 3y = 6. (6) Find the equations of normal to y = x3 – 3x that is parallel to 2x + 18y – 9 = 0. (7) Let P be a point on the curve y = x3 and suppose that the tangent line at P intersects the curve again at Q. Prove that the slope at Q is four times the slope at P. (8) Prove that the curve 2x2 + 4y2 = 1 and 6x2 – 12y2= 1 cut each other at right angles. (9) At what angle θ do the curves y = ax and y = bx intersect (a ≠ b) ? (10) Show that the equation of the normal to the curve x = a cos3 θ ; y = a sin3θ at ‘θ’ is x cos θ – y sin θ = a cos 2θ. (11) If the curve y2 = x and xy = k are orthogonal then prove that 8k2 = 1.
18
5.6 Mean value theorems and their applications : In this section our main objective is to prove that between any two points of a smooth curve there is a point at which the tangent is parallel to the chord joining two points. To do this we need the following theorem due to Michael Rolle. 5.6.1 Rolle’s Theorem : Let f be a real valued function that satisfies the following three conditions : (i) f is defined and continuous on the closed interval [a, b] (ii) f is differentiable on the open interval (a, b) (iii) f (a) = f (b) Then there exists atleast one point c ∈ (a,b) such that f ′(c) = 0 Some observations : Rolle’s theorem is applied to the position function s = f(t) of a moving object. If the object is in the same place at two different instants t = a and t = b then f(a) = f(b) satisfying hypothesis of Rolle’s theorem. Therefore the theorem says that there is some instant of time t = c between a and b where f ′(c) = 0 i.e., the velocity is 0 at t = c. Note that this is also true for an object thrown vertically upward (neglecting air resistance). Rolle’s Theorem applied to theory of equations : If a and b are two roots of a polynomial equation f(x) = 0, then Rolle’s Theorem says that there is atleast one root c between a and b for f ′(x) = 0. Rolle’s theorem implies that a smooth curve cannot intersect a horizontal line twice without having a horizontal tangent in between. Rolle’s theorem holds trivially for the function f(x) = c, where c is a constant on [a,b]. The converse of Rolle’s Theorem is not true ie., if a function f satisfies f ′(c) = 0 for c ∈ (a,b) then the conditions of hypothesis need not hold. Example 5.21 : Using Rolle’s theorem find the value(s) of c. (i) f(x) = 1 − x2 , −1 ≤ x ≤ 1 (ii) f(x) = (x − a) (b − x), a ≤ x ≤ b, a ≠ b. 1 (iii) f(x) = 2x3 − 5x2 − 4x + 3, 2≤ x ≤ 3
19
Solution : (i) The function is continuous in [−1,1] and differentiable in (−1,1). f(1) = f (−1) = 0 all the three conditions are satisfied. 1 − 2x −x = f ′(x) = 2 2 1 −x 1 − x2 f ′(x) = 0 ⇒ x = 0. (Note that for x = 0, denominator = 1 ≠ 0) Thus the suitable point for which Rolle’s theorem holds is c = 0. (ii) f(x) = (x − a) (b − x), a ≤ x ≤ b, a ≠ b. f (x) is continuous on [a,b] and f ′(x) exists at every point of (a,b). f(a) = f(b) = 0 All the conditions are satisfied. ∴ f ′(x) = (b − x) − (x − a) a+b f ′(x) = 0 ⇒ − 2x = − b − a ⇒ x = 2 a+b The suitable point ‘c’ of Rolle’s theorem is c = 2 1 (iii) f(x) = 2x3 − 5x2 − 4x + 3, 2 ≤ x ≤ 3 1 1 f is continuous on 2 , 3and differentiable in 2 , 3 f(½) = 0 = f(3). All the conditions are satisfied. f ′(x) = 6x2 − 10x − 4 1 f ′(x) = 0 ⇒ 3x2 − 5x− 2 = 0 ⇒ (3x + 1) (x −2) = 0 ⇒ x = − 3 or x = 2. 1 1 x = − 3 does not lie in 2, 3 ∴x = 2 is the suitable ‘c’ of Rolle’s theorem Remark : Rolle’s theorem cannot be applied if any one of the conditions does not hold. Example 5.22 : Verify Rolle’s theorem for the following : (i) f(x) = x3 − 3x + 3 0 ≤ x ≤ 1 (ii) f(x) = tan x, 0 ≤ x ≤ π (iii) f(x) = | x |, −1 ≤ x ≤ 1 (iv) f(x) = sin2 x, 0 ≤ x ≤ π (v) f(x) = ex sin x, 0 ≤ x ≤ π (vi) f(x) = x (x − 1) (x − 2), 0 ≤ x ≤ 2
20
Solution : (i) f(x) = x3 − 3x + 3 0 ≤ x ≤ 1 f is continuous on [0,1] and differentiable in (0,1) f(0) = 3 and f(1) = 1 ∴ f (a) ≠ f (b) ∴ Rolle’s theorem, does not hold, since f (a) = f (b) is not satisfied. Also note that f ′(x) = 3x2 − 3 = 0 ⇒ x2 = 1 ⇒ x = ±1 There exists no point c ∈ (0,1) satisfying f ′(c) = 0. (ii) f(x) = tan x, 0 ≤ x ≤ π π f ′(x) is not continuous in [0,π] as tan x tends to + ∞ at x = 2, ∴ Rolles theorem is not applicable. (iii) f(x) = | x |, −1 ≤ x ≤ 1 f is continuous in [−1,1] but not differentiable in (−1,1) since f ′(0) does not exist. ∴ Rolles theorem is not applicable. (iv) f(x) = sin2 x, 0 ≤ x ≤ π f is continuous in [0,π] and differentiable in (0,π). f(0) = f (π) = 0 (ie.,) f satisfies hypothesis of Rolle’s theorem. f ′(x) = 2 sin x cos x = sin 2x π 3π f ′(c) = 0 ⇒ sin 2c = 0 ⇒ 2c = 0, π, 2π, 3π, ... ⇒ c = 0, 2, π, 2 , ... π π since c = 2 ∈ (0,π), the suitable c of Rolle’s theorem is c = 2. (v) f(x) = ex sin x, 0 ≤ x ≤ π ex and sin x are continuous for all x, therefore the product ex sin x is continuous in 0 ≤ x ≤ π. f ′(x) = ex sin x + ex cos x = ex (sin x + cos x) exist in 0 < x < π ⇒ f ′(x) is differentiable in (0,π). f(0) = e0 sin 0 = 0 f(π) = eπ sin π = 0 ∴ f satisfies hypothesis of Rolle’s theorem Thus there exists c∈ (0, π) satisfying f ′(c) = 0 ⇒ ec(sin c + cos c) = 0 ⇒ ec = 0 or sin c + cos c = 0
21
ec = 0 ⇒ c = − ∞ which is not meaningful here. sin c 3π ⇒ sin c = − cos c ⇒ cos c =−1 ⇒ tan c = − 1 = tan 4 3π ⇒ c = 4 is the required point. (vi) f(x) = x (x − 1) (x − 2), 0 ≤ x ≤ 2, f is continuous in [0,2] and differentiable in (0,2) f(0) = 0 = f(2), satisfying hypothesis of Rolle’s theorem Now f ′(x) = (x − 1) (x − 2) + x (x −2) + x (x −1) = 0 1 ⇒ 3x2 − 6x + 2 = 0 ⇒ x = 1 ± 3 1 The required c in Rolle’s theorem is 1 ± ∈ (0,2) 3 Note : There could exist more than one such ‘c’ appearing in the statement of Rolle’s theorem. Example 5.23 : Apply Rolle’s theorem to find points on curve y = − 1 + cos x, where the tangent is parallel to x-axis in [0, 2π]. Solution : y f(x) is continuous in [0,2π] and 2π π x differentiable in (0,2π) (0,0) f(0) = 0 = f(2π) satisfying hypothesis -1 of Rolle’s theorem. -2 Now f ′(x) = − sin x = 0 ⇒ sin x = 0 (π,-2) x = 0, π, 2π, . . . Fig. 5.13 x = π, is the required c in (0,2π). At x = π, y = −1 + cos π = −2. ⇒ the point (π,−2) is such that at this point the tangent to the curve is parallel to x-axis. EXERCISE 5.3 (1) Verify Rolle’s theorem for the following functions : (i) f(x) = sin x, 0≤x≤π (ii) f(x) = x2, (iii) f(x) = | x − 1|,
0≤ x ≤1 0≤x≤2 3 3 (iv) f(x) = 4x3 − 9x, − 2 ≤ x ≤ 2
22
(2) Using Rolle’s theorem find the points on the curve y = x2+1, −2 ≤ x ≤ 2 where the tangent is parallel to x − axis.
5.6.2 Mean Value Theorem (Law of the mean due to Lagrange) : Many results in this section depend on one central fact called law of the mean or mean value theorem due to Joseph – Louis Lagrange. Theorem :Let f(x) be a real valued function that satisfies the following conditions : (i) f(x) is continuous on the closed interval [a,b] (ii) f(x) is differentiable on the open interval (a,b) Then there exists at least one point c ∈ (a,b) such that f(b) − f(a) f ′(c) = …(1) b−a Some Observations : Note that if f(a) = f(b) then the law of the mean reduces to the Rolle’s theorem. Interpretation of law of the mean when applied to an equation of motion s = f(t) : The quantity ∆s = f(b) − f(a) is the change in s corresponding to ∆t = b – a and R.H.S. of (1) is ∆s f(b) − f(a) = = average velocity from t = a to t = b. b−a ∆t The equation then tells us that there is an instant ‘c’ between a and b at which the instantaneous velocity f ′(c) is equal to the average velocity. For example, if a car has traveled 180 kms in 2 hours then the speedometer must have read 90 kms/hr at least once.
23
y C
1
( x)
2
y =f
The slope f ′(c)of the curve at C (c, f(c)) f(b) − f(a) of the is the same as the slope b−a and chord joining the points A (a, f(a)) B (b, f(b)). Geometrically means that if the function f is continuous on [a,b] and differentiable on (a,b) then there is atleast one number c in (a,b) where the tangent to the curve is parallel to the chord through A and B.
B
A x
(0,0)
Fig. 5.14
Remarks (1) : Since the value of c satisfies the condition a < c < b, it follows c−a that (c − a) < (b − a) or (< 1) = θ, (say). b−a i.e.,
c−a = θ ⇒ c − a = θ (b − a), 0 < θ < 1. b−a
But then c = a + θ (b − a) ∴ the law of the mean can be put in the form f(b) − f(a) = (b − a) f ′(c) = (b − a) f ′[a + θ (b − a)], 0 < θ < 1 and this is used in calculating approximate values of functions. (2) Letting b − a = h, the above result can be written as f(a + h) = f(a) + hf ′(a + θh), 0 < θ < 1 (3) If we let a = x, h = ∆x, law of the mean becomes f(x + ∆x) = f(x) + ∆x f ′(x + θ∆x) for some θ such that 0 < θ < 1. Example 5.24 : Verify Lagrange’s law of the mean for f(x) = x3 on [−2,2] Solution : f is a polynomial, hence continuous and differentiable on [− 2, 2]. f(2) = 23 = 8 ; f (−2) = (−2)3 = −8 f ′(x) = 3x2 ⇒ f ′(c) = 3c2 By law of the mean there exists an element c ∈ (− 2, 2) such that f(b) − f(a) 8 − (−8) =4 ⇒ 3c2 = 4 b−a 4 2 i.e., c2 = 3 ⇒ c = ± 3 2 2 The required ‘c’ in the law of mean are and − as both lie in [−2,2]. 3 3 f ′(c) =
Example 5.25 : A cylindrical hole 4 mm in diameter and 12 mm deep in a metal block is rebored to increase the diameter to 4.12 mm. Estimate the amount of metal removed. Solution : The volume of cylindrical hole of radius x mm and depth 12 mm is given by
24
12mm
V = f(x) = 12 πx2 ⇒ f ′(c) = 24πc. To estimate f(2.06) − f(2) : 4mm By law of mean, f(2.06) − f(2) = 0.06 f ′(c) = 0.06 (24 πc), 2 < c < 2.06 Take c = 2.01 f(2.06) − f(2) = 0.06 × 24 π × 2.01 = 2.89 π cubic mm. Fig. 5.15 Note : Any suitable c between 2 and 2.06 other than 2.01 also will give other estimates. Example 5.26 : Suppose that f(0) = − 3 and f ′(x) ≤ 5 for all values of x, how large can f(2) possibly be? Solution : Since by hypothesis f is differentiable, f is continuous everywhere. We can apply Lagrange’s Law of the mean on the interval [0,2]. There exist atleast one ‘c’∈(0, 2) such that f(2) − f(0) = f ′(c) ( 2 − 0) f(2) = f(0) + 2 f ′(c) = −3 + 2 f ′(c) Given that f ′(x) ≤ 5 for all x. In particular we know that f ′(c) ≤ 5. Multiplying both sides of the inequality by 2, we have 2f ′(c) ≤ 10 f(2) = −3 + 2 f ′(c) ≤ −3 + 10 = 7 i.e., the largest possible value of f(2) is 7. Example 5.27 : It took 14 sec for a thermometer to rise from −19°C to 100°C when it was taken from a freezer and placed in boiling water. Show that somewhere along the way the mercury was rising at exactly 8.5°C/sec. Solution : Let T be the temperature reading shown in the thermometer at any time t. Then T is a function of time t. Since the temperature rise is continuous and since there is a continuous change in the temperature the function is differentiable too. ∴ By law of the mean there exists a ‘t0’ in (0, 14) such that T(t2) − T(t1) = T ′(t0) t2 − t1 Here T ′(t0) is the rate of rise of temperature at C. 25
Here t2 − t1 = 14, T(t2) = 100 ; T(t1) = − 19 100 + 19 119 = 14 = 8.5C/sec T ′(t0) = 14 EXERCISE 5.4 (1) Verify Lagrange’s law of mean for the following functions : (i) f(x) = 1 − x2, [0,3]
1 (ii) f(x) = x , [1,2]
(iii) f(x) = 2x3 + x2 − x − 1, [0,2]
(iv) f(x) = x2/3, [−2,2]
(v) f(x) = x3 − 5x2 − 3x , [1,3] (2) If f(1) = 10 and f ′(x) ≥ 2 for 1 ≤ x ≤ 4 how small can f(4) possibly be? (3) At 2.00 p.m a car’s speedometer reads 30 miles/hr., at 2.10 pm it reads 50 miles / hr. Show that sometime between 2.00 and 2.10 the acceleration is exactly 120 miles /hr2.
Generalised Law of the Mean : If f(x) and g(x) are continuous real valued functions on [a,b] and f and g are differentiable on (a,b) with g ′(x) ≠ 0 everywhere on (a,b) then there exist f ′(c) f(b) − f(a) atleast one value of x, say x = c, between a and b such that = g(b) − g(a) g′(c) Remarks : (1) This theorem is also known as Cauchy’s generalised law of the mean. (2) Lagrange’s law of the mean is a particular case of Cauchy law of the mean for the case g(x) = x for all x ∈ [a,b] (3) Note that g(b) ≠ g(a), for, suppose g(b) = g(a), then by Rolle’s theorem, g′(x) = 0 for some x in (a,b) contradicting hypothesis of the generalized law of the mean. Extended Law of the mean : If f(x) and its first (n − 1) derivatives are continuous on [a,b] and if f(n)(x) exists in (a,b), then there exist atleast one value of x, x = c say, in (a,b) such that f ′′(a) f ′(a) f (n−1)( a) f (n)(c) (b−a)n−1+ n! (b−a)n...(1) f(b)=f(a)+ 1! (b−a)+ 2! (b−a)2+...+ (n −1)!
26
Remarks : (1) If in the extended law of the mean b − a = h then b = a + h and (1) becomes f ′′(a) f ′(a) f (n−1)( a) n − 1 f (n)(c) n h + n! h ...(2) f(a + h) = f(a) + 1! h + 2! h2 + ... (n −1)! for some c ∈ (a, a + h) and this is known as Taylor’s theorem. (2) When b is replaced by the variable x then (1) becomes f ′(a) f (n−1)( a) f (n)(c) (x − a)n − 1 + n! (x − a)n f(x) = f(a) + 1! (x − a) +... (n −1)! for some c ∈ (a, x) (3) If n becomes sufficiently large (i.e., ; as n → ∞) in Taylors theorem, then (2) becomes f ′′(a) f ′(a) f (n)(a) ...(3) f(a + h) = f(a) + 1! h + 2! h2 + . . . + n! hn + ... provided f is differentiable any number of times. This series of expansion of f(a + h) about the point a is usually known as Taylor’s Series. (4) If in the extended law of the mean a is replaced by 0 and b is replaced with the variable x, (1) becomes, f ′′(0) f ′(0) f (n−1)(0) n − 1 f (n) (c) n x + n! x ___(4) f(x) = f(0) + 1! x + 2! x2 +...+ (n−1)! for some c ∈ (0,x) and is known as Maclaurin’s theorem. (5) If n is sufficiently large (i.e., n → ∞) in Maclaurin’s theorem, then it f ′′(0) f ′(0) becomes f(x) = f(0) + 1! x + 2! x2 + . . . provided f is differentiable any number of times, This series expansion of f(x) about the point 0 is usually known as Maclaurin’s Series. π Illustration : The Taylor’s series expansion of f(x) = sin x about x = 2 is obtained by the following way. π π f(x) = sin x ; f 2 = sin 2 = 1 π π f ′(x) = cos x ; f ′ 2 = cos 2 = 0 π f ′′(x) = − sin x ; f ′′ 2 = −1 π f ′′′(x) = − cos x ; f ′′′ 2 = 0
27
π ∴ f(x) = sin x = f 2 +
π f ′ 2
π 1! x −2 +
π f″ 2
2
x − π 2! 2 + ... 2
(−1) π π = 1 + 0 x −2 + 2! x − 2 + ... 4
1 1 π 2 π sin x = 1 − 2! x −2 + 4! x − 2 − ... Example 5.28 : Obtain the Maclaurin’s Series for 1) ex
3) arc tan x or tan−1x
2) loge(1 + x)
Solution : (1)
f(x) = ex
;
f(0) = e0 = 1
f ′(x) = ex
; f ′(0) = 1
f ′′(x) = e
;
x
f ′′(0) = 1
! 1.x 1 1 f(x) = ex = 1 + 1! + 2! x2 + 3! x3 … x2 x3 x = 1 + 1! + 2! + 3! + ... holds for all x f(x) = loge(1 + x) : 1 f ′(x) = 1 + x ; −1 f ′′(x) = ; (1 + x)2 +1.2 f ′′′(x) = ; (1 + x)3 −1.2.3 ; f ′′′′(x) = (1 + x)4
(2)
f(0) = loge1 = 0 f ′(0) = 1 f ′′(0) = −1 f ′′′(0) = 2! f ′′′′(0) = − (3!)
1 3! 1 2! f(x) = loge(1 + x) = 0 + 1! x − 2! x2 + 3! x3 − 4! x4 − ... + …. x2 x3 x4 x − 2 + 3 − 4 + .... −1 < x ≤ 1.
28
(3)
f(x) = tan−1x f ′(x) =
; f(0) = 0
1 = 1 −x2 + x4 – x6…. ; f ′(0) = 1 = 1! 1 + x2
f ′′(x) = − 2x + 4x3 – 6x5 …. ; f ′′(0) = 0 f ′′′(x) = − 2 + 12x2 – 30x4 …. ; f ′′′(0) = −2 = −(2!) f iv(x) = 24x − 120x3 …. ; f iv(0) = 0 f v(x) = 24 − 360x2 …. ; f v(0) = 24 = 4! 1 0 2 0 4! tan−1 x = 0 + 1! x + 2! x2 − 3! x3 + 4! x4 + 5! x5 + … 1 1 = x − 3 x3 + 5 x5 − … holds in | x | ≤ 1. EXERCISE 5.5 Obtain the Maclaurin’s Series expansion for : 1 π π (2) cos2x (3) 1 + x (4) tan x, − 2 < x < 2 (1) e2x
5.7 Evaluating Indeterminate forms : Suppose f(x) and g(x) are defined on some interval [a,b], satisfying Cauchy’s generalized law of the mean and vanish at a point x = a of this interval f(x) such that f(a) = 0 and g(a) = 0, then the ratio g(x) is not defined for x = a 0 and gives a meaningless expression 0 but has a very definite meaning for values of x ≠ a. Evaluating the limit x → a of this ratio is known as evaluating 0 indeterminate forms of the type 0. 3x −2 If f(x) = 3x − 2 and g(x) = 9x + 7, then 9x + 7 is an indeterminate form ∞ of the type as the numerator and denominator becomes ∞ in the limiting ∞ case, x tends to ∞.
29
We also have other limits
lim lim ex lim lim (x − ex), x→ 0 xx, x1/x , x→ ∞ x x→ ∞ x→ ∞
1 /(x−1) lim which lead to other indeterminate forms of the types and x→ 1 x
0 . ∞, ∞ − ∞, 00, ∞0 and 1∞ respectively. These symbols must not be taken literally. They are only convenient labels for distinguishing types of behaviour at certain limits. To deal with such indeterminate forms we need a tool that facilitates the evaluation. This tool was devised by John Bernoulli for calculating the limit of a fraction whose numerator and denominator approach zero. This tool today is known as l’Hôpital’s rule after Guillaume Francois Antoinede l’Hôpital. l’Hôpital’s rule : Let f and g be continuous real valued functions defined on the closed interval [a,b], f, g be differentiable on (a,b) and g′(c) ≠ 0. f ′(c) = L it follows that g′(c) x→ c
Then if lim f(x) = 0, lim g(x) = 0 and if lim x→ c
x→ c
f(x) lim g(x) = L.
x→ c
Remarks : (1) Using l’Hôpital’s method, evaluation of the limits of indeterminate forms works faster than conventional methods. For instance, consider sin x lim x . This limit we know is 1, which we obtained through x→ 0 geometrical constructions, a laborious method. sin x cos x But lim x = lim 1 = cos 0 = 1 x→ 0 x→ 0 (2) Note that l’Hôpital’s rule can be applied only to differentiable functions for which the limits are in the indeterminate form. For, x+1 1 x+1 1 lim is 3 while if l’Hôpital’s rule is applied lim x + 3 = 1 = 1. x + 3 x→ 0 x→ 0 Here f(x) = x + 1 g(x) = x + 3 are both differentiable but not in the indeterminate form (3) The conclusion of l’Hôpital’s rule is unchanged if lim f(x) = 0 and x→ a
lim g(x) = 0 and replaced by lim f(x) = ± ∞ and lim f(x) = ± ∞. x→ a
x→ a
30
x→ a
(4) All other indeterminate forms mentioned above can also be reduced to ∞ 0 0 or ∞ by a suitable transformation. We need the following result in some problems Composite Function Theorem : Result : If lim g(x) = b and f is continuous at b, x→ a
then lim f(g(x) = f lim g(x) x→ a
x→a
x Example 5.29 : Evaluate : lim tan x x→ 0 x 0 Solution : lim tan x is of the type 0 . x→ 0 x 1 1 ∴ lim tan x = lim 2 =1=1 x→ 0 x→ 0 sec x 1 sin x if exists Example 5.30 : Find lim −11 x → + ∞ tan x 1 Solution : Let y = x As x → ∞, y → 0 1 sin x sin y 0 = lim lim −1 = 0 1 −1 y → 0 tan y x → + ∞ tan x 1 cos y = lim 1 = 1 = 1 y→0 2 1 + y log(sin x) Example 5.31 : lim 2 π (π − 2x) x→ /2
0 Solution : It is of the form 0 log(sin x) lim lim 2 = π (π − 2x) π
x→ /2
1 sin x cos x 2(π − 2x) × (−2)
x→ /2
31
= lim
cotx 0 = − 4(π − 2x) 0
= lim
− cosec2x −1 = −4×−2 8
π x→ /2
π x→ /2
Note that here l’Hôpital’s rule, applied twice yields the result. x2 Example 5.32 : Evaluate : lim ex x→∞ x2 ∞ Solution : lim x is the type ∞ e x→∞ x2 2x 2 2 lim x = lim x = lim x = ∞ =0 e e e x→∞ x→∞ x→∞ 1 Example 5.33 : Evaluate : lim cosec x − x x→ 0 1 Solution : lim cosec x − x is of the type ∞ − ∞. x→ 0 1 1 x − sin x 0 1 lim cosec x − = lim sin x − x = lim x sinx = 0 x x→ 0 x→ 0 x→ 0 1 − cos x 0 sinx = 0 type = lim sin x + x cos x x x − x sin x cos + cos x→ 0 x→ 0 0 =2 = 0 lim
sin x
Example 5.34 : Evaluate : lim (cot x) x→ 0 sin x
Solution : lim (cot x)
is of the type ∞0.
x→ 0 sin x
Let y = (cot x) ⇒ log y = sin x log (cot x) lim (log y) = lim sin x log (cot x) x→ 0
x→ 0
log (cot x) ∞ is of the type cosec x ∞ x→ 0
= lim
32
Applying l’Hôpital’s rule, 1 (− cosec2 x) cotx log (cot x) lim lim cosec x = x→ 0 −cosec x cot x x→ 0 sin x 1 0 = lim cos x × cos x = 1 = 0 x→ 0 i.e., lim log y = 0 x→ 0
By Composite Function Theorem, we have 0 = lim log y = log lim y ⇒ lim y = e0 = 1
x→ 0
x→ 0
x→ 0
Caution : When the existence of lim f(x) is not known, log lim f(x) is x→ a
x→ a
meaningless. sinx
Example 5.35 : Evaluate lim x x→ 0 +
Solution :
lim x
sinx
is of the form 00.
x→ 0 + sinx
Let y = x
⇒ log y = sin x log x.
Note that x approaches 0 from the right so that log x is meaningful log x i.e., log y = cosec x log x −∞ lim log y = lim which is of the type . cosec x ∞ x→ 0 + x→ 0 + Applying l’Hôpital’s rule, 1 x log x lim = lim cosec x x→ 0 + x→ 0 +−cosec x cot x − sin2x 0 = lim x cos x of the type 0 x→ 0 +
33
2 sin x cos x = 0 x x→ 0 + sin x − cos x
= lim ie.,
lim logy = 0 x→ 0 +
By Composite Function Theorem, we have 0 = lim log y = log lim y ⇒ lim y = e0 = 1 x→ 0 + x→ 0 + x→ 0 + Example 5.36 : The current at time t in a coil with resistance R, inductance L and subjected −Rt E to a constant electromotive force E is given by i = R 1− e L . Obtain a suitable formula to be used when R is very small. Solution :
−Rt
E 1− e L i lim = R→0 R R→0 lim
0 (is of the type 0.)
−Rt
=
lim R→0
t E× L e L 1
Et Et = L ⇒ lim i = L is the suitable formula. R→0
EXERCISE 5.6 Evaluate the limit for the following if exists, sin πx tan x − x (1) lim (2) lim 2 −x x→ 2 x→ 0 x − sinx sin −1x x x→ 0
x n − 2n x→ 2 x − 2 1 1 − 2 tan−1 x x2 (6) lim 1 x→∞ x (4) lim
(3) lim
(5)
lim x→∞
2 sin x 1/x logex x
(7)
lim x→ ∞
(9)
lim x2 logex. x→0+
cotx lim cot 2x x→0 1 x−1 (10) lim x x→1 (8)
34
(11)
lim
π − x→ /2
lim
cos x (tanx)
lim (12) x→0+ xx
1
/x
(13) x → 0 (cos x)
5.8 Monotonic Functions : Increasing, Decreasing Functions Differential calculus has varied applications. We have already seen some applications to geometrical, physical and practical problems in sections 5.2, 5.3 and 5.4 In this section, we shall study some applications to the theory of real functions. In sketching the graph of a y y =f(x) Positive function it is very useful to B Negative Gradient know where it raises and where Gradient (slope) Positive (slope) D it falls. The graph shown in Gradient (slope) Fig. 5.16 raises from A to B, C falls from B to C, and raises again from C to D. A The function f is said to be ψ increasing on the interval [a,b], φ x decreasing on [b,c], and c d O a x1 x2 b increasing again on [c,d]. We Fig. 5.16 use this as the defining property of an increasing function. Definition : A function f is called increasing on an interval I if f(x1) ≤ f(x2) whenever x1 < x2 in I. It is called decreasing on I if f(x1) ≥ f(x2) whenever x1 < x2 in I. A function that is completely increasing or completely decreasing on I is called monotonic on I. In the first case the function f preserves the order. i.e., x1 < x2 ⇒ f(x1) ≤ f(x2) and in the later case the function f reverses the order i.e., x1 < x2 ⇒ f(x1) ≥ f(x2). Thanks to the order preserving property, increasing functions are also known as order preserving functions. Similarly, the decreasing functions are also known as order reversing functions. Illustrations : (i) Every constant function is an increasing function. (ii) Every identity function is an increasing function. 35
(iii) The function f(x) = sin x is not an increasing function on R; but π f(x) = sin x is increasing on 0, 2 . (iv) The function f(x) = 4 – 2x is decreasing π (v) The function f(x) = sin x is decreasing in the interval 2, π Note that f is increasing is equivalent to (− f) is decreasing. Do you agree that each constant function is both increasing and decreasing? Caution : It is incorrect to say that if a function is not increasing, then it is decreasing. It may happen that a function is neither increasing nor decreasing. For instance, if we consider the interval [0,π], the function sin x is neither π π increasing nor decreasing. It is increasing on 0, 2 and decreasing on 2, π. There are other functions that are even worse. They are not monotonic on any subinterval also. But most of the functions that we consider are not so bad. Usually, by looking at the graph of the function one can say whether the function is increasing or decreasing or neither. The graph of an increasing function does not fall as we go from left to right while the graph of a decreasing function does not rise as we go from left to right. But if we are not given the graph, how do we decide whether a given function is monotonic or not ? Theorem 1 gives us a criterion to do just that. Theorem 1 : Let I be an open interval. Let f : I → R be differentiable. Then (i) f is increasing if and only if f ′(x) ≥ 0 for all x in I. (ii) f is decreasing if and only if f ′(x) ≤ 0 for all x in I. Proof : (i) Let f be increasing and x ∈ I. Since f is differentiable f ′(x) exists and lim f(x + h) – f(x) is given by f ′(x) = h→0 . If h > 0, then x + h > x and since f is h increasing, f(x + h) ≥ f(x). Hence f(x + h) – f(x) ≥ 0. If h < 0, then x + h < x and f (x + h) ≤ f(x). Hence f(x + h) − f(x) ≤ 0 So either f(x + h) – f(x) and h are both non-negative or they are both non – positive. f(x + h) – f(x) is non-negative for all non-zero values of h and Therefore h f(x + h) – f(x) must also be non-negative. Thus, f ′(x) ≥ 0 lim h h→0
36
Conversely, let f ′(x) ≥ 0, for all x in I. Let x1 < x2 in I. We shall prove that f(x1) ≤ f(x2). f(x2) – f(x1) = f ′(c) , for x1 < c < x2 By the Law of mean, x – x 2 1 f(x2) – f(x1) ≥ 0. Also x2 – x1> 0 ( ∴ x1 < x2) Since, f ′(c) ≥ 0, we have x – x 2 1 Thus f(x2) – f(x1) ≥ 0 or f(x1) ≤ f(x2). Hence f is increasing (ii) can be proved in a similar way. It can also be deduced by applying result (i) to the function (– f). Geometrical interpretation : The above theorem expresses the following geometric fact. If on an interval I = [a,b] a function f(x) increases, then the tangent to the curve y = f(x) at each point on this interval forms an acute angle ϕ with the x-axis or (at certain points) is horizontal (See Fig.5.16), the tangent of this angle is not negative. Therefore f ′(x) = tan ϕ ≥ 0. If the function f(x) decreases on the interval [b,c] then the angle of inclination of the tangents form an obtuse angle (or, at some points, the tangent is horizontal) ; the tangent of this angle is not positive f ′(x) = tan ψ ≤ 0. From the class of increasing functions we can separate out functions which are strictly increasing. The following definition gives the precise meaning of the term strictly increasing function. Definition : f : I → R is said to be strictly increasing if x1 < x2 implies that f(x1) < f(x2). We can similarly say that a function defined on I is strictly decreasing if x1 < x2 implies f(x1) > f(x2) For example, a constant function is not strictly increasing, nor is it strictly decreasing (Fig. 5.17). The greatest integer function f(x)= x too, is increasing (Fig. 5.18), but not strictly increasing, where as the function f(x) = x is strictly increasing (Fig. 5.19). y = x
y
y
1 f(x) =1 0
Fig. 5.17
f(
x)
3 -2 1 1 -2
Fig. 5.18
37
3
x
O
Fig. 5.19
x
Theorem 2 : (i) Let f ′ be positive on I. Then f is strictly increasing on I. (ii) Let f ′ be negative on I. Then f is strictly decreasing on I. The proof of the theorem is easy and is left as an exercise. Corollary : f is strictly monotonic on the interval I, if f ′ is of the same sign through out I. You may have noticed that there is a difference between the statement of Theorem 1 and Theorem 2. “f is increasing if and only if f ′ is non – negative” “ If f ′ > 0, then f is strictly increasing”. Can we have if and only if in Theorem 2 also ? The answer is no as shown in the following example. Illustration : Define f : R→ R by f(x) = x3. Suppose x1 < x2, Then x2 – x1 > 0 and x12 + x22 > 0 This implies x23 – x13 = (x2 – x1) (x22 + x12 + x1 x2) 1 = (x2 – x1) 2 [(x12 + x22)+ (x1 + x2)2] > 0 ⇒ x13 < x23 Thus whenever x1 < x2, f(x1) < f(x2). Hence f(x) = x3 is strictly increasing. But its derivate f ′(x) = 3x2 and f ′(0) = 0. Hence its derivate f ′ is not strictly positive. Note: If a function changes its signs at different points of a region (interval) then the function is not monotonic in that region. So to prove the non- monotonicity of a function, it is enough to prove that f ′ has different signs at different points. Example 5.37 : Prove that the function f (x) = sin x + cos2x is not monotonic on π the interval 0, 4 . Let f(x) = sin x + cos 2x Solution : Then f ′(x) = cos x – 2sin 2x Now f ′(0) = cos 0 – 2 sin 0 = 1 – 0 = 1 > 0 π π π and f ′4 = cos 4 – 2 sin 2 4 1 = –2×1<0 2
38
π π Thus f ′ is of different signs at 0 and 4 Therefore f is not monotonic on 0, 4 Example 5.38 : Find the intervals in which f(x) = 2x3 + x2 −20x is increasing and decreasing. Solution : f ′(x) = 6x2 + 2x – 20 = 2(3x2 + x − 10) = 2 (x + 2) ( 3x −5) Now f ′(x) = 0 ⇒ x = − 2, and x = 5/3. The values − 2 and 5/3 divide the real line (the domain of f(x)) into intervals (−∞, −2), (− 2, 5/3)and (5/3, ∞) . −∞
0 5/3
-2
∞
Fig. 5.20 Interval − ∞ < x < –2 − 2 < x < 5/3
x +2 – +
3x – 5 –
f ′(x) +
Interval of inc / dec Increasing on (– ∞, –2]
−
−
decreasing on [− 2, 5/3]
5/3 < x < ∞
+
+
+
increasing on [5/3, ∞)
Note (i) : If the critical numbers are not included in the intervals, then the intervals of increasing (decreasing) becomes strictly increasing (strictly decreasing) Note : (ii) The intervals of inc / dec can be obtained by taking and checking a sample point in the sub-interval. Example 5.39 : Prove that the function f(x) = x2 − x + 1 is neither increasing nor decreasing in [0,1] Solution :
f (x) = x2 − x + 1
f ′(x) = 2x − 1 1 1 1 f ′(x) ≥ 0 for x ≥ 2 i.e., x ∈ 2 , 1 ∴ f(x) is increasing on 2 , 1 1 1 1 Also f ′(x) ≤ 0 for x ≤ 2 ⇒ x ∈ 0, 2. Also f ′(x) is decreasing on 0, 2 Therefore in the entire interval [0,1] the function f(x) is neither increasing nor decreasing. Example 5.40 : Discuss monotonicity of the function f(x) = sin x, x ∈ [0, 2π]
39
π 3π f (x) = sin x and f ′(x) = cos x = 0 for x = 2 , 2 in [0,2π] Now π 3π f ′(x) ≥0 for 0 ≤ x ≤ 2 and 2 ≤ x ≤ 2π. Therefore f(x) = sin x is increasing on 0, π and 3π, 2π i.e., sin x is increasing on 0, π ∪ 3π , 2π 2 2 2 2
Solution :
π 3π Also, f ′(x) ≤ 0 for 2 ≤ x ≤ 2 . Therefore f(x) = sin x is decreasing on π , 3π 2 2 x −2 Example 5.41 : Determine for which values of x, the function y = x + 1 , x ≠ −1 is strictly increasing or strictly decreasing. Solution : dy (x + 1) 1 − (x −2) 1 3 = > 0 for all x ≠ − 1. 2 dx = (x + 1) (x +1)2
x −2 y = x + 1 , x ≠ −1
∴ y is strictly increasing on R − {−1}. Example 5.42 : Determine for which values of x, the function f(x) = 2x3 − 15x2 + 36x + 1 is increasing and for which it is decreasing. Also determine the points where the tangents to the graph of the function are parallel to the x axis. Solution :
f ′(x) = 6x2 − 30x + 36 = 6(x − 2) (x − 3)
f ′(x) = 0 ⇒ x = 2, 3. Therefore the points 2 and 3 divide the real line into (− ∞, 2), (2, 3) (3, ∞). Interval
x −2
x–3
f ′(x)
Intervals of inc / dec
−∞
–
–
+
increasing on (– ∞, 2]
2
+
−
−
decreasing on [2, 3]
+
+
+
increasing on [3, ∞)
3
The points where the tangent to the graph of the function are parallel to the x − axis are given by f ′(x)= 0, ie., when x = 2, 3 Now f(2) = 29 and f(3) = 28. Therefore the required points are (2, 29) and (3, 28)
40
Example 5.43 : Show that f(x) = tan−1 (sin x + cos x), x > 0 is a strictly increasing function π in the interval 0, 4 . Solution :
f(x) = tan−1(sin x + cos x). 1 cos x − sin x f ′(x) = (cos x − sin x) = 2 + sin 2x > 0 2 1 + (sin x + cos x)
π since cos x−sin x > 0 in the interval 0, 4 and 2 + sin 2x > 0) π ∴ f(x) is strictly increasing function of x in the interval 0 , 4
EXERCISE 5.7 (1) Prove that ex is strictly increasing function on R. (2) Prove that log x is strictly increasing function on (0, ∞) (3) Which of the following functions are increasing or decreasing on the interval given ? 1 1 (i) x2 – 1 on [0,2] (ii) 2x2 + 3x on − 2 , 2 (iii) e−x on [0,1]
(iv) x(x − 1) (x + 1) on [−2, −1]
π (v) x sin x on 0, 4
(4) Prove that the following functions are not monotonic in the intervals given. (i) 2x2 + x − 5 on [−1,0]
(ii) x (x − 1) (x + 1) on [0,2]
(iii) x sin x on [0,π]
π (iv) tan x + cot x on 0, 2
(5) Find the intervals on which f is increasing or decreasing. (i) f(x) = 20 − x − x2
(ii)
f(x) = x3 − 3x + 1
(iii) f(x) = x3 + x + 1
(iv)
(v) f(x) = x + cos x in [0, π]
(vi)
f(x) = x −2sin x, [0, 2π] f(x) = sin4 x + cos4 x in [0, π/2]
41
Inequalities : Example 5.44 : Prove that ex > 1 + x for all x > 0. f ′(x) = ex − 1 > 0 for x > 0
Let f(x) = ex − x − 1 ⇒
Solution :
i.e., f is strictly increasing function. ∴ for x > 0, f(x) > f(0) i.e., (ex − x − 1) > (e0 − 0 − 1) ; ex > x + 1 Example 5.45 : Prove that the inequality (1 + x)n > 1+nx is true whenever x > 0 and n > 1. Solution : Consider the difference f(x) = (1 + x)n − (1 + nx) Then f ′(x) =
n(1 + x)n−1 − n = n[(1 + x)n−1 − 1]
Since x > 0 and n − 1 > 0, we have (1 + x)n−1 > 1, so f ′(x) > 0. Therefore f is strictly increasing on [0, ∞). For x > 0 ⇒ f(x) > f(0) i.e., (1 + x)n − (1 + nx) > (1 + 0) − (1 + 0) i.e., (1 + x)n − (1 + nx) > 0
i.e., (1 + x)n > (1 + nx)
π Example 5.46 : Prove that sin x < x < tan x, x∈0, 2
y
π f ′(x) = 1 − cos x > 0 for 0 < x < 2
x =π/2 y= x
Let f(x) = x − sin x
y =tan x
Solution :
∴ f is strictly increasing. y =sin x
For x > 0, f(x) > f(0) ⇒ x − sin x > 0 ⇒ x > sin x
… (1)
π/2
0
Let g(x) = tan x − x π g′(x) = sec2x − 1 = tan2x > 0 in 0, 2
Fig. 5.21
∴ g is strictly increasing For x > 0, f(x) > f(0) ⇒ tan x − x > 0 ⇒ tan x > x … (2) From (1) and (2)
sin x < x < tan x
42
x
EXERCISE 5.8 (1) Prove the following inequalities : x3 x2 (ii) sin x > x − 6 , x > 0 (i) cos x > 1 − 2 , x > 0 (iii) tan−1 x < x for all x > 0
(iv) log (1 + x) < x for all x > 0.
5.9 Maximum and Minimum values and their applications :
(slope)
R
P
Positiv e
Gradie nt
ve Negati nt Gradie ) e p lo (s
Positiv e Grad ient
y
Some of the most important applications of differential calculus are optimization problems, in which we are required to find the optimal (best) way of doing something. In many cases these problems can be reduced to finding the maximum or minimum values of a function. Many practical problems require us to minimize a cost or maximize an area or somehow find the best possible outcome of a situation.
(slope )
“For since the fabric of the Universe is most perfect and the work of a most wise creator, nothing at all takes place in the Universe in which some rule of maximum or minimum does not appear” Leonard Euler
Q
x
O
Fig. 5.22
Let us first explain exactly what we mean by maximum and minimum values. In fig 5.22 the gradient (rate of change) of the curve changes from positive between O and P to negative between P and Q and positive again between Q and R. At point P, the gradient is zero and as x increases, the gradient (slope) of the curve changes from positive just before P to negative just after. Such a point is called a maximum point and appears as the ‘crest of a wave’.
y Maximum Point
O
Maximum Point Point of Inflexion
x Minimum Point
Fig. 5.23
43
At point Q, the gradient is also zero and as x increases the gradient of the curve changes from negative just before Q to positive just after. Such a point is called a minimum point and appears as ‘the bottom of a valley’. Points such as P and Q are given the general name, turning points.
y
f (d) f (a) aO b
c
d
e
x
Fig. 5.24
It is possible to have a turning point, the gradient on either side of which is the same. Such a point is given the special name of a point of inflection as shown in Fig 5.23.
Definition : A function f has an absolute maximum at c if f(c) ≥ f(x) for all x in D, where D is the domain of f. The number f(c) is called maximum value of f on D. Similarly f has an absolute minimum at c if f(c) ≤ f(x) for all x in D and the number f(c) is called the minimum value of f on D. The maximum and minimum values of f are called extreme values of f.: Fig.5.24 shows the graph of a function f with absolute maximum at d and absolute minimum at a. Note that (d, f(d)) is the highest point on the graph and (a, f(a)) is the lowest point. In Fig. 5.24 if we consider only values of x near b, for instance, if we restrict our attention to the interval (a,c) then f(b) is the largest of those values of f(x) and is called a local maximum value of f. Likewise f(c) is called a local minimum value of f because f(c) ≤ f(x) for x near c, in the interval (b,d). The function f also has a local minimum at e. In general we have the following definition. Definition : A function f has a local maximum (or relative maximum) at c if there is an open interval I containing c such that f(c) ≥ f(x) for all x in I. Similarly, f has a local minimum at c if there is an open interval I containing c such that f(c) ≤ f(x) for all x in I. Illustrations : (1) The function f(x)=cos x takes on its (local and absolute) maximum value of 1 infinitely many times since cos 2nπ = 1 for any integer and −1 ≤ cos x ≤ 1 for all x. Like wise cos (2n + 1)π = −1 is its (local and absolute) minimum value, n is any integer.
44
(2) If f(x) = x2, then f(x) ≥ f(0) because x2 ≥ 0 for all x. Therefore f(0) = 0 is the absolute (and local) minimum value of f. This corresponds to the fact that the origin is the lowest on the parabola y = x2 See Fig.5.25 However, there is no highest point on the parabola and so this function has no maximum value.
y
y = x2
x O Min. value =0; No Max.
Fig. 5.25 y 8 6 y =x3
(3) If f(x) = x3 then from the graph
4
of f(x) shown in Fig 5.26, we see
2 -4
that this function has neither an
-2
0
absolute maximum value nor an
-2
absolute minimum value. In fact it
-4
has no local extreme values either.
-6
2
4
x
No Minimum No Maximum
-8
Fig. 5.26 y
(4) Consider the function 4
3
2
f(x) = 3x − 16x + 18x ; −1 ≤ x ≤ 4. The graph is shown in Fig. 5.27. We can see that f(1) = 5 is a local maximum, whereas the absolute maximum is f(−1)=37. Also f(0) = 0 is a local minimum and f(3)= −27 is both local and absolute minimum. We have seen that some functions have extreme values, while others do not. The following theorem gives conditions under which a function is guaranteed to possess extreme values.
45
(-1,37)
40
(4,32)
30 20 10 (1,5) -1 0 -10
1 2 (2,-8)
3
-20 -30
(3,-27)
Fig. 5.27
x
The Extreme value theorem : If f is continuous on a closed interval [a,b] then f attains an absolute maximum value f(c) and an absolute minimum value f(d) at some number c and d in [a,b] The next two examples show that a function need not possess extreme values if either of the hypotheses (continuity or closed interval) is omitted from the extreme value theorem. y
(5) Consider the function x2
, 0≤x<1 0 , 1 ≤ x ≤ 2 The function is defined on the closed interval [0,2] but has no maximum value. Notice that the range of f is the interval [0,1). The function takes on value close to 1 but never attains the value 1.
1
f(x) =
0
1
x
Fig. 5.28
This is because the hypotheses of f to be continuous fails. Note that x = 1 is a point of discontinuity, for, Lim x→ 1 −
Lim Lim f(x) = x→ 1 − (x2) = 1 ; x→ 1 + f(x) = 0
(6) The function f(x) =x2, 0 < x < 2 is continuous on the interval (0,2) but has neither a maximum nor a minimum value. The range of f is the interval (0,4). The values 0 and 4 are never taken on by f. This is because the interval (0,2) is not closed.
y 4 f (x) =x2, 0
0
2
x
Fig. 5.29 If we alter the function by including either end point of the interval (0,2) then we get one of the situations shown in Fig. 5.30, Fig. 5.31, Fig. 5.32 In particular the function f(x) = x2, 0 ≤ x ≤ 2 is continuous on the closed interval [0,2]. So the extreme value theorem says that the function has an absolute maximum and an absolute minimum.
46
4
4
4
f2 (x) =x2, 0≤x<2
f1 (x) =x2, 0
2
Fig. 5.30
x
f3 (x) =x2, 0≤x≤2
No Maximum Minimum f2 (0) =0
Maximum f1 (2) =4 No Minimum
O
y
y
y
0
x
2
Maximum f3 (2) =4 Minimum f3 (0) =0
0
2
x
Fig. 5.31
Fig. 5.32 Inspite of the above examples we point out that there are functions which are neither continuous nor differentiable but still attains minimum and maximum values. For instance, consider 1 , x is irrational f(x) = 0 , x is rational (This function is known as characteristic function on the set of irrational numbers) This function is nowhere differentiable and everywhere discontinuous. But the maximum value is 1 and the minimum value is 0. The extreme value theorem says that a continuous function on a closed interval has a maximum value and minimum value, but it does not tell us how to find their extreme values. y Fig. 5.33 shows the graph of a (c, f (c)) function f with a local maximum at c and a local minimum at d. It appears that at the maximum and (d, f (d)) minimum points the tangent line is horizontal and therefore has slope zero. We know that the derivative x is the slope of the tangent line, so it c O d Fig. 5.33 appear that f ′(c) = 0 and f ′(d) = 0. The following theorem shows that this is always true for differentiable functions. Fermat’s Theorem : If f has a local extremum (maximum or minimum) at c and if f ′(c) exists then f ′(c) = 0. The following examples caution us that we cannot locate extreme values simply by setting f ′(x) = 0 and solving for x.
47
y
(7) The function f(x) = | x | has its (local and absolute) minimum value at 0, but that value cannot be found by setting f ′(x) = 0 because f ′(x) does not exist.
y =- x
y =x
O
x
y =|x|
Fig. 5.34 (8) The function f(x) = 3x − 1, 0 ≤ x ≤ 1 has its maximum value when x = 1 but f ′(1) = 3 ≠ 0. This does not contradict Fermat’s Theorem. Since f(1) = 2 is not a local maximum. Note that the number 1 is not contained in an open interval in the domain of f.
y (1,2)
2
y =3x – 1 0≤x≤1
1 -1
0
1
x
2
-1 (0,-1)
Fig. 5.35
Remark : The above examples demonstrate that even when f ′(c) = 0 there need not be a maximum or minimum at c. Further more, there may be an extreme value even when f ′(c) ≠ 0 or when f ′(c) does not exist. y
(9) If f(x) = x3. Then f ′(x) = 3x2, so f ′(0) = 0. But f has no maximum or minimum at 0 as you can see from its graph. (observe that x3 > 0 for x > 0 and x3 < 0 for x < 0). The fact that f ′(0) = 0 simply means that the curve y = x3 has a horizontal tangent at (0,0). Instead of having a maximum or minimum at (0, 0) the curve crosses its horizontal tangent there.
8 6
y =x3
4 2 -2 0 2 -2
4
x
-4 -6 -8
Fig. 5.36 Fermats’ theorem does suggest that we should atleast start looking for extreme values of f at the numbers c where f ′(c) = 0 or f ′(c) does not exist. Definition : A critical number of a function f is a number c in the domain of f such that either f ′(c) = 0 or f ′(c) does not exist.
48
Stationary points are critical numbers c in the domain of f, for which f ′(c)= 0. 3
/ Example 5.47 : Find the critical numbers of x 5 (4 − x)
Solution :
3 8 / / f(x) = 4 x 5 − x 5
f ′(x) = =
12 −2/5 8 3/5 x − 5 5 x 4 −2/5 (3 − 2x) 5 x
3 Therefore f ′(x) = 0 if 3 − 2x = 0 i.e., if x = 2 . f ′(x) does not exist when 3 x = 0. Thus the critical numbers are 0 and 2 . Note that if f has a local extremum at c, then c is a critical number of f, but not vice versa. To find the absolute maximum and absolute minimum values of a continuous function f on a closed interval [a,b] : (1) Find the values of f at the critical numbers, of f in (a,b). (2) Find the values of f(a) and f(b) (3) The largest of the values from steps 1 and 2 is the absolute maximum value, the smallest of these values is the absolute minimum value. Example 5.48 : Find the absolute maximum and minimum values of the 1
function. f(x) = x3 − 3x2 + 1 , − 2 ≤ x ≤ 4 1
[
]
Solution : Note that f is continuous on ; − 2 , 4 f(x) = x3 − 3x2 + 1
f ′(x) = 3x2 − 6x = 3x (x − 2) -1/2 0
1
2
3
4
Fig. 5.37 Since f ′(x) exists for all x, the only critical numbers of f are x = 0, x = 2. 1 Both of these critical numbers lie in the interval − 2 , 4 . Value of f at these critical numbers are f(0) = 1 and f(2) = −3.
49
The values of f at the end points of the interval are 1 1 1 3 1 2 f −2 = −2 −3 −2 +1=8
( ) ( )
( )
and f(4) = 43 − 3 × 42 + 1 = 17 Comparing these four numbers, we see that the absolute maximum value is f(4) = 17 and the absolute minimum value is f(2) = − 3. Note that in this example the absolute maximum occurs at an end point, whereas the absolute minimum occurs at a critical number. Example 5.48(a): Find the absolute maximum and absolute minimum values of f(x) =x − 2sin x, 0 ≤ x ≤ 2π. Solution : f(x) =x − 2 sin x, is continuous in [0, 2π] f ′(x) = 1 − 2 cos x 1 5π π f ′(x) = 0 ⇒ cos x = 2 ⇒ x = 3 or 3 The value of f at these critical points are π π π π f 3 = 3 − 2 sin 3 = 3 − 3 5π 5π 5π f 3 = 3 − 2 sin 3 5π = 3 + 3 ≈ 6.968039 The values of f at the end points are f(0) = 0 and f(2π) = 2 π ≈ 6.28 π π Comparing these four numbers, the absolute minimum is f 3 = 3 − 3 and 5π 5π the absolute maximum is f 3 = 3 + 3 . In this example both absolute minimum and absolute maximum occurs at the critical numbers. Let us now see how the second derivatives of functions help determining the turning nature (of graphs of functions) and in optimization problems. The second derivative test : Suppose f is continuous on an open interval that contains c. (a) If f ′(c) = 0 and f ′′(c) > 0, then f has a local minimum at c. (b) If f ′(c) = 0 and f ′′(c) < 0, then f has a local maximum at c.
50
Example 5.49 : Discuss the curve y = x4 − 4x3 with respect to local extrema. Solution :
f(x) = x4 − 4x3 f ′(x) = 4x3 − 12x2 , f ′′(x) = 12x2 − 24x
To find the critical numbers we set f ′(x) = 0 and obtain x = 0 and x = 3. To use the second derivative test we evaluate the sign of f ′′ at these critical numbers. f ′′(0) = 0, f ′′(3) = 36 > 0. Since f ′(3) = 0 and f ′′(3) > 0, f(3) = − 27 is a local minimum value and the point (3, −27) is a minimum point. Since f ′′(0) = 0 the second derivative test gives no information about the critical number 0. But since f ′(x) < 0 for x < 0 and also for 0 < x < 3, the first derivative test tells us that f does not have a local extremum at 0. We summarise the above discussion as follows : Procedure for finding and distinguishing stationary points. dy (i) Given y = f(x) determine dx (i.e., f ′(x)) dy (ii) Let dx = 0 and solve for the critical numbers x. (iii) Substitute the values of x into the original function y = f(x) to find the corresponding y-coordinate values. This establish the co-ordinates of the stationary points. To determine the nature of the stationary points, d2y and substitute into it the values of x found in (ii). dx2 If the result is : (a) positive − the point is a minimum one (b) negative − the point is a maximum one (c) zero − the point cannot be an extremum (minimum or maximum) OR (iv) Find
(v) Determine the sign of the gradient (slope f ′(x) of the curve just before and just after the stationary points. If the sign change for the gradient of the curve is (a) positive to negative − this point is a maximum one (b) negative to positive − this point is a minimum one Example 5.50 : Locate the extreme point on the curve y = 3x2 − 6x and determine its nature by examining the sign of the gradient on either side.
51
Solution : Following the above procedure dy (i) Since y = 3x2 − 6x, dx = 6x − 6 dy (ii) At a stationary point, dx = 0, hence x = 1 (iii) When x = 1, y = 3(1)2 − 6(1) = − 3. Hence the coordinates of the stationary point is (1, − 3). dy If x is slightly less than 1, say 0.9, then dx = 6(0.9) − 6 = − 0.6 < 0. dy If x is slightly greater than 1, say 1.1 then dx = 6(1.1) − 6 = 0.6 > 0. Since the gradient (slope of the curve) changes its sign from negative to positive (1, − 3) is a minimum point. Example 5.51 : Find the local minimum and maximum values of f(x) = x4 − 3x3 + 3x2 − x f(x) = x4 − 3x3 + 3x2 − x
Solution :
f ′(x) = 4x3 − 9x2 + 6x − 1 At a turning point, f ′(x) = 0 gives 4x3 − 9x2 + 6x − 1 = 0 1 (x − 1)2 (4x − 1) = 0 ⇒ x = 1, 1, 4 1 1 − 27 When x = 1, f(1) = 0 and when x = 4, f 4 = 256
1 − 27 Hence the coordinates of the stationary points are (1, 0) and 4 , 256
f ′′(x) = 12x − 18x + 6 = 6(2x − 3x + 1) = 6(x − 1) (2x − 1) 2
2
When x = 1, f ′′(1) = 0. Thus the second derivative test gives no information about the extremum nature of f at x = 1. 1 1 9 1 − 27 When x = 4 , f ′′ 4 = 4 > 0, hence 4 , 256 is a minimum point.
Caution : No function will attain local maximum / minimum at the end points of its domain.
52
EXERCISE 5.9 (1) Find the critical numbers and stationary points of each of the following functions. (i) f(x) = 2x − 3x2
(ii) f(x)
= x3 − 3x + 1
(iii) f(x) = x4/5 (x − 4)2
(iv) f(x)
=
x+1 x +x+1 2
(v) f(θ) = sin2 2θ in [0, π] (vi) f(θ) = θ + sin θ in [0, 2π] (2) Find the absolute maximum and absolute minimum values of f on the given interval : (i)
f(x) = x2 − 2x + 2,
[0,3]
(ii)
f(x) = 1 − 2x − x2,
[−4,1]
(iii) f(x) = x3 − 12x + 1, (iv) f(x) = (v)
9 − x2 ,
x f(x) = x + 1,
[−1,2] [1,2]
(vi) f(x) = sin x + cos x, (vii) f(x) =
[−3,5]
0, π 3
x − 2 cos x,
[−π, π]
(3) Find the local maximum and minimum values of the following : (i) x3 − x
(ii) 2x3 + 5x2 − 4x
(iii) x4 − 6x2
(iv) (x2 − 1)3
(v) sin2 θ ,
(vi) t + cos t
[0, π]
5.10 Practical problems involving maximum and minimum values : The methods we have learnt in this section for finding extreme values have practical applications in many areas of life. A business person wants to minimise costs and maximise profits. We also solve such problems as maximising areas, volumes and profits and minimising distances, times and costs. In solving such practical problems, the greatest challenge is often to convert the word problem into maximum – minimum problem by setting up the function that is to be maximised or minimised.
53
As a problem solving technique we suggest the following principles. (1) Understand the problem : The first step is to read the problem carefully until it is clearly understood. Ask yourself what is the unknown? What are the given quantities? What are the given conditions? (2) Draw diagram : In most problems it is useful to draw a diagram and identify the given and required quantities on the diagram. (3) Introduce notation : Assign a symbol to the quantity to be maximised or minimised, say Q. Also select symbols (a,b,c …,x, y, z) for the other unknown quantities and lable the diagram with these symbols. (4) Express Q in terms of some other symbols from step 3. (5) If Q has been expressed as a function of more than one variable in step 4, use the given information to find relationship (in the form of equation) among these variables. Then use these equations to eliminate all but one of these variables in the expression for Q.Thus Q will be given as a function of one variable x, say, Q = f(x). Write the domain of this function. (6) Use the methods discussed to find the absolute maximum or minimum value of f. Remarks : (1) If the domain is a closed interval then we apply the absolute max/min property to maximize / minimize the given function (see 5.52, 5.58). (2) If the domain is an open interval then we apply either first derivative test (5.53) or second test for finding local max / min. Instead of first derivative one can also apply second derivative test if the second test exist. Similarly instead of second derivative test one can also apply first derivative test. (3) All these cases ultimately lead us to the absolute max / min only. Example 5.52 : A farmer has 2400 feet of fencing and want to fence of a rectangular field that borders a straight river. He needs no fence along the river. What are the dimensions of the field that has the largest area ? y Solution : We wish to maximize the area A x x of the rectangle. Let x and y be the width and length of the rectangle (in feet). Then we express A in terms of x and y as A = xy Fig. 5.38
54
We want to express A as a function of just one variable, so we eliminate y by expressing it in terms of x. To do this we use the given information that the total length of the fencing is 2400 ft. Therefore 2x + y = 2400 Hence
y = 2400 − 2x and the area is A= x (2400 – 2x) = 2400 x − 2x2
Note that x ≥ 0 and x ≤ 1200 (otherwise A < 0). So the function that we wish to maximize is A (x) = 2400 x − 2x2, 0 ≤ x ≤ 1200. A′(x) = 2400 − 4x, so to find the critical numbers we solve the equation 2400 − 4x = 0 which gives x = 600. The maximum of A must occur either at this critical number or at an end point of the interval. Since A(0) = 0, A(600) = 7,20,000 and A(1200) = 0, thus the maximum value is A (600) = 720,000. When x = 600, y = 2400 − 1200 = 1200 Hence the rectangular field should be 600 ft wide and 1200 ft long. Note : This problem also be done by using second derivative test (local). In this case x > 0 and y > 0. Example 5.53 : Find a point on the parabola y2 = 2x that is closest to the point (1,4) y Solution : Let (x,y) be the point on the parabola y2 = 2x. The distance between the points (1,4) and (x,y) is d =
4
(1,4) y2 = 2x (x,y)
(x −1)2 + (y − 4)2 . 0
y2 (x,y) lies on y = 2x ⇒ x = 2 , y2 so d2= f(y) = ( 2 − 1)2 + (y − 4)2 2
1
2
x
Fig. 5.39
(Note that the minimum of d occurs at the same point as the minimum of d2)
y2 Now f ′(y) = 2 2 − 1 (y) + 2 (y − 4) = y3 − 8 = 0 at a critical point. y3 − 8 = 0 ⇒ y = 2 (since y2 + 2y + 4 = 0 is not possible)
55
Observe that f ′(y) < 0 when y < 2 and f ′(y) > 0 when y > 2, so by the first derivate test, for absolute extrema, the absolute minimum occurs when y2 y = 2. The corresponding value of x is x = 2 = 2. Thus the point on y2 = 2x closest to (1,4) is (2,2). Note : This problem also be done by using second derivative test Example 5.54 : Find the area of the largest rectangle that can be inscribed in a semi circle of radius r. Solution : x2 +y2 =r2 Let θ be the angle made by OP P(r cos θ, r sin θ) with the positive direction of x–axis. r Then the area of the rectangle A is θ x A(θ) = (2 r cosθ) (r sinθ) Fig. 5.40 = r2 2 sin θ cos θ = r2 sin 2θ Now A(θ) is maximum when sin 2θ is maximum. The maximum value of π π π sin 2θ = 1 ⇒ 2θ = 2 or θ = 4 . (Note that A′ (θ) = 0 when θ = 4 ) π π Therefore the critical number is 4. The area A4 = r2. Note : The dimensions of the largest rectangle that can be inscribed in a r semicircle are 2r , 2 π π A ′(θ) = 2r2 cos 2θ = 0 ⇒ 2θ = 2 ; θ = 4 Aliter : π π A ′′(θ) = −4r2 sin 2 θ < 0, for θ = 4 ⇒ θ = 4 gives the π maximum point and the maximum point is 4 , r2 From the above problem, we understand that the method of calculus gives the solution faster than the algebraic method. Example 5.55 : The top and bottom margins of a poster are each 6 cms and the side margins are each 4 cms. If the area of the printed material on the poster is fixed at 384 cms2, find the dimension of the poster with the smallest area. Solution : Let x and y be the length and breadth of printed area, then the area xy = 384
56
Dimensions of the poster area are (x + 8) and (y + 12) respectively. Poster area A = (x + 8) (y + 12) = xy +12x + 8y + 96 = 12x + 8y + 480 384 = 12x + 8 x + 480 1 A′ = 12 − 8 × 384 × 2 x
x +8 6 cms 4
x y
4
y +12
6 cms
Fig. 5.41
1 A″ = 16 × 384 × 3 x A′ = 0 ⇒ x = ± 16 But x > 0 ∴ x = 16 when x = 16, A′′ > 0 ∴ when x = 16, the area is minimum ∴ y = 24 ∴ x + 8 = 24, y + 12 = 36 Hence the dimensions are 24cm and 36 cm. Example 5.56 : Show that the volume of the largest right circular cone that can 8 be inscribed in a sphere of radius a is 27 (volume of the sphere). Solution : Given that a is the radius of the sphere and let x be the base radius of a the cone. If h is the height of the cone, c then its volume is α y 1 2 V = 3 πx h O x 1 …(1) = 3 π x2 (a + y) Fig. 5.42 where OC = y so that height h = a + y. From the diagram x2 + y2 = a2
(2)
Using (2) in (1) we have 1 V = 3 π (a2 − y2) (a + y)
57
For the volume to be maximum : 1 V ′=0 ⇒ 3 π [a2 − 2ay − 3y2] = 0 ⇒ 3y = +a or y = −a a ⇒ y = 3 and y = − a is not possible a 2 Now V″ = − π 3 (a + 3y) < 0 at y = 3 a ∴ the volume is maximum when y = 3 and the maximum volume is 8a2 1 8 4 3 8 1 π × 9 (a + 3 a) = 27 (3 πa ) = 27 (volume of the sphere) 3 Example 5.57 : A closed (cuboid) box with a square base is to have a volume of 2000 c.c. The material for the top and bottom of the box is to cost Rs. 3 per square cm. and the material for the sides is to cost Rs. 1.50 per square cm. If the cost of the materials is to be the least, find the dimensions of the box. Solution : Let x, y respectively denote the length of the side of the square base and depth of the box. Let C be the cost of the material Area of the bottom = x2 Area of the top = x2 Combined area of the top and bottom = 2x2 Area of the four sides = 4xy Cost of the material for the top and bottom = 3(2x2) Cost of the material for the sides = (1.5) (4xy) = 6xy Total cost C = 6x2 + 6xy
…(1) 2
Volume of the box V = (area) (depth) = x y=2000 …(2) 12000 Eliminating y from (1) & (2) we get C(x) = 6x2 + x …(3) where x > 0, ie., x ∈ (0,+ ∞) and C(x) is continuous on (0, + ∞). 12000 C ′ (x) = 12x − x2 C ′ (x) = 0 ⇒ 12x3 − 12000 = 0 ⇒ 12(x3 − 103) = 0 ⇒ x = 10 or x2 + 10x + 100 = 0
58
x2 + 10x + 100 = 0 is not possible ∴ The critical numbers is x = 10. 24000 24000 Now C ″(x) = 12 + ; C ″(10) = 12 + 1000 = 36 > 0 3 x ∴
C is minimum at (10,C(10)) = (10, 1800) ∴ the base length is 10cm and 2000 depth is y = 100 = 20 cm. Example 5.58 :
8-x
A man is at a point P on a bank of a straight river, 3 km wide, and wants to reach point Q, 8 km downstream on the opposite bank, as quickly as possible. He could row his boat directly across the river to point R and then run to Q, or he could row directly to Q, or he could row to some point S between Q and R and then run to Q. If he can row at 6 km/h and run at 8 km/h where should he land to reach Q as soon as possible ? Solution : 3km P R Let x be the distance from R to S. Then the √( x2 +9 x running distance is 8 − x and the distance ) distance PS = x2 + 9 . We know that time = rate . S Then the rowing time Rt =
x2 + 9 6
and the running time rt =
(8 −x) 8
Q
Fig. 5.43 Therefore the total time T = Rt + rt =
(8 −x) x2 + 9 + 8 , 0 ≤ x ≤ 8. 6
Notice that if x = 0, he rows to R and if x = 8 he rows directly to Q. x 1 T ′(x) = 0 ⇒ T ′(x) = − 8 = 0 for critical points. 2 6 x +9 4x = 3
x2 + 9
16x2 = 9 (x2 + 9) 7x2 = 81
59
9 9 since x = − is not admissible. 7 7 9 The only critical number is x = . We calculate T at the end point of the 7 9 domain 0 and 8 and at x = . 7 ⇒ x=
T(0) = 1.5, T
9 7 73 = 1 + 8 ≈ 1.33, and T(8) = 6 ≈ 1.42 7
Since the smallest of these values of T occurs when x = should land the boat at a point
9 , the man 7
9 km (≈ 3.4 km) down stream from his starting 7
point. (1) (2) (3) (4) (5) (6)
EXERCISE 5.10 Find two numbers whose sum is 100 and whose product is a maximum. Find two positive numbers whose product is 100 and whose sum is minimum. Show that of all the rectangles with a given area the one with smallest perimeter is a square. Show that of all the rectangles with a given perimeter the one with the greatest area is a square. Find the dimensions of the rectangle of largest area that can be inscribed in a circle of radius r. 5 Resistance to motion, F, of a moving vehicle is given by, F = x + 100x. Determine the minimum value of resistance.
5.11 Concavity (convexity) and points of inflection : Figure 5.44 (a), (b) shows the graphs of two increasing functions on [a, b]. Both graphs join point A to point B but they look different because they bend in different directions. How can we distinguish between these two types of behaviour? In fig. 5.44 (c), (d) tangents to these curves have been drawn at several points. In (a) the curve lies above the tangents and f is called concave upward (convex downward) on [a, b]. In (b) the curve lies below the tangents and g is called concave downward (convex upward) on [a, b]
60
y
y
B
B
f
g
A 0
A
a
b
x
Fig. 5.44 (a)
y
b
x
Fig. 5.44 (b)
y
B
B
A
A
x
0
a
0
x
0
Fig. 5.44(c)
Fig. 5.44 (d)
Definition : If the graph of f lies above all of its tangents on an interval I, then it is called concave upward (convex downward) on I. If the graph of f lies below all of its tangents on I, it is called concave downward (convex upward) on I. Let us see how the second derivative helps to determine the intervals of concavity (convexity). Looking at Fig.5.44(c), you can see that, going from left to right, the slope of the tangent increases. This means that the derivative f ′(x) is an increasing function and therefore its derivative f ′′(x) is positive. Likewise in Fig.5.44 (d) the slope of the tangent decreases from left to right, so f ′(x) decreases and therefore f ′′(x) is negative. This reasoning can be reversed and suggests that the following theorem is true. The test for concavity (convexity) : Suppose f is twice differentiable on an interval I. (i) If f ′′(x) > 0 for all x in I, then the graph of f is concave upward (convex downward) on I. (ii) If f′′(x) < 0 for all x in I, then the graph of f is concave downward (convex upward) on I.
61
Definition : A point P on a curve is called a point of inflection if the curve changes from concave upward (convex downward) to concave downward (convex upward) or from concave downward (convex upward) to concave upward (convex downward) at P. That is the point that separates the convex part of a continuous curve from the concave part is called the point of inflection of the curve. It is obvious that at the point of inflection the tangent line, if it exists, cuts the curve, because on one side the curve lies under the tangent and on the other side, above it. The following theorem says under what situation a critical point of f′ becomes a point of inflection. Theorem : Let a curve be defined by an equation y = f(x). If f ′′(x0) = 0 or f ′′(x0) does not exist and if the derivative f ′′(x) changes sign when passing through x = x0, then the point of the curve with abcissa x = x0 is the point of inflection. Equivalently the point (x0, f(x0)) is a point of inflection of the graph of f if there exists a neighbourhood (a, b) of x0 such that f ′′(x) > 0 for every x in (a, x0) and f ′′(x) < 0 for every x in (x0, b) or vice versa. y of x0, f ′′(a) andyf ′′(b) differ in sign.y That yis in the neighbourhood A
A
O
x
O
B
B x
O
x
O
x
Fig. 5.45 Remark : We caution the reader that points of inflections need not be critical points and critical points need not be points of inflections. However x = x0 is a critical point such that f ′(x) does not change its sign as f(x) passes through x0, then x0 is a point of inflection and for points of inflections x0, it is necessary that f ′′(x0) = 0. If f ′′(x) does not change its sign even if f ′′(x0) = 0 then x0 cannot be a point of inflection. Thus the conjoint of the above discussion is that for points of inflections x0, f ′′(x0) = 0 and in the immediate neighbourhood (a, b) of x0, f
′′(a) and f ′′(b) must differ in sign.
62
If x = x0 is a root of odd order − simple, triple, etc. of the function f ′(x) = 0, then x = x0 yields a maximum or minimum. If x = x0 is a root of even order, x = x0 yield a point of inflection with a horizontal tangent. These concepts are made clear in the following illustrative example y = x3. y′ = 3x2 and y ′′ = 6x. Here y′(0) = 0 and y′′(0) = 0 and x = 0 happens to be a critical point of both y and y′. Clearly y′ (x) > 0 for x < 0 and x > 0. Thus y′ does not change its sign as f(x) passes through x = 0. That is y′ (− 0.1) > 0 and y′(0.1) > 0 i.e., in the neighbourhood (− 0.1, 0.1) of 0, y′ does not change its sign. Thus the first derivative test confirms that (0, 0) is a point of inflection. y Again y′′(0) = 0, y′′(− 0.1) < 0 y =x3 and y′′(0.1) > 0. Here y′′ changes its Concave sign as y(x) passes through x = 0. In upward Convex this case the second derivative downward (concavity) test also confirms that (0, 0 0) is a point of inflection. Note that (0, x Convex 3 0) separates the convex part of y = x upward from the concave part. Note also that y′(x) = 3x2 and x = 0 is a double root of y′(x) = 0. The root order test also confirms that (0, 0) is a point of inflection with x-axis as the horizontal tangent at (0, 0) Example 5.59 : Determine the domain of concavity y = 2 − x2.
Concave downward
Fig. 5.46 (convexity)
of
the
y = 2 − x2 y′ = − 2x and y′′ = − 2 < 0 for x ∈ R Here the curve is everywhere concave downwards (convex upwards). Example 5.60 : Determine the domain of convexity of the function y = ex. Solution : y = ex ; y′′ = ex > 0 for x Hence the curve is everywhere convex downward. Solution :
63
curve
Example 5.61 : Test the curve y = x4 for points of inflection. Solution :
y 16
y = x4
y =x4
12
y′′ = 12x2 = 0 for x = 0
and y′′ > 0 for x < 0 and x > 0 8 Therefore the curve is concave 4 upward and y′′ does not change sign 0 as y(x) passes through x = 0. Thus the -2 -1 1 2 curve does not admit any point of Fig. 5.47 inflection. Note : The curve is concave upward in (− ∞, 0) and (0, ∞).
x
Example 5.62 : Determine where the curve y = x3 − 3x + 1 is cancave upward, and where it is concave downward. Also find the inflection points. Solution : f(x) = x3 − 3x + 1
-∞
x =0
f ′(x) = 3x − 3 = 3(x − 1) 2
∞
2
Fig. 5.48
Now f ′′(x) = 6x Thus f ′′(x) > 0 when x > 0 and f ′′(x) < 0 when x < 0. The test for concavity then tells us that the curve is concave downward on (− ∞, 0) and concave upward on (0, ∞). Since the curve changes from concave downward to concave upward when x = 0, the point (0, f(0)) i.e., (0, 1) is a point of inflection. Note that f ′′(0) = 0 Example 5.63 : Discuss the curve y = x4 − 4x3 with respect to concavity and points of inflection. Solution : f(x) = x4 − 4x3 ⇒ f ′(x) = 4x3 − 12x2 f ′′(x) = 12x2 − 24x = 12x (x − 2) Since f ′′(x) = 0 when x = 0 or 2, we divide the real line into three intervals.
-∞
0
Fig. 5.49 (− ∞, 0), (0, 2), (2, ∞) and complete the following chart.
64
2
∞
f ′′(x) = 12x (x − 2) + − +
Inerval (− ∞, 0) (0, 2) (2, ∞)
concavity upward downward upward
The point (0, f(0)) i.e., (0, 0) is an inflection point since the curve changes from concave upward to concave downward there. Also (2, f(2)) i.e., (2, − 16) is an inflection point since the curve changes from concave downward to concave upward there. Note : The intervals of concavity can be obtained by taking and checking a sample point in the sub-interval. Example 5.64 : Find the points of inflection and determine the intervals of convexity and concavity of the Gaussion curve y = e−x 2
2
Solution : y′ = − 2xe−x ; y′′ = 2e−x (2x2 − 1) (The first and second derivatives exist everywhere). Find the values of x for which y′′ = 0 2
2
2e−x (2x2 − 1) = 0 1 1 x=− , or x = 2 2
-∞
-1/√2
0
1/√2
∞
Fig. 5.50 1 1 when x < − we have y′′ > 0 and when x > − we have y′′ < 0 2 2 The second derivative changes sign from positive to negative when passing 1 1 through the point x = − . Hence, for x = − , there is a point of inflection 2 2
on the curve; its co-ordinates are −
1 − , e 2 2 1
1 1 we have y′′ < 0 and when x > we have y′′ > 0 . Thus 2 2 1 there is also a point of inflection on the curve for x = ; its co-ordinates are 2 When x <
1 1 − , e 2. (Incidentally, the existence of the second point of inflection follows 2
directly from the symmetry of the curve about the y-axis). Also from the signs of the second derivatives, it follows that
65
for − ∞ < x < − for − for
1 the curve is concave upward ; 2
1 1
1 < x < ∞ the curve is concave upward. 2
Example 5.65 : Determine y = x3 − 3x + 2
the
points
of
inflection
if
any,
of
the
function
y = x3 − 3x + 2 dy 2 dx = 3x − 3 = 3(x + 1) (x − 1)
Solution :
d2y = 6x = 0 ⇒ x = 0 dx2 Now
d2y (− 0.1) = 6(− 0.1) < 0 and dx2
d2y (0.1) = 6(0.1) > 0. In the neighbourhood (− 0.1, 0.1) dx2 of 0, y′′ (− 0.1) and y′′(0.1) are of opposite signs. Therefore (0, y (0)) i.e., (0, 2) is a point of inflection. Note : Note that x = 0 is not a critical point since y′ (0) = − 3 ≠ 0. Example 5.66 : Test for points of inflection of the curve y = sinx, x ∈ (0, 2π) Solution :
y′ = cosx y′′ = − sinx = 0 ⇒ x = nπ, n = 0, ±1, ± 2, ...
since x ∈(0, 2π), x = π corresponding to n = 1. Now y′′ (.9π) = − sin (.9π) < 0 and y′′(1.1π) = − sin (1.1 π) > 0 since sin (1.1π) is negative The second derivative test confirms that (π, f(π)) = (π, 0) is a point of inflection.
66
Note : Note that x = π is not a stationary point since y′(π) = cos π = − 1 ≠ 0. In fact y = sin x admits countable number of points of inflections in the range (− ∞, ∞), each of which is given by (nπ, 0), n = 0, ± 1, ±2, … and in none of the cases, y′(nπ) = (− 1)n vanishes. This shows that points of inflections need not be stationary points. EXERCISE 5.11 Find the intervals of concavity and the points of inflection of the following functions : (1)
f(x) = (x − 1)1/3
(2)
f(x) = x2 − x
(3)
f(x) = 2x3 + 5x2 − 4x
(4)
f(x) = x4 − 6x2
(5)
f(θ) = sin 2θ in (0, π)
(6)
y
= 12x2 − 2x3 − x4
67
Testing a differentiable function for maximum and minimum with a first derivative This gives us the following diagram of possible cases. Signs of derivative f ′(x) when passing through Character of critical point critical point x0 x < x0 x = x0 x > x0 +
f ′ (x0) = 0 or is
−
Maximum point
−
discontinuous f ′ (x0) = 0 or is
+
Minimum point
+
discontinuous f ′(x0) = 0 or is
+
Neither maximum nor minimum (function increases). But is a point of inflection. Neither maximum nor minimum (function decreases) But is a point of inflection.
discontinuous −
f ′(x0) = 0 or is
−
discontinuous
Second derivative test This gives us the following diagram of possible cases. Signs of derivative f ′′(x) at the critical point of f(x) or f ′(x)
Character of the point
x = x0 f ′(x0)
f ′′(x0)
0
−
0
+
x < x0
Maximum point Minimum point
f ′′(x0)
Critical point of f Critical point of f x > x0
Point of Inflection Point of inflection Unknown Unknown
+
0 or ≠ 0
0
−
−
0 or ≠ 0
0
+
+ −
0 or ≠ 0 0 or ≠ 0
0 0
+ −
68
6. DIFFERENTIAL CALCULUS APPLICATIONS-II 6.1 Differentials : Errors and Approximation dy dx to denote the derivative of y with respect to x but we have regarded it as a single entity and not as a ratio. In this section we give the quantities dy and dx separate meanings in such a way that their ratio is equal to the derivative. We also see that these quantities, called differentials, are useful in finding the approximate values of functions. We have used the Liebnitz notation
Definition 1 : Let y = f(x) be a differentiable function. Then the quantities dx and dy are called differentials. The differential dx is an independent variable that is dx can be given any real number as the value. The differential dy is then defined in terms of dx by the relation dy = f ′(x) dx (dx ≈ ∆x) Note : (1) The differentials dx and dy are both variables, but dx is an independent variable, where as dy is a dependent variable – it depends on the values of x and dx. If dx is given a specific value and x is taken to be some specific number in the domain of f, then the numerical value of dy is determined. (2) If dx ≠ 0 we can divide both sides of dy = f ′(x) dx by dx to obtain dy dy dx = f ′(x). Thus dx now is the ratio of differentials. Example 6.1 : If y = x3 + 2x2 (i)find dy (ii) find the value of dy when x = 2 and dx = 0.1 Solution : (i) If f(x) = x3 + 2x2, then f ′(x) = 3x2 + 4x, so dy = (3x2 + 4x) dx (ii) Substituting x = 2 and dx = 0.1, we get dy = (3 × 22 + 4 × 2)0.1 = 2.
69
6.1.1 Geometric meaning of differentials : Let P(x,f(x)) and Q(x + ∆x, f(x +∆x)) be points on the graph of f and set dx = ∆x. The corresponding change in y is ∆y = f(x + ∆x) − f(x) The slope of the tangent line PR is the derivate f ′(x). Thus the directed distance from S to R is f ′(x) dx = dy.
y
R Q
y =f (x) P
S
∆y
dy
dx = ∆x 0
x
x + ∆x
x
Fig. 6.1 Therefore dy represents the amount that the tangent line rises or falls whereas ∆y represents the amount that the curve y = f(x) rises or falls when x changes by an amount dx. dy ∆y ∆y dy lim Since dx = ∆x → 0 , we have ≈ ….(1) when ∆x is small. ∆x ∆x dx Geometrially, this says that the slope of the secant line PQ is very close to the slope of the tangent line at P when ∆x is small. If we take dx = ∆x, then (1) becomes ∆y ≈ dy ….(2) which says that if ∆x is small, then the actual change in y is approximately equal to the differential dy. Again this is geometrically evident in the case illustrated by Fig. 6.1. The actual change in y is referred as absolute error. The actual error in y is ∆y ≈ dy. ∆y Actual change in y The quantity y = Actual value of y is called relative error and the ∆y quantity y × 100 is called percentage error. The approximation given by (2) can be used in computing approximate values of functions. Suppose that f(a) is a known number and an approximate value is calculated for f(a + ∆x) where dx is small, since f(a + ∆x) = f(a) + ∆y, (2) gives, f(a + ∆x) ≈ f(a) + dy….(3) Example 6.2 : Compute the values of ∆y and dy if y = f(x) = x3 + x2 − 2x + 1 where x changes (i) from 2 to 2.05 and (ii) from 2 to 2.01 Solution : (i) We have f(2) = 23 + 22 − 2(2) + 1 = 9 f(2.05) = (2.05)3 + (2.05)2 − 2(2.05) + 1 = 9.717625. and ∆y = f(2.05) − f(2) = 0.717625. In general dy = f ′(x) dx = (3x2 + 2x − 2)dx When x = 2, dx = ∆x = 0.05 and dy = [(3(2)2+2(2)−2] 0.05 = 0.7
70
(ii)
f(2.01) = (2.01)3 − (2.01)2 − 2(2.01) + 1 = 9.140701 ∴ ∆y = f(2.01) − f(2) = 0.140701
When dx = ∆x = 0.01, dy = [3(2)2 + 2(2) − 2]0.01 = 0.14 Remark : The approximation ∆y ≈ dy becomes better as ∆x becomes smaller in Example 6.2. Also dy was easier than to compute ∆y. For more complicated functions it may be impossible to compute ∆y exactly. In such cases the approximation by differentials is especially useful. Example 6.3 : Use differentials to find an approximate value for Solution : Let y = f(x) =
3
1 3
1 x = x. Then dy = 3 x.
−2 3
3
65.
dx
Since f(64) = 4. We take x = 64 and dx = ∆x = 1 −2
1 1 1 This gives dy = 3 (64) 3 (1) = 3(16) = 48 ∴
3
1 65 = f(64 + 1) ≈ f(64) + dy = 4 + 48 ≈ 4.021
3 Note : The actual value of 65 is 4.0207257... Thus the approximation by differentials is accurate to three decimal places even when ∆x = 1. Example 6.4 : The radius of a sphere was measured and found to be 21 cm with a possible error in measurement of atmost 0.05 cm. What is the maximum error in using this value of the radius to compute the volume of the sphere ? 4
Solution : If the radius of the sphere is r, then its volume is V = 3 π r3. If the error in the measured value of r is denoted by dr = ∆r, then, the corresponding error in the calculated value of V is ∆V. which can be approximated by the differential dV = 4πr2 dr. When r = 21 and dr=0.05, this becomes dV = 4π(21)2 0.05 ≈ 277. The maximum error in the calculated volume is about 277 cm3. Note : Although the possible error in the above example may appear to be rather large, a better picture of the error is given by the relative error, which is computed by dividing the error by the total volume. ∆V dV 277 V ≈ V ≈ 38,808 ≈ 0.00714
71
dr
0.05
Thus a relative error of r = 21 ≈ 0.0024 in the radius produces a relative error of about 0.007 in the volume. The errors could also be expressed as percentage errors of 0.24% in the radius and 0.7% in the volume. Example 6.5 : The time of swing T of a pendulum is given by T = k l where k is a constant. Determine the percentage error in the time of swing if the length of the pendulum l changes from 32.1 cm to32.0 cm. 1
Solution :
If T = k l
= k l2
1 − dT k Then dl = k 2 × l 2 = 2 l and dl = 32.0 − 32.1 = −0.1 cm Error in T = Approximate change in T. dT k (−0.1) ∆T ≈ dT = dl dl = 2 l k (−0.1) 2 l ∆T Percentage error = T × 100 % = × 100 % k l −0.1 −0.1 = 2l × 100 % = 2(32.1) × 100% = − 0.156% Hence the percentage error in the time of swing is a decrease of 0.156%. Aliter : T = k l 1 Taking log on both sides, log T = log k + 2 log l 1 1 1 Taking differential on both sides, T dT = 0 + 2 l × dl ∆T 1 1 1 i.e, T ≈ T dT = 0 + 2 l × dl 1 dl ∆T T × 100 = 2 × l × 100 1 (−0.1) = 2 × 32.1 × 100 = − 0.156% ie., the percentage error in the time of swing is a decrease of 0.156. 1
Caution : Differentiation is carried out with the common understanding that the function involved admit logarithmic differentiation.
72
Example 6.6 : A circular template has a radius of 10 cm (± 0.02). Determine the possible error in calculating the area of the templates. Find also the percentage error. dA Solution : Area of circular template A = πr2, hence dr = 2πr, Approximate change in area ∆A ≈ (2πr)dr. When r = 10 cm and dr = 0.02 ∆A = (2π 10) (0.02) ≈ 0.4π cm2 i.e, the possible error in calculating the template area is approximately 1.257 cm2 0.4π Percentage error ≈ × 100 = 0.4% π(10)2 Example 6.7 : Show that the percentage error in the nth root of a number is 1 approximately n times the percentage error in the number . 1 n
Solution : Let x be the number. Let y = f(x) = (x) 1 Then log y = n log x 1 1 1 Taking differential on both sides, we have y dy = n × x dx 1 1 1 ∆y i.e., y ≈ y dy = n . x dx ∆y 1 dx ∴ y × 100 ≈ n x × 100 1 = n times the percentage error in the number. Example 6.8 : Find the approximate change in the volume V of a cube of side x meters caused by increasing the side by 1% Solution : The volume of the cube of side x is, V = x3 ; dV = 3x2 dx When dx = 0.01x,
dV = 3x2 × (0.01x) = 0.03 x3 m3.
EXERCISE 6.1 (1) Find the differential of the functions 4
(i) y = x5
(ii) y =
x−2 (iv) y = 2x + 3
(v) y = sin 2x
73
x
(iii) y =
x4 + x2 + 1
(vi) y = x tan x
(2) Find the differential dy and evaluate dy for the given values of x and dx. (i)
1 y = 1 − x2 , x = 5, dx = 2
(ii)
y = x4 − 3x3+ x −1, x = 2, dx = 0.1.
(iii)
y = (x2 + 5)3, x = 1, dx = 0.05
(iv)
y =
(v)
π y = cos x, x = 6 dx = 0.05
1 − x , x = 0, dx = 0.02
(3) Use differentials to find an approximate value for the given number (i)
1 (ii) 10.1
36.1
(iii) y =
3
1.02 +
4
(iv) (1.97)6
1.02
(4) The edge of a cube was found to be 30 cm with a possible error in measurement of 0.1 cm. Use differentials to estimate the maximum possible error in computing (i) the volume of the cube and (ii) the surface area of cube. (5) The radius of a circular disc is given as 24 cm with a maximum error in measurement of 0.02 cm. (i) Use differentials to estimate the maximum error in the calculated area of the disc. (ii) Compute the relative error ?
6.2 Curve Tracing : The study of calculus and its applications is best understood when it is studied through the geometrical representation of the functions involved. In order to investigate the nature of a function (graph) it is not possible to locate each and every point of the graph. But we can sketch the graph of the function and know its nature by certain specific properties and some special points. To do this we adopt the following strategies. (1) Domain, Extent, Intercepts and origin : (i) Domain of a function y = f(x) is determined by the values of x for which the function is defined.
74
(ii) Horizontal (vertical) extent of the curve is determined by the intervals of x (y) for which the curve exists. (iii) x = 0 yields the y − intercept and y = 0 yields the x – intercept (iv) If (0,0) satisfies the given equation then the curve will pass through the origin. (2) Symmetry : Find out whether the curve is symmetrical about any line with the help of the following rules : The curve is symmetrical about (i) the x-axis if its equation is unaltered when y is replaced by − y (ii) the y-axis if its equation is unaltered when x is replaced by − x. (iii) the origin if it is unaltered when x is replaced by − x and y is replaced by − y simultaneously. (iv) the line y = x if its equation is unchanged when x and y are replaced by y and x. (v) the line y = − x if its equation is unchanged when x and y are replaced by − y and − x. (3) Asymptotes (parallel to the co-ordinate axes only) : If y → c, c finite [x → k, k finite] whenever x → ± ∞ [y → ± ∞] then the line y = c [x = k] is an asymptote parallel to x − axis [y – axis]. (4) Monotonicity : Determine the intervals of x for which the curve is decreasing or increasing using the first derivates test. (5) Special points (Nature of bending) : Determine the intervals of concavity and inflection points using the first and second derivatives test. Illustrative Example : Example 6.9 : Trace the curve y = x3 + 1 Solution : (1) Domain, Extent, intercepts and origin : The function is defined for all real values of x and hence the domain is the entire interval (−∞, ∞). Horizontal extent is −∞ < x < ∞ and vertical extent is − ∞ < y < ∞. Clearly x = 0 yields the y intercept as + 1 and y = 0 yields the x intercepts as −1. It is obvious that the curve does not pass through (0,0).
75
(2) Symmetry Test : The symmetry test shows that the curve does not possess any of the symmetry properties. (3) Asymptotes : As x → c (for c finite) y does not tend to ± ∞ and vice versa. Therefore the curve doest not admit any asymptote. (4) Monotonicity : The first derivative test shows that the curve is increasing throughout (−∞,∞) since y′ ≥ 0 for all x. y
(5) Special points : The curve is concave downward in (−∞, 0) and concave upward in (0, ∞) since
8 6
y′′ = 6x < 0 for x < 0
y =x3 +1
4
y′′ = 6x > 0 for x > 0 and y′′ = 0 for x = 0 yields (0,1) as the inflection point
-6 -4 -2
2 0 -2
2
4
x
-4 -6 -8
Fig. 6.2 Example 6.10 : Trace the cure y2 = 2x3. Solution : (1) Domain, extent, Intercept and Origin : When x ≥ 0, y is well defined. As x → ∞, y → ± ∞, The curve exists in first and fourth quadrant only The intercepts with the axes are given by : x = 0, y = 0 and when y = 0, x = 0 Clearly the curve passes through origin. (2) Symmetry : By symmetry test, we have, the curve is symmetric about x – axis only. (3) Asymptotes : As x → + ∞, y → ± ∞, and vice versa. ∴ the curve does not admit asymptotes. (4) Monotonicity : For the branch y = 2 x3/2 of the curve is increasing since dy 3/2 dx > 0 for x > 0 and the branch y = − 2x of the curve is decreasing dy since dx < 0 for x > 0
76
(5) Special points : (0,0) is not a point of inflection. y
This curve is called a semi – cubical parabola. Note : (0, 0) admits a pair of tangents which coincide, resulting in a special point, called cusp.
(0,0)
x
Fig. 6.3 Example 6.11 : Discuss the curve y2 ( 1 + x) = x2 (1 − x) for (i) existence (ii) symmetry (iii) asymptotes (iv) loops Solution : (i) Existence : The function is not well defined when x >1 and x ≤ −1 and the curve lies between −1 < x ≤ 1. (ii) Symmetry : The curve is symmetrical about the x − axis only. (iii) Asymptotes : x = −1 is a vertical asymptote to the curve parallel to y − axis. (iv) Loops : (0,0) is a point through which the curve passes twice and hence a loop is formed between x = 0 and x = 1. x =-1
y
(0,0) (-1,0)
(1,0)
x
Fig. 6.4 Example 6.12 : Discuss the curve a2 y2 = x2 (a2 − x2), a > 0 for (i) existence (ii) symmetry (iii) asymptotes (iv) loops Solution : (i) Existence : The curve is well defined for (a2 − x2) ≥ 0 i.e., x2 ≤ a2 i.e., x ≤ a and x ≥ − a
77
(ii) Symmetry : The curve is symmetrical about x-axis, y – axis, and hence about the origin. (iii) Asymptotes : It has no asymptote. (iv) Loops : For −a < x < 0 and 0 < x < a, y2 > 0 ⇒ y is positive and negative ∴ a loop is formed between x = 0 and x = a and another loop is formed between x = −a and x = 0. y
(0,0)
-a
a
x
Fig. 6.5 2
Example 6.13 : Discuss the curve y = (x − 1) (x − 2)2. for (i) existence (ii) symmetry (iii) asymptotes (iv) loops Solution : (i) Existence : The curve is not defined for x − 1 < 0, ie., whenever x < 1, the R.H.S. is negative ⇒ y2 < 0 which is impossible. The curve is defined for x ≥ 1. (ii) Symmetry : The curve is symmetrical about x-axis. (iii) Asymptote : The curve does not admit asymptotes. (iv) Loops : Clearly a loop is formed between (1, 0) and (2, 0). y 2 1 0
1
2
3
x
Fig. 6.6 EXERCISE 6.2 (1) Trace the curve y = x3 Discuss the following curves for (i) existence (ii) symmetry (iii) asymptotes (iv) loops (3) y2 (2 + x) = x2 (6 − x) (2) y2 = x2 (1 − x2) (5) y2 = (x − a) (x − b)2 ; a, b > 0, a > b. (4) y2 = x2 (1 − x)
78
6.3 Partial Differentiation : A nation’s economy (E) depends on many factors. An yield (Y) of a crop also depends on various factors such as rain, soil, manure etc., Similarly the character (C) of a child is formed by its parent’s characters, environment etc., In plane geometry, area (A) and volume (V) also depend on the dimensions like length, breadth and height. In all the above cases either economy or yield or character or area or volume depends on more than one variable (factor). If any small change is effected in any of the variables (factors), it becomes necessary to know what changes will be caused in the respective dependent variable E or Y or C or A or V. These small changes can take place in all the variables (independent) simultaneously or in some of them while others are not subjected to any change. The study of these changes in the dependent variable while a corresponding change is made in one or more of the independent variables, keeping the remaining independent variables fixed leads to what is known as partial differentiation. g h For clarity, let us consider the ∆x area (A) of a rectangle of length x f c d and breadth y. Then A = xy = f(x,y). x Note that ‘A’ depends on two A independent variables x and y. a b e y A = xy = area of abcd ∆y Fig. 6.7 Suppose a small change is made in y ie., y + ∆y instead of y, then the new area A′ = x(y + ∆y). Note that x is fixed still there is change in the area A. Similarly, if we interchange roles of x and y in the above we get new area abgh = A′′ = (x + ∆x)y. Note that change in both x and y will also cause change in area A. In this case the area is (x +∆x) (y + ∆y) = area of aeih. But we shall restrict ourselves to the discussion of the change in one variable fixing the rest. We may consider functions of two or three independent variables only. We can also discuss the continuity problems and the limit process for functions depending on more than one variable similar to that of their counterpart in single variable differential calculus.
79
Partial Derivatives : Let (x0,y0) be any point in the domain of definition of f(x,y). Let u = f(x, y) We define partial derivative of u with respect to x at the point (x0,y0) as the ordinary derivative of f(x,y0) with respect to x at the point x = x0. i.e.,
∂u d = dx f(x,y0) x=x ∂x (x , y ) 0 0 0 lim f(x0 + h, y0) − f(x0,y0) = h→0 , (denoted by fx or ux at (x0, y0)) h
provided the limit exists. Similarly, partial derivatives of u = f(x,y) with respect to y at the point (x0,y0) is ∂u d = dy f(x0,y) y=y ∂y (x , y ) 0 0 0 lim f(x0, y0 + h) − f(x0,y0) = h→0 (denoted by fy or uy at (x0, y0)) h provided the limit exists. A function is said to be differentiable at a point (at all points on a domain) if its partial derivatives exist at that point (at all points of a domain). The process of finding partial derivatives is called partial differentiation. Remark : Throughout we shall consider only continuous functions of two or three variables possessing continuous first order partial derivatives. Second Order Partial Derivatives : When we differentiate a function u = f(x,y) twice we obtain its second order derivatives, defined by, ∂ ∂f ∂2f 2 = ∂x ∂x ∂x
;
∂2f ∂ ∂f and 2 = ∂y ∂y ∂y
∂ ∂f ∂ ∂f ∂2f ∂2f = = = denoted respectively ∂x ∂y ∂y ∂x ∂x ∂y ∂y ∂x as fxx or uxx, fyy or uyy and fxy = fyx or uxy = uyx Note that since the function and its partial derivaties are continuous the order of differentiation is immaterial (A result due to Euler)
80
Chain rule (function of a function rule) of two variables : u = f (x,y) dependent variable
If u = f(x,y) is differentiable and x and y are differentiable functions of t, then u is a differentiable function of t and
∂u / ∂y
∂u / ∂x x
du ∂f dx ∂f dy dt = ∂x dt + ∂y dt
y dy / dt
dx / dt
Tree diagram to remember the chain rule : (2 variables)
t independent variable
dx / dt
Tree diagram to remember the chain rule : (3 – variables)
dy / dt
∂u / ∂y
Fig. 6.8 Chain rule (function of a function rule) of three variables : If u = f(x,y, z) is differentiable u dependent variable and x, y, z are differentiable functions of t, then u is a differentiable ∂u / ∂z ∂u / ∂x function of t and y du ∂f dx ∂f dy ∂f dz x z dt = ∂x dt + ∂y dt + ∂z dt dz / dt
t independent variable
Fig. 6.9 Chain rule for partial derivatives : If w = f(u,v), u = g(x,y), ; v = h (x,y) then ∂w ∂w ∂u ∂w ∂v = + ∂x ∂u ∂x ∂v ∂x
;
∂w ∂w ∂u ∂w ∂v = + ∂y ∂u ∂y ∂v ∂y w = f (u,v)
w = f (u,v) ∂w / ∂v
∂w / ∂u u
u
v
x
v ∂v / ∂y
∂u / ∂y
∂v / ∂x
∂u / ∂x
∂w / ∂v
∂w / ∂u
Fig. 6.10
81
y
Homogeneous functions : A function of several variables is said to be homogeneous of degree n if multiplying each variables by t (where t > 0) has the same effect as multiplying the original function by tn. Thus, f(x,y) is homogeneous of degree n if f(tx, ty) = tn f(x,y) Euler’s Theorem : ∂f ∂f +y = nf If f(x,y) is a homogeneous function of degree n, then x ∂y ∂x Remark : Euler’s theorem can be extended to several variables. Example 6.14 : Determine :
∂2u ∂u ∂u ∂2u ∂2u ∂2u , , 2, 2, and ∂y∂x ∂x ∂y ∂x ∂y ∂x ∂y
if u(x,y) = x4 + y3 + 3x2 y2 + 3x2y ∂u ∂u = 4x3 + 6xy2 + 6xy ; = 3y2 + 6x2y + 3x2 Solution : ∂y ∂x ∂2u ∂2u 2 2 = 12x + 6y + 6y ; = 6y + 6x2 ∂x2 ∂y2 ∂2 u ∂2u = 12 xy + 6x ; = 12xy + 6x ∂x ∂y ∂y∂x Note that
∂2 u ∂2u = due to continuity of u and its first order partial ∂x ∂y ∂y∂x
derivatives. Example 6.15 : If u = log (tan x + tan y + tan z), prove that ∑ sin 2x
∂u = 2 ∂x
sec2x ∂u = tanx + tany + tanz ∂x
Solution :
2 sin x cos x . sec2x 2 tan x ∂u = tan x + tan y + tan z = tan x + tan y + tan z ∂x 2 tan y ∂u = tan x + tan y + tan z similarly, sin 2y ∂y sin 2x
sin 2z L.H.S. = ∑ sin 2x
2 tan z ∂u = tan x + tan y + tan z ∂z 2 (tan x + tan y + tan z) ∂u = tan x + tan y + tan z = 2 = R.H.S ∂x
82
Example 6.16 : If U =(x − y) (y − z) (z − x) then show that Ux + Uy + Uz = 0 Solution :
Ux = (y − z) {(x − y) (− 1) + (z − x).1}
= (y − z) [(z − x) − (x − y)] Similarly Uy = (z − x) [(x − y) − (y − z)] Uz = (x − y) [(y − z) − (z − x)] Ux + Uy + Uz = (y − z) [(z − x) − (z − x)] + (x − y) [− (y − z) + (y − z)] + (z − x) [(x − y) − (x − y)] =0 Example 6.17 : Suppose that z = ye Solution :
x2
dz where x = 2t and y = 1 − t then find dt
∂z dx ∂z dy dz dt = ∂x dt + ∂y dt ∂z dy ∂z x2 x2 dx = ye 2x ; = e ; dt = 2 ; dt = −1 ∂y ∂x dz x2 x2 = y 2x e (2) + e (−1) dt
x2 x2 4t2 4t2 = 4 xy e − e = e [(8t (1 − t) − 1)] = e (8t − 8t2 −1) (Since x = 2t and y = 1 − t) x ∂w ∂w and Example 6.18 : If w = u2 ev where u = y and v = y log x, find ∂x ∂y Solution : We know
∂w ∂u ∂w ∂v ∂w ∂w ∂u ∂w ∂v ∂w = + ; and = + ∂u ∂x ∂v ∂x ∂y ∂u ∂y ∂v ∂y ∂x ∂w ∂w = 2uev ; = u2ev ; ∂u ∂v 1 ∂u −x ∂u =y ; = 2 ∂x ∂y y y ∂v =x ; ∂x
∴
∂v ∂y
= log x.
∂w 2uev y x = y + u2ev x = xy 2 (2 + y) ∂x y
83
∴
∂w −x = 2uev 2 + u2ev log x ∂y y x x2 = 3 xy [ylog x − 2], (since u = y and v = y log x) y
dw Example 6.19 : If w = x + 2y + z2 and x = cos t ; y = sin t ; z = t. Find dt dw ∂w dx ∂w dy ∂w dz Solution : We know dt = + + ∂x dt ∂y dt ∂z dt ∂w dx = 1 ; dt = − sin t ∂x dy ∂w = 2 ; dt = cos t ∂y dz ∂w = 2 z ; dt = 1 ∂z dw ∴ dt = 1 ( − sin t) + 2 cos t + 2z = − sin t + 2 cos t + 2 t 1
Example 6.20 : Verify Euler’s theorem for f(x,y) = Solution :
2 2
x + y2
1 = t f(x,y) = t−1 f(x, y)
1
f(tx, ty) =
2
2 2
t x +t y
∴ f is a homogenous function of degree −1 and by Euler’s theorem, x
∂f ∂f +y = −f ∂y ∂x
Verification :
1 fx = − 2
2x
(x2 +y2)
3
−x
=
/2
(x2 +y2)
3 /2
−y
Similarly, fy =
(x2 +y2)
3 /2
x2 + y2
xfx + yfy = −
2
(x
3 2 /2
+y Hence Euler’s theorem is verified.
)
84
=
−1 x2 + y2
= − f.
Example 6.21 : If u is a homogenous function of x and y of degree n, prove that ∂2u ∂u ∂2u + y 2 = (n − 1) ∂y ∂x ∂y ∂y Solution : Since U is a homogeneous function in x and y of degree n, Uy is homogeneous function in x and y of degree n − 1. Applying Euler’s theorem for Uy we have, x
x(Uy)x + y (Uy)y = (n −1) Uy i.e., xUyx + y Uyy = (n −1) Uy i.e., x
∂2u ∂2u ∂u + y 2 = (n − 1) ∂x ∂y ∂y ∂y
Example 6.22 : Using Euler’s theorem, prove that x
∂u 1 ∂u + y = tan u if ∂y 2 ∂x
x−y x + y
u = sin−1
Solution: R.H.S. is not homogeneous and hence 1 x−y ⇒ f is homogeneous of degree 2 . define f = sin u = x+ y ∂f 1 ∂f ∴ By Euler’s theorem, x + y = 2 f ∂y ∂x 1 ∂ ∂ i.e., x . (sin u) + y (sin u) = 2 sin u ∂x ∂y ∂u 1 ∂u x . cos u + y . cos u = 2 sin u ∂y ∂x ∂u ∂u 1 x + y = 2 tan u ∂x ∂y EXERCISE 6.3 ∂2u ∂2u = for the following functions : ∂x ∂y ∂y ∂x x y (ii) u = 2 − 2 (i) u = x2 + 3xy + y2 y x
(1) Verify
x (iv) u = tan−1 y .
(iii) u = sin 3x cos 4y
85
x2 + y2 , show that x
(2) (i) If u = x y
∂u ∂u +y =u ∂x ∂y
y
x y ∂u ∂u x (ii) If u = e sin y + e cos x , show that x +y = 0. ∂x ∂y dw (3) Using chain rule find dt for each of the following : (i) w = e xy where x = t2, y = t3 (ii) w = log (x2 + y2) where x = et, y = e− t x where x = cos t, y = sin t. (iii) w = (x2 + y2) (iv) w = xy + z where x = cos t, y = sin t, z = t (4) (i) Find
∂w ∂w and if w = log (x2 + y2) where x = r cos θ, y = r sin θ ∂θ ∂r
(ii) Find
∂w ∂w and if w = x2 + y2 where x = u2 − v2, y = 2uv ∂v ∂u
(iii) Find
∂w ∂w and if w = sin−1 xy where x = u + v, y = u − v. ∂v ∂u
(5) Using Euler’s theorem prove the following :
x3 + y3 ∂u ∂u prove that x +y = sin 2u. ∂x ∂y x− y
(i) If u = tan−1
∂u ∂u x +y = 3u. (ii) u = xy2 sin y , show that x ∂x ∂y (iii) If u is a homogeneous function of x and y of degree n, prove that ∂2u ∂u ∂2u x 2 +y = (n − 1) ∂x ∂x ∂y ∂x (iv) If V = zeax + by and z is a homogenous function of degree n in x and ∂V ∂V y prove that x +y = (ax + by + n)V. ∂x ∂y
86
7. INTEGRAL CALCULUS AND ITS APPLICATIONS 7.1. Introduction : In class XI, we have studied the direct evaluation of definite integrals as the limit of integral sums. Even when the integrands are very simple, direct evaluation of definite integrals as the limit of integral sum involves great difficulties. Sometimes this method involves cumbersome computations. There is a formula called Second Fundamental Theorem on Calculus that yields a practical and convenient method for computing definite integrals in case where the anti-derivative of the integrand is known. This method which was discovered by Newton and Leibnitz utilises ‘the profound relationship’ that exists between integration and differentiation. In this chapter we have the following five sections dealing with the concept and applications of definite integrals. (i) To solve simple problems using second fundamental theorem of calculus. (ii) Properties of definite integral. (iii) Reduction formulae (iv) Area under the curve and volume of solid of revolution about an axis. (v) Length of the curve and the surface area of a solid of revolution about an axis.
7.2. Simple definite integrals : First fundamental theorem of calculus : x Theorem 7.1 : If f(x) is a continuous function and F(x) = ⌠ ⌡ f(t)dt, then we a have the equation F′(x) = f(x).
Second fundamental theorem of calculus : Theorem 7.2 : If f(x) is a continuous function with domain a ≤ x ≤ b, then b
⌠ ⌡ f(x)dx = F(b) − F(a) where F is any anti-derivative of f. a π/2 sin x dx Example 7.1 : Evaluate ⌠ ⌡ 1 + cos2x 0
87
π/2 sin x dx Let I = ⌠ ⌡ 1 + cos2x 0
Solution:
t = cos x 0 π/2 1 0
x Let t = cos x dt = − sin x dx (or) sin x dx = − dt t 0 − dt π π 0 ∴I = ⌠ = − [tan−1 t] 1 = − 0 − 4 = 4 ⌡ 1 + t2 1 1 Example 7.2 : Evaluate ⌠ x ex dx
⌡ 0
Solution: Using the method of integration by parts
Here
⌠ udv = uv − ⌠ ⌡v du ⌡ 1
x ⌠ x e dx = ⌡
1 (xex)0
dv = ex dx
1 − ⌠ ex dx
0
v = ex
⌡ 0
= e−
1 (ex)0
= e − (e − 1) =1 a Example 7.3 : Evaluate ⌠
⌡
a2 − x2 dx
0 a Solution:
⌠ ⌡
x a2 − x2 dx = 2
a2 x a2 − x2 + 2 sin−1 a
a
0
a2 a = 0 + 2 sin−1 a − (0 + 0) a2 π πa2 a2 = 2 sin−1(1) = 2 2 = 4
88
u=x du = dx
0
π/2 Example 7.4 : Evaluate ⌠ e2x cos x dx ⌡ 0 Solution:
eax We know ∫eax cos bx dx = 2 (a cos bx + b sin bx) a + b 2 π/2 π/2 e2x ∴ ⌠ e2x cos x dx = 2 (2 cos x + sin x) ⌡ 2 + 12 0 0 eπ e0 = 5 (0 + 1) − 5 (2 + 0)
eπ 2 1 = 5 − 5 = 5 (eπ − 2) EXERCISE 7.1 Evaluate the following problems using second fundamental theorem : π/2 π/2 1 (2) ⌠ cos3x dx (3) ⌠ 9 − 4x2 dx (1) ⌠ sin2x dx
⌡
⌡
0 π/4 (4) ⌠ 2 sin2x sin 2x dx
0 1
dx
(5) ⌠
⌡
⌡
⌡
4 − x2 0 1 (sin−1x)3
0 2
dx (7) ⌠ 2 ⌡ x + 5x + 6 1 1 (10) ⌠ x2 ex dx
(8) ⌠
⌡
1 − x2
dx
0 π/2 (11) ⌠ e3x cos x dx
⌡
⌡
0
0
0 π/2 sin x dx (6) ⌠ ⌡ 9 + cos2x 0 π/2 (9) ⌠ sin 2x cos x dx
⌡
0 π/2 (12) ⌠ e−x sin x dx
⌡ 0
7.3 Properties of Definite Integrals : b b Property (1) : ⌠ f(x)dx = ⌠ f(y) dy ⌡ ⌡ a a Proof : Let F be any anti-derivative of f b ∴ ⌠ f(x) dx = [F(b) − F(a)]
⌡ a
89
… (i)
b
⌠ f(y) dy = [F(b) − F(a)] ⌡ a b From (i) and (ii)
… (ii)
b
f(x) dx = ⌠ f(y) dy ⌠ ⌡ ⌡ a
a
That is, integration is independent of change of variables provided the limits of integration remain the same. b a Property (2) : ⌠ f(x)dx = − ⌠ f(x) dx ⌡ ⌡ a b Proof : Let F be any anti-derivative of f b ∴ ⌠ f(x) dx = [F(b) − F(a)]
⌡
… (i)
a a
f(x) dx = [F(a) − F(b)] = − [F(b) − F(a)] ⌠ ⌡
… (ii)
b b
From (i) and (ii)
a f(x) dx = − ⌠ ⌠ f(x) dx ⌡ ⌡ a b
That is, if the limits of definite integral are interchanged, then the value of integral changes its sign only. b b Property (3) : ⌠ f(x)dx = ⌠ f(a + b − x) dx
⌡
⌡
a
a
Proof :
Let u = a + b − x
u=a+b−x
∴ du = − dx
x
a
b
or dx = − du u b b a b b ∴⌠ f(a + b − x)dx = − ⌠ f(u) du = ⌠ f(u) du = ⌠ f(x) dx
a
⌡
⌡
⌡
⌡
a
b
a
a
90
a a Property (4) : ⌠ f(x)dx = ⌠ f(a − x) dx
⌡
⌡
0
Proof :
0 Let u = a − x u=a−x x o a ∴ du = − dx u a o or dx = − du a a a o ∴⌠ f(a − x)dx = − ⌠ f(u) du = ⌠ f(u) du = ⌠ f(x) dx
⌡
⌡
⌡
⌡
a 0 0 0 Property (5) (Without proof) : If f(x) is integrable on a closed interval containing the three numbers a, b and c, then b c b ⌠ f(x) dx = ⌠ f(x) dx + ⌠ f(x) dx
⌡
⌡
⌡
a a regardless of the order of a, b and c.
c
2a a a Property (6) : ⌠ f(x)dx = ⌠ f(x) dx + ⌠ f(2a − x) dx ⌡ ⌡ ⌡ 0 0 0 2a a 2a Proof : Consider ⌠ f(x)dx = ⌠ f(x) dx + ⌠ f(x) dx
⌡
⌡
⌡
0
0
a
Put x = 2a − u in the second integral on the R.H.S., and dx = − du 2a o ⌠ f(x)dx = − ⌠ f(2a − u) du
⌡
… (1) u = 2a − x x a 2a u a o
⌡
a
a a
= ⌠ f(2a − u) du ⌡ 0 a = ⌠ f(2a − x) dx
⌡ 0
91
b b ‡ ⌠ f(x) dx = ⌠ f(y) dy ⌡ ⌡ a a
2a a a Hence (1) becomes ⌠ f(x) dx = ⌠ f(x) dx + ⌠ f(2a − x) dx ⌡ ⌡ ⌡ 0 0 0 2a a Property (7) : if f(2a − x) = f(x) ⌠ f(x)dx = 2 ⌠ f(x) dx
⌡
⌡
0
0
=0 if f(2a − x) = − f(x) Proof : We know that by property 2a a a … (1) ⌠ f(x)dx = ⌠ f(x) dx + ⌠ f(2a − x) dx
⌡
⌡
⌡
0 0 0 If f(2a − x) = f(x) then (1) becomes 2a a a a ⌠ f(x)dx = ⌠ f(x) dx + ⌠ f(x) dx = 2 ⌠ f(x) dx
⌡
⌡
⌡
⌡
0 0 0 0 If f(2a − x) = − f(x) then (1) becomes a a 2a ⌠ f(x)dx = ⌠ f(x) dx − ⌠ f(x) dx = 0
⌡
⌡
⌡
0
0
0
Hence proved. a a Property (8) : (i) ⌠ f(x)dx = 2⌠ f(x) dx,
⌡
⌡
−a a
0
if f is an even function.
(ii) ⌠ f(x) dx = 0 ⌡
if f is an odd function.
−a a a 0 Proof : Consider ⌠ f(x)dx = ⌠ f(x) dx + ⌠ f(x) dx
⌡
⌡
⌡
−a
−a
0
Let x = − t in the first integral of the R.H.S. Then dx = − dt
92
… (1)
x t
x=−t 0 −a a 0
a o f(− t) (− dt) + f(x) dx = ⌠ ⌠ ⌠ f(x) dx ⌡ ⌡ ⌡ a 0 −a a 0 = − ⌠ f(− t) dt + ⌠ f(x) dx a
∴ (1) becomes
⌡
⌡
a
0
a
a
= ⌠ f(− t) dt + ⌠ f(x) dx ⌡ ⌡ 0 0 a a a ∴ ⌠ f(x) dx = ⌠ f(− x) dx + ⌠ f(x) dx
⌡
⌡
⌡
⌡
−a
0
⌡
0 0 −a Case (ii) : If ‘f’ is an even function, then (2) becomes a a a ⌠ f(x) dx = ⌠ f(x) dx + ⌠ f(x) dx
⌡ 0
a = 2 ⌠ f(x) dx ⌡ 0 Case (iii) : If ‘f’ is an odd function then (2) becomes a a a (− f(x) dx + f(x) dx = ⌠ ⌠ ⌠ f(x) dx
⌡
⌡
⌡
−a
0
0 a
a = − ⌠ f(x) dx + ⌠ f(x) dx = 0 ⌡ ⌡ 0 0 Hence proved. π/4 Example 7.5 : Evaluate ⌠ x3 sin2x dx. ⌡ Solution:
− π/4 Let f(x) = x3 sin2x = x3 (sin x)2 ∴ f(− x) = (− x)3 (sin (− x))2 = (− x)3 (− sin x)2 = − x3 sin2x = − f(x)
93
… (2)
f(− x) = − f(x)
∴
∴ f(x) is an odd function. π/4 3 2 ⌠ x sin x dx. = 0 (by property)
⌡
− π/4 Example 7.6 : 1 3−x Evaluate ⌠ log 3 + x dx ⌡ −1 3−x Let f(x) = log 3 + x 3 + x = log (3 + x) − log (3 − x) ∴ f(− x) = log 3 −x = − [log (3 − x) − log (3 + x)] 3−x = − log 3 + x = − f(x) Thus f(− x) = − f(x) ∴ f(x) is an odd function. 1 3−x ∴ ⌠ log 3 + x dx = 0 ⌡ −1 Example 7.7 : π/2 Evaluate : ⌠ x sin x dx Solution:
⌡
− π/2 Solution: Let f(x) = x sin x f(− x) = (− x) sin (− x) = x sin x (‡ sin (− x) = − sin x) ∴ f(x) is an even function. π/2 π/2 ⌠ x sin x dx = 2 ⌠ x sin x dx
⌡
⌡
− π/2
0
= 2 {x (− cos x)}
94
π/2 − ⌠ (− cos x) dx ⌡ 0 0
π/2
Using the method of integration by parts π/2 π/2 = 2 0 + ⌠ cos x dx = 2 [sin x] ⌡ 0 0 = 2 [1 − 0] = 2
π/2 Example 7.8 : Evaluate ⌠ sin2x dx ⌡ − π/2
Let f(x) = sin2x = (sin x)2 f(− x) = (sin (− x))2 = (− sin x)2 = sin2x = f(x) Hence f(x) is an even function. π/2 π/2 1 π/2 ∴ ⌠ sin2x dx = 2 ⌠ sin2x dx = 2 × 2 ⌠ (1 − cos 2x) dx
Solution:
⌡
⌡
⌡
− π/2
0
0
sin 2x = x − 2
π/2 0
π =2
π/2 f(sin x) Example 7.9 : Evaluate ⌠ f(sin x) + f(cos x) dx ⌡ 0 π/2 f(sin x) Solution: Let I = ⌠ f(sin x) + f(cos x) dx ⌡ 0 π f sin 2 − x π/2 = ⌠ dx ⌡ π π f sin − x + f cos 2 − x 0 2 π/2 f (cos x) ∴ I = ⌠ f(cos x) + f (sin x) dx ⌡ 0 π/2 f(sin x) + f(cos x) π/2 π (1) + (2) gives 2 I = ⌠ f(cos x) + f(sin x) dx = ⌠ dx = [x]π/2 =2 0 ⌡ ⌡ o o π ∴ I=4
95
… (1)
… (2)
1 Example 7.10 : Evaluate ⌠ x(1 − x)n dx
⌡ 0
Solution:
1 Let I = ⌠ x(1 − x)n dx ⌡ 0 1 n = ⌠ (1 − x) [1 − (1 − x)] dx
⌡ 0 1
a a ‡ ⌠ f(x) dx = ⌠ f(a − x) dx ⌡ ⌡ 0 o
1 = ⌠ (1 − x) xn dx = ⌠ (xn − xn + 1) dx ⌡ ⌡ 0 0 1
xn + 1 xn + 2 1 n + 2 − (n + 1) 1 = n + 1 − n + 2 = n + 1 − n + 2 = (n + 1) (n + 2) 0
1
1
x(1 − x)n dx = (n + 1) (n + 2) ⌠ ⌡ 0 π/2 Example 7.11 : Evaluate ⌠ log (tan x)dx ⌡ 0 π/2 Solution: Let I = ⌠ log (tan x)dx
⌡
… (1)
0 π/2
π = ⌠ log tan 2 − x dx ⌡ 0 π/2 I = ⌠ log (cot x) dx
⌡
0 π/2
(1) + (2) gives
2I = ⌠ [log (tan x) + log (cot x)] dx ⌡ 0
96
… (2)
π/2 π/2 = ⌠ [log (tan x) . (cot x)] dx = ⌠ (log 1) dx = 0 ⌡ ⌡ 0 0 (‡ log 1 = 0) ∴ I=0 π/3 dx Example 7.12 : Evaluate ⌠ ⌡ 1 + cot x π/6 Solution:
π/3 dx Let I = ⌠ 1 + cot x ⌡ π/6 π/3 I= ⌠ ⌡
sin x dx sin x + cos x
… (1)
π/6 π 3
=⌠ ⌡ π 6
π π sin 3 + 6 − x dx
π π sin 3 + 6 − x +
π π cos 3 + 6 − x
b b ‡ ⌠ f(x) dx = ⌠ f(a + b − x) dx ⌡ ⌡ a a π 3 =⌠ ⌡ π 6
π sin 2 − x
π sin 2 − x +
π cos 2 − x
π/3 I= ⌠ ⌡
cos x dx cos x + sin x
π/3 2I = ⌠ ⌡
sin x + cos x dx cos x + sin x
π/6
(1) + (2) gives
π/6
97
dx
… (2)
π/3 π/3 π π π =3 − 6 = 6 2I = ⌠ dx = [x] ⌡ π/6 π/6 π ∴ I = 12 EXERCISE 7.2 Evaluate the following problems using properties of integration. π/2 1 π/4 (1) ⌠ sin x cos4 x dx (2) ⌠ x3 cos3x dx (3) ⌠ sin3x cos x dx
⌡
⌡
−1 π/2
−π/4 π/2
cos3x dx ⌠ ⌡
(4)
(5) ⌠ sin2x cosx dx ⌡
− π/2 1 1 (7) ⌠ log x − 1 dx
⌡
⌡
− π/2 3 (8) ⌠
0 π/3
⌡
x dx x+ 3−x
0
0 π/4
(6) ⌠ x sin2x dx ⌡ − π/4 1 (9) ⌠ x (1 − x)10 dx
⌡ 0
dx (10) ⌠ ⌡ 1 + tan x π/6
7.4 Reduction formulae : A formula which expresses (or reduces) the integral of the nth indexed function interms of that of (n − 1)th indexed (or lower indexed) function is called a reduction formula. Reduction formulae for ∫ sinnx dx. ⌠ cosnx dx (n is a positive integer) :
⌡
1 n−1 Result 1 : If In = ∫ sinnx dx then In = − n sinn−1x cos x + n In − 2 1 n−1 Result 2 : If In = ⌠ cosnx dx then In = n cosn−1x sin x + n In − 2
⌡
Result 3 : π/2 n π/2 n ⌠ sin x dx = ⌠ cos x dx =
⌡
⌡
0
0
n −n 1 . nn −− 32 . nn −− 54 ... 23 . 1 when n is odd n − 1 . n − 3 . n − 5 ... 1 . π n n − 2 n − 4 2 2 when n is even 98
Note : For the proofs of these above three results, refer Solution Book.
⌠sin5x dx Example 7.13 : Evaluate : ⌡ Solution :
n If In = ⌠ ⌡sin x dx, then we have
n−1 1 In = − n sinn−1x cos x + n In−2
… (I)
⌠sin5x dx = I5 ∴⌡ 4 1 (when n=5 in I) = − 5 sin4x cos x + 5 I3 4 1 2 1 = − 5 sin4x cos x + 5 − 3 sin2x cosx + 3 I1 (when n=3 in I) 1 4 4 8 5 2 ⌠ … (II) ⌡sin x dx = − 5 sin x cos x − 15 sin x cosx + 15 I1 1 I1 = ⌠ ⌡sin x dx = − cos x + c
1 4 4 8 5 2 ∴ ⌠ ⌡sin x dx = − 5 sin x cos x − 15 sin x cos x − 15 cos x + c 6 Example 7.14 : Evaluate : ⌠ ⌡sin x dx n Solution : If In = ⌠ ⌡sin x dx, then we have
1 n−1 In = − n sinn − 1x cos x + n In − 2
… (I)
6 ∴⌠ ⌡sin x dx = I6
1 5 (when n=6 in I) = − 6 sin5x cos x + 6 I4 1 5 1 3 = − 6 sin5x cos x + 6 − 4 sin3x cosx + 4 I2 (when n=4 in I) 1 5 5 ⌠sin6x dx = − 6 sin5x cos x − 24 sin3x cos x + 8 I2 (when n=2 in I) ⌡ 1 5 5 1 1 = − 6 sin5x cos x − 24 sin3x cos x + 8 − 2 sin x cosx + 2 I0 5 1 5 5 6 5 3 ⌠ ⌡sin x dx = − 6 sin x cos x − 24 sin x cos x − 16 sin x cos x + 16 I0
99
0 I0 = ⌠ ⌡sin x dx = ⌠ ⌡dx = x 5 1 5 5 ⌠sin6x dx = − 6 sin5x cos x − 24 sin3x cos x − 16 sin x cos x + 16 x ∴⌡
Example 7.15 : Evaluate : π/2 (i) ⌠ sin7x dx ⌡ 0
π/2 (ii) ⌠ cos8x dx ⌡ 0
2π x (iii) ⌠ sin9 4 dx ⌡ 0
π/6 (iv) ⌠ cos73x dx ⌡ 0 Solution : (i)We have π/2 n n − 1 n − 3 ... 2 sin x dx = n . ⌠ 3 when ‘n’ is odd n−2 ⌡ 0 π/2 7 6 4 2 16 sin x dx = 7 . 5 . 3 = 35 ⌠ ⌡ 0 π/2 n n − 1 . n − 3 . n − 5 ... 1 . π (ii) ⌠ cos x dx = n n − 2 n − 4 2 2 when ‘n’ is even ⌡ 0 π/2 7 5 3 1 π 35π ∴ ⌠ cos8x dx = 8 . 6 . 4 . 2 . 2 = 256 ⌡ 0 2π 9 x sin 4 dx (iii) ⌠ ⌡ 0 t=x/4 x x 0 2π Put 4 = t t 0 ∴ dx = 4dt π/2 2π
π/2
0
0
x 512 8 6 4 2 sin9 4 dx = 4 ⌠ sin9t dt = 4. 9 . 7 . 5 . 3 . = 315 ⌠ ⌡ ⌡
100
(iv)
π/6 7 ⌠ cos 3x dx
⌡ 0
t = 3x Put 3x = t 3dx = dt dx = 1/3 dt π/6 7 1 cos 3x dx = 3 ⌠ ⌡ 0
x
0
π/6
t
0
π/2
π/2 7 1 6 4 2 16 cos t dt = 3 7 . 5 . 3 . = 105 ⌠ ⌡ 0
π/2 Example 7.16 : Evaluate : ⌠ sin4x cos2x dx ⌡ 0 Solution : π/2 4 π/2 4 2 2 ⌠ sin x cos x dx= ⌠ sin x (1 − sin x) dx
⌡
⌡
0
0 π/2 π/2 π/2 = ⌠ (sin4x − sin6x) dx = ⌠ sin4x dx− ⌠ sin6x dx ⌡ ⌡ ⌡ 0 0 0 3 1 π 5 3 1 π π = 4 . 2 . 2 − 6 . 4 . 2 . 2 = 32 Two important results : The following two results are very useful in the evaluation of certain types of integrals. (1) If u and v are functions of x, then ∫udv = uv − u′v1 + u′′v2 − u′′′v3 + ... + (− 1)n unvn + ... where u′, u′′, u′′′ ... are successive derivatives of u and v1, v2, v3 ... are repeated integrals of v The above formula is well known as Bernoulli’s formula. Bernoulli’s formula is advantageously applied when u = xn (n is a positive integer). n ∞ (2) If n is a positive integer, then ⌠ xne−ax dx = n+1 ⌡ a 0
101
Note : The above formula is known as a particular case of Gamma Integral. Example 7.17 : Evaluate : 1 ∞ ∞ (ii) ⌠ x e− 4x dx (iii) ⌠ x5e−4x dx (iv) ⌠ e−mxx7 dx (i) ∫x3e2x dx
⌡
⌡
⌡
0
0
0 dv = e2x dx
Solution : (1) ∫x3e2x dx Using Bernoulli’s formula ∫udv = uv − u′v1 + u′′v2 ... We get
u = x3
v = 1/2 e2x
u′ = 3x2
v1 = 1/4 e2x
u′′ = 6x
v2 = 1/8 e2x
u′′′ = 6
v3 = 1/16 e2x
1 1 1 1 x3 e2x dx = (x3) 2 e2x − (3x2) 4 e2x + (6x) 8 e2x − (6) 16 e2x ⌠ ⌡ 1 3 3x 3 = 2 e2x x3 − 2x2 + 2 − 4
u=x
dv = e−4x dx 1 v = − 4 e−4x
u′ = 1
1 v1 = 16 e−4x
1 (ii) ⌠ x e− 4x dx
⌡
0 Using Bernoulli’s formula we get 1 xe ⌠ ⌡
− 4x
1 1 dx = (x) − 4 e−4x − (1) 16 e−4x
0
1
0
1 1 = − 4 e−4 − 0 − 16 (e−4 − e0) 1 5 = 16 − 16 e−4 ∞ (iii) ⌠ x5e−4x dx ⌡ 0
5 ∞ Using Gamma Integral ⌠ x5e−4x dx = 6 ⌡ 4 0
7 ∞ (iv) ⌠ e−mxx7 dx = 8 (Using Gamma Integral) ⌡ m 0
102
EXERCISE 7.3 (1) Evaluate :
4
(ii) ⌠ cos5x dx
(i) ⌠sin x dx
⌡
⌡
π/2 (2) Evaluate : (i) ⌠ sin6x dx ⌡ 0 π/4 8 (3) Evaluate : (i) ⌠ cos 2x dx
π/2 (ii) ⌠ cos9x dx ⌡ 0 π/6 7 (ii) ⌠ sin 3x dx
⌡
⌡
0 1
0 ∞ 6 −x/2 (ii) ⌠ x e dx
(4) Evaluate : (i) ⌠ x e−2x dx
⌡
⌡
0
0
7.5 Area and Volume :
x =a
x =b
In this section, we apply the definite integral to compute measure of area, length of arc and surface area. In our treatment it is understood that area, volume etc. is a number without any unit of measurement attached to it. 7.5.1 Area of bounded regions : Theorem : Let y = f(x) be a y continuous function defined on B [a, b], which is positive (f(x) lies on y =f(x) A or above x-axis) on the interval [a, b]. Then, the area bounded by the curve y = f(x), the x-axis and the ordinates x = a and x = b is given by C
b
Area = ⌠ f(x)dx or ⌠ ydx ⌡ ⌡ a a
⌡ a
y
A
(i.e., The area below the x-axis is negative)
103
C
D
x =b
b b Area = ⌠ (− y) dx = ⌠ (− f(x) dx) a
x
Fig. 7.1
If f(x) ≤ 0 (f(x) lies on or below x-axis) for all x in a ≤ x ≤ b then area is given by
⌡
D
x =a
b
y =f(x)
Fig. 7.2
B
x
y= (3/ 2) (
x+
2)
Example 7.18 : Find the area of the region bounded by the line 3x − 2y + 6 = 0, x = 1, x = 3 and x-axis. y Since the line 3x − 2y + 6 = 0 lies above the x-axis in the interval [1, 3], (i.e., y > 0 for x ∈ (1,3)) the required area 3 3 3 A = ⌠ ydx = 2 ⌠ (x + 2) dx ⌡ ⌡ 1 1 x 0 1 2 3 3 2 3 x = 2 2 + 2x Fig. 7.3 1 3 1 3 = 2 2 (9 − 1) + 2(3 − 1) = 2 [4 + 4] Area = 12 sq. units Example 7.19: Find the area of the region bounded by the line 3x − 5y − 15 = 0, x = 1, x = 4 and x-axis. y The line 3x − 5y − 15 = 0 lies 1 4 below the x-axis in the interval x = 1 x and x = 4 5) 4 -1 x 3 ∴Required area = ⌠ (− y) dx 5 )(
⌡
(1/ y=
1
Fig. 7.4 4 1 3 = ⌠ − 5 (3x − 15) dx = 5 ⌡ 1
3 x2 (5 − x) dx = 5x − ⌠ 5 2 4
⌡ 1
4
1
3 1 = 5 5(4 − 1) − 2 (16 − 1) 3 15 9 = 5 15 − 2 = 2 sq. units. Example 7.20: Find the area of the region bounded y = x2 − 5x + 4, x = 2, x = 3 and the x-axis.
104
For all x, 2 ≤ x ≤ 3 the curve lies below the x-axis. 3 Required area = ⌠ (− y) dx
y
⌡
0
2 3
1
2
3
x
4
-1
= ⌠ − (x2 − 5x + 4) dx ⌡ -2 2 3 x3 x2 Fig. 7.5 = − 3 − 5 2 + 4x 2 45 13 8 20 − 13 = − 9 − 2 + 12 − 3 − 2 + 8 = − 6 = 6 sq. units Area between a continuous curve and y-axis : y Let x = f(y) be a continuous y =d B function of y on [c, d]. The area D bounded by the curve x = f(y) and the x =f(y) abscissae y = c, y = d to the right of d A C y-axis is given by ⌠ xdy y =c x ⌡ Fig. 7.6 y c y =d
B
If the curve lies to the left of y-axis between the lines y = c and d y = d, the area is given by ⌠ (− x) dy.
x =f(y)
⌡
A
y =c
x
c
D
y
y= 2x +1
Fig. 7.7
Example 7.21: Find the area of the region bounded by y = 2x + 1, y = 3, y = 5 and y – axis. Solution : The line y = 2x + 1 lies to the right of y-axis between the lines y = 3 and y = 5. d ∴ The required area A = ⌠ xdy
y =5 y =3
⌡ c
x
0
Fig. 7.8
105
C
5 y−1 1 5 = ⌠ 2 dy = 2 ⌠ (y − 1)dy ⌡ ⌡ 3 3 5 1 25 9 1 y2 = 2 2 − y = 2 2 − 2 − (5 − 3) 3 1 = 2 [8 − 2] = 3 sq. units
y= 2x
+4
Example 7.22: Find the area of the region bounded y = 2x + 4, y = 1 and y = 3 and y-axis. The curve lies to the left of y-axis between the lines y = 1 and y = 3 ∴ Area is given by 3 y =3 A = ⌠ (− x) dy
⌡ 1 3
y =1
y−4 = ⌠ − 2 dy ⌡
3 2
1
Fig. 7.9
3 1 3 1 1 y2 = 2 ⌠ (4 − y)dy = 2 4y − 2 = 2 [8 − 4] = 2 sq. units. 1 ⌡ 1 Remark : f (x) If the continuous curve f crosses b the x-axis, then the integral ⌠ f(x) dx x
⌡
1
a
a gives the algebraic sum of the areas between the curve and the axis, counting area above as positive and below as negative. c d b f(x) dx = ⌠ f(x) dx + ⌠ (− f(x)) dx + ⌡ ⌡ ⌠
⌡ a
x
1 0
a ↓ above axis
c
c
Fig. 7.10 b f(x) dx ⌠ ⌡ d
↓ below axis
106
↓ above axis
f (x)
d
b
x
5 Example 7.23: (i) Evaluate the integral ⌠ (x − 3)dx
⌡
1 (ii) Find the area of the region bounded by the line y + 3 = x, x = 1 and x = 5 Solution : 5 5 x2 25 1 (i) = 2 − 15 − 2 − 3 = 12 − 12 = 0 … I (x − 3) dx = − 3x ⌠ 2 ⌡ 1 1 (ii) The line y = x − 3 crosses x-axis at x = 3 From the diagram it is clear that A1 y −3 x = lies below x-axis. y 3 A2 ∴ A1 = ⌠ (− y) dx.
⌡
1
1
3
O
As A2 lies above the x-axis 5 A2 = ⌠ ydx
5
x
A1
⌡
Fig. 7.11
3 5 3 5 ∴ Total area = ⌠(x − 3)dx = ⌠ − (x − 3) dx + ⌠ (x − 3) dx ⌡ ⌡ ⌡ 1 1 3 = (6 − 4) + (8 − 6) = 2+2 = 4 sq. units … (II) Note : From I and II it is clear that the integral f(x) is not always imply an area. The fundamental theorem asserts that the anti-derivative method works even when the function f(x) is not always positive. Example 7.24: Find the area bounded by the curve y = sin 2x between the ordinates x = 0, x = π and x-axis. Solution : The points where the curve y = sin 2x meets the x-axis can be obtained by putting y = 0. sin 2x = 0 ⇒ 2x = nπ , n ∈ Z
107
n x = 2 π.
π π i.e., x = 0, ± 2, ± π, ± 3 2…
π ∴ The values of x between x = 0 are x = π are x = 0, 2, π π The limits for the first arch are 0 and 2 and the curve lies above x-axis. π The limits for the second arch are 2 and π and the curve lies below x-axis. ∴Required area
y
π/2 π A = ⌠ sin 2x dx + ⌠ (− sin 2x)dx ⌡ ⌡ 0 π/2
π/2
π
x
0 y =sin 2x
π/2
− cos2x cos2x π = 2 0 + 2 π/2
Fig. 7.12
1 = 2 [−cos π + cos 0 + cos 2π − cos π] 1 = 2 [1 + 1 + 1 + 1] = 2 sq. units. Example 7.25: Find the area between the curves y = x2 − x − 2, x-axis and the lines x = − 2 and x = 4
∴ A2 = − ⌠ y dx ⌡
−1 Hence required area
-2
2 A2
Fig. 7.13
108
x =4
x-2
-2
2
A1
y= 2 x –
The part A2 lies below x-axis.
y
x =-2
Solution : y = x2 − x − 2 = (x + 1) (x − 2) This curve intersects x-axis at x = − 1 and x=2 Required area = A1 + A2 + A3
A3 4
x
4 −1 2 = ⌠ y dx + ⌠ (− y)dx + ⌠ y dx ⌡ ⌡ ⌡ 2 −2 −1 4 −1 2 = ⌠ (x2 − x − 2) dx + ⌠ − (x2 − x − 2)dx + ⌠ (x2 − x − 2) dx ⌡ ⌡ ⌡ 2 −2 −1 11 9 26 = 6 + 2 + 3 = 15 sq. units
⌡
(b, f (b))
f
x =b
(a, f (a))
x =a
General Area Principle : Let f and g be two continuous curves, with f lying above g. then the area R between f and g, from x = a to x = b, is given by b R = ⌠ (f − g)dx
A bo
ve
:y
= x+ 1
a g (a, g (a)) (b, g (b)) No restriction on f and g where they lie. Both may be lie above or Fig. 7.14 below the x-axis or g lies below and f lies above the x-axis. Example 7.26: Find the area between the line y=x + 1 and the curve y = x2 − 1. Solution : To get the points of intersection of the curves we should solve the equations y = x + 1 and y = x2 − 1. y we get, x2 − 1 = x + 1 x2 − x − 2 = 0 ⇒ (x − 2) (x + 1) = 0 ∴ x = −1 or x = 2 ∴ The line intersects the curve at x = − 1 and x = 2. x b f(x) g(x) -1 1 2 Required area = ⌠ above − below dx Below : y =x2 - 1 ⌡ a Fig. 7.15 2 = ⌠ [(x + 1) − (x2 − 1)]dx
⌡
−1
109
2 2 x2 x3 = ⌠ [2 + x − x2]dx = 2x + 2 − 3 −1 ⌡ −1 1 1 8 9 = 4 + 2 − 3 − − 2 + 2 + 3 = 2 sq. units Example 7.27: Find the area bounded by the curve y = x3 and the line y = x. Solution : The line y = x lies above the curve y = x3 in the first quadrant and y = x3 lies above the line y = x in the third quadrant. To get the points of intersection, solve the curves y = x3, y = x ⇒ x3 = x . We get x = {0, ± 1} 1 0 The required area = A1 + A2 = ⌠ [g(x) − f(x)]dx + ⌠ [f(x) − g(x)]dx
⌡
⌡
−1
0
1 0 = ⌠ (x3 − x)dx + ⌠ (x − x3)dx ⌡ ⌡ 0 −1 0 1 x4 x2 x2 x4 = 4 − 2 + 2 − 4 −1 0 1 1 1 1 = 0 − 4−0 − 2+2 − 0−4 − 0
-1 y = x3
y = x3 x 1
(-1, -1)
1 1 1 1 1 = − 4 + 2 + 2 − 4 = 2 sq. units.
y=
x
0
y=
(1, 1)
x
y
Fig. 7.16
Example 7.28: Find the area of the region enclosed by y2 = x and y = x − 2 Solution : The points of intersection of the parabola y2 = x and the line y = x − 2 are (1, − 1) and (4, 2) y To compute the region [shown in figure (6.17)] by integrating with x =y2 (4, 2) respect to x, we would have to split the region into two parts, because the x equation of the lower boundary y =x - 2 changes at x = 1. However if we (1, -1) integrate with respect to y no splitting is necessary. Fig. 7.17
110
2 Required area = ⌠ (f(y) − g(y) dy ⌡ −1 2 2 y3 y2 = ⌠ [(y + 2) − y2]dy = 2 + 2y − 3 −1 ⌡ −1 4 1 8 1 = 2 − 2 + (4 + 2) − 3 + 3 3 9 9 = 2 + 6 − 3 = 2 sq. units.
y= √(
16
–
x2 )
Example 7.29: Find the area of the region common to the circle x2 + y2 = 16 and the parabola y2 = 6x y Solution : The points of intersection 2 2 2 y =√6x of x + y = 16 and y = 6x are D C (2,2√3) (2, 2 3) and (2, − 2 3) Required area is OABC Due to symmetrical property, the B x required area O 2 OABC = 2 OBC i.e., 2{[Area bounded by y2 = 6x, x = 0, x = 2 and x-axis] + [Area A 2 2 bounded by x + y = 16, x = 2, x = 4 and x-axis]} Fig. 7.18
2 = 2⌠ ⌡
0
4 6x dx + 2⌠ ⌡ 2
x3/22 x = 2 6 3/2 + 2 2 0
16 − x2 dx 42 x 42 − x2 + 2 sin−1 4
8 12 8π 3 − 2 12 + 8π − 3 4 = 3 (4π + 3) =
111
4 2
Example 7.30: Compute the area between the curve y = sin x and y = cosx and the lines x = 0 and x = π y Solution : To find the points of y= intersection solve the two equations. sin 1 x π 1 π Sin x = cos x = ⇒x=4 x 0 2 3π/2 π/4 π/2 -1 y= 5π −1 co s ⇒x= 4 sin x = cos x = x 2 Fig. 7.19 π From the figure we see that cos x > sin x for 0 ≤ x < 4 and sin x > cos x for π 4
⌡
0 a b = 4⌠ a ⌡ 0
0
a2 − x2 dx
Fig. 7.20
112
a
4b a = a ⌠ ⌡ 0
4b x a2 − x2 dx = a 2
a2 x a2 − x2 + 2 sin−1 a
a 0
4b a2 π 4b a2 = a 0 + 2 sin−1(1) − 0 = a 2 2 = π ab sq. units. a π/2 By using parametric form i.e., 4 ⌠ y dx = 4 ⌠ b sin θ (− a sin θ) dθ, we ⌡ ⌡ 0 0 get the same area. Example 7.32: Find the area of the curve y2 = (x − 5)2 (x − 6) (i) between x = 5 and x = 6 (ii) between x = 6 and x = 7 Solution : (i) y2 = (x − 5)2 (x − 6) ∴y = (x − 5) x − 6 This curve cuts the x-axis at x = 5 and at x = 6
⌡
y= (x –
a 7
= 2 ⌠ (x − 5) x − 6 dx ⌡ 6 (Since the curve is symmetrical about x-axis) 7 = 2 ⌠ (t + 1) t dt ⌡ 6 1 = 2 ⌠ (t3/2 + t1/2)dt
⌡ 0
113
5
6
Fig. 7.21 Take t = x − 6 dt = dx t = x−6 x 6 7 t 0 1
x =7
5)√ (x –
6)
When x takes any value between 5 and 6, y2 is negative. ∴ The curve does not exist in the interval 5 < x < 6. Hence the area between the curve at x = 5 and x = 6 is zero. b (ii) Required area = ⌠ ydx
7
x
1
2 2 6 + 10 32 t5/2 t3/2 = 2 5 + 3 = 2 5 + 3 = 2 15 = 15 sq. units 2 2 0 Example 7.33: Find the area of the loop of the curve 3ay2 = x(x − a)2 Solution : y Put y = 0 ; we get x = 0, a It meets the x-axis at x = 0 and x = a ∴ Here a loop is formed between the points (0, 0) and (a, 0) about x-axis. Since the curve is symmetrical about x-axis, the area of the loop is twice the area of the portion above the x-axis. a Required area = 2 ⌠ y dx
(a,0)
Fig. 7.22
⌡ 0
2 a x (x − a) [x3/2 − a x]dx dx = − 3a 3a ⌠ ⌡ ⌡ 0 0 a 8a2 8 3 a2 2 2 5/2 2a 3/2 = x − x =− 3 0 15 3 = 45 3a 5
a = −2⌠
∴ Required area =
8 3 a2 45 sq. units.
Example 7.34: Find the area bounded by x-axis and an arch of the cycloid x = a (2t − sin 2t), y = a (1 − cos 2t) Solution : The curves crosses x-axis when y = 0. ∴ a(1 − cos 2t) = 0 ∴ cos 2t = 1
;
2t = 2nπ, n ∈ z
∴ t = 0, π, 2π, … ∴ One arch of the curve lies between 0 and π
114
x
b Required area = ⌠ y dx ⌡ a π = ⌠ a(1 − cos 2t) 2a (1 − cos 2t) dt
⌡ 0
y = a(1 − cos2t) x = a (2t − sin 2t) dx = 2a(1 − cos 2t) dt
π π π = 2a2 ⌠ (1 − cos 2t)2dt = 2a2 ⌠ (2 sin2t)2dt = 8a2 ⌠ sin4t dt ⌡ ⌡ ⌡ 0 0 0 π/2 = 2 × 8a2 ⌠ sin4t dt ⌡ 0
a 2a ‡ ⌠ f(x) dx = 2⌠ f(2a − x)dx ⌡ ⌡ 0 0
3 1 π = 16a2 4 × 2 × 2 = 3πa2 sq. units.
7.5.2 Volume of solids of revolution : Let f be a non-negative and continuous curve on [a, b] and let R be the region bounded above by the graph of f, below by the x-axis and on the sides by the lines x = a and x = b [Fig 6.23 (a)]. f(x) f(x) x
a
b
Fig. 7.23 (b)
Fig. 7.23(a)
When this region is revolved about the x-axis, it generates a solid having circular cross sections (Fig. 7.23(b)]. Since the cross section at x has radius f(x), the cross-sectional area is A(x) = π [f(x)]2 = πy2 The volume of the solid is generated by moving the plane circular disc [Fig.6.23(b)] along x-axis perpendicular to the disc.
115
b b Therefore volume of the solid is V = ⌠ π [f(x)]2dx = ⌠ π y2 dx
⌡
⌡
a
a y
(ii) If the region bounded by the graph of x = g(y), the y-axis and on the sides by the lines y = c and y = d (Fig. 7.24) then the volume of the solid generated is given by
g(y)
d d V = ⌠ π [g(y)]2dx = ⌠ π x2 dy
⌡
⌡
c
c Fig. 7.24
Example 7.35: Find the volume of the solid that results when the ellipse x2 y2 + = 1 (a > b > 0) is revolved about the minor axis. a2 b2 Solution : Volume of the solid is obtained by revolving the right side of the curve x2 y2 + = 1 about the y-axis. a2 b2 Limits for y is obtained by putting x = 0 ⇒ y2 = b2 ⇒ y = ± b a2 From the given curve x2 = 2 (b2 − y2) b ∴ Volume is given by
y b -a
a -b
Fig. 7.25
b a2 2 2 a2 2 y3 V = ⌠ π x dy = ⌠ π 2 (b − y ) dy = 2π 2 b y − 3 0 ⌡ ⌡ b b c −b d
2
b
= 2π
a2 3 b3 4π 2 b − 3 = 3 a b cubic units b2
Example 7.36: Find the volume of the solid generated when the region enclosed by y = x, y = 2 and x = 0 is revolved about the y-axis.
116
Solution : Since the solid is generated by revolving about the y-axis, rewrite y = x as x = y2. Taking the limits for y, y = 0 and y = 2 (putting x = 0 in x = y2, we get y = 0) d Volume is given by V = ⌠ π x2dy
y
⌡
y =√x y =2
x
Fig. 7.26 2 πy5 32 π 4 = ⌠ π y dy = 5 0 = 5 cubic units. ⌡ 0 c 2
EXERCISE 7.4 (1) Find the area of the region bounded by the line x − y = 1 and (i) x-axis, x = 2 and x = 4 (ii) x-axis, x = − 2 and x = 0 (2) Find the area of the region bounded by the line x − 2y − 12 = 0 and (i) y-axis, y = 2 and y = 5 (ii) y-axis, y = − 1 and y = − 3 (3) Find the area of the region bounded by the line y = x − 5 and the x-axis between the ordinates x = 3 and x = 7. (4) Find the area of the region bounded by the curve y = 3x2 − x and the x-axis between x = − 1 and x = 1. (5) Find the area of the region bounded by x2 = 36y, y-axis, y = 2 and y = 4. (6) Find the area included between the parabola y2 = 4ax and its latus rectum. x2 y2 (7) Find the area of the region bounded by the ellipse 9 + 5 = 1 between the two latus rectums. (8) Find the area of the region bounded by the parabola y2 = 4x and the line 2x − y = 4. (9) Find the common area enclosed by the parabolas 4y2 = 9x and 3x2 = 16y (10) Find the area of the circle whose radius is a Find the volume of the solid that results when the region enclosed by the given curves : (11 to 14) (11) y = 1 + x2, x = 1, x = 2, y = 0 is revolved about the x-axis. (12) 2ay2 = x(x − a)2 is revolved about x-axis, a > 0. 117
(13) y = x3, x = 0, y = 1 is revolved about the y-axis. x2 y2 (14) 2 + 2 = 1 is revolved about major axis a > b > 0. a b (15) Derive the formula for the volume of a right circular cone with radius ‘r’ and height ‘h’. (16) The area of the region bounded by the curve xy = 1, x-axis, x = 1. Find the volume of the solid generated by revolving the area mentioned about x-axis.
7.6. Length of the curve :
(i) If the function f(x) and its derivative f ′(x) are continuous on [a, b] then the arc length L of the curve y = f(x) from x = a to x = b is defined b dy 2 1 + dx dx to be L = ⌠ ⌡ a (ii) Similarly for a curve expressed in the form x = g(y), where g is continuous on [c, d], the arc length L from y = c to y = d is given by d dx 2 L= ⌠ 1 + dy dy ⌡ c (iii) When the equation of the curve y = f(x) is represented in parametric form x = φ(t), y = Ψ(t), α ≤ t ≤ β where φ(t) and Ψ(t) are continuous function with continuous derivatives and φ′(t) does not vanish in the β given interval then L = ⌠ (φ′(t))2 + (Ψ′(t))2 dt
⌡ α
7.7 Surface area of a solid : (i) If the function f(x) and its derivatives f ′(x) are continuous on [a, b], then the surface area of the solid of revolution obtained by the revolution about x-axis, the area bounded by the curve y = f(x) the two ordinates x = a, x = b and x-axis is b dy 2 S.A. = 2π ⌠ y 1 + dx dx ⌡ a
118
f(x) x
Fig. 7.27
y
(ii) Similarly for the curve expressed in the form x = g(y) where g′(y) is continuous on [c, d], the surface area of the solid of revolution obtained by the revolution about y-axis, the area bounded by the curve x = g(y) the two abscissa y = c, y = d and y axis is
g(y)
dx 2 1 + dy dy
d S.A. = 2π ⌠ y
⌡ c
Fig. 7.28 (iii) When the equation of the curve y = f(x) is represented in parametric form x = g(t), y = h(t), α ≤ t ≤ β where g(t) and h(t) are continuous function with continuous derivatives and g′(t) does not vanish in the t=β interval, then S.A. = 2π ⌠ y (g′(t))2 + (h′(t)) 2 dt.
⌡
t=α Example 7.37: Find the length of the curve 4y2 = x3 between x = 0 and x = 1 y Solution : 2 3 4y2 =x3 4y = x Differentiating with respect to x dy 8y dx = 3x2 x =1 x
dy 3x2 dx = 8y
dy 2 1 + dx =
1+
9x4 64y2
Fig. 7.29
4
4
9x 9x 9x 1+ 1 + 16 2 = 3 = 16 × 4y 16x The curve is symmetrical about x-axis. The required length 1 1 dy 2 9x 1/2 1 + dx dx = 2⌠ 1 + 16 dx L = 2⌠ =
1+
⌡
⌡
0
0
119
9x 1 + 16 1 64 9x 3/21 = 2 × 9 3 = 27 1 + 16 0 16 × 2 0 3/2
64 125 61 = 27 64 − 1 = 27 x 2/3 y 2/3 Example 7.38: Find the length of the curve a + a = 1
y
Solution : x = a cos3t, y = a sin3t is the parametric form of the given astroid, where 0 ≤ t ≤ 2π dx 2 dt = − 3a cos t sin t ;
a
-a
a
dy 2 dt = 3a sin t cos t
-a
x
x2/3 + y2/3 =a2/3 Astroid
Fig. 7.30 2
2
dx + dy = dt dt
9a2 cos4t sin2t + 9a2 sin4t cos2t = 3a sin t cos t
Since the curve is symmetrical about both axes, the total length of the curve is 4 times the length in the first quadrant. π But t varies from 0 to 2 in the first quadrant. π/2 ∴ Length of the entire curve = 4 ⌠ ⌡ 0
2
2
dx + dy dt dt dt
π/2 π/2 = 4 ⌠ 3a sin t cos t dt = 6a ⌠ sin 2t dt ⌡ ⌡ 0 0 cos 2tπ/2 = − 3a [cos π − cos 0] 2 0
= 6a . −
= − 3a [− 1 − 1] = 6a
120
Example 7.39: Show that the surface area of the solid obtained by revolving the arc of the curve y = sin x from x = 0 to x = π about x-axis is 2π [ 2 + log (1 + 2)] Solution : y = sin x dy Differentiating with respect to x dx = cos x.
dy 2 1 + dx =
∴
1 + cos2x
b Surface area = ⌠ 2πy
⌡
dy 2 1 + dx dx
a when the area is rotated about the x-axis. π S = ⌠ 2π sin x
⌡
1 + cos2x dx
Put cos x = t
t = cos x
− sin x dx = dt
0 −1 = ⌠ 2π ⌡ 1 t = 4π 2
1 1 + t2 (− dt) = 4π ⌠
⌡
x
0
π
t
1
−1
1 + t2 (dt)
0 1 1 + t2 + 2 log
(t +
1 1 + t2) 0
= 2π [ 2 + log (1 + 2)] − 0 = 2π [ 2 + log (1 + 2)] Example 7.40: Find the surface area of the solid generated by revolving the cycloid x = a(t + sin t), y = a(1 + cos t) about its base (x-axis). Solution : y = 0 ⇒ 1 + cos t = 0 cos t = − 1 ⇒ t = − π, π x = a (t + sin t) ; y = a (1 + cos t) dx dy dt = a (1 + cos t) dt = − a sin t 2
2
dx + dy = dt dt
t a2 (1 + cos t)2 + a2 sin2t = 2a cos 2
121
π t Surface area = ⌠ 2πa (1 + cos t) 2a cos 2 dt ⌡ −π π
π t t t = ⌠ 2π a . 2 cos2 2 . 2 a cos 2 dt = 16π a2 ⌠ cos3 2 dt ⌡ ⌡ 0 −π π/2 t = 16πa2 ⌠ 2cos3 x dx Take 2 = x ⌡ 0 2 = 32πa2I3 = 32πa2 × 3 64 = 3 πa2 sq. units. EXERCISE 7.5 (1) Find the perimeter of the circle with radius a. (2) Find the length of the curve x = a(t − sin t), y = a(1 − cos t) between t = 0 and π. (3) Find the surface area of the solid generated by revolving the arc of the parabola y2 = 4ax, bounded by its latus rectum about x-axis. (4) Prove that the curved surface area of a sphere of radius r intercepted between two parallel planes at a distance a and b from the centre of the sphere is 2πr (b − a) and hence deduct the surface area of the sphere. (b > a).
122
8. DIFFERENTIAL EQUATIONS 8.1. Introduction : One of the branches of Mathematics conveyed clearly in the principal language of science called “Differential equations”, plays an important role in Science, Engineering and Social Sciences. Let us analyse a few of the examples cited below. (1) Suppose that there are two living species which depend for their survival on a common source of food supply. This fact results in a competition in consuming the available food. The phenomenon, is commonly noticed in the plant life having common supply of water, fertilizer and minerals. However, whenever the competition between two species begins, the growth rate of one is retarded and we can note that the rate of retardation is naturally proportional to the size of the other species present at time t. This situation can be expressed as a Mathematical model whose solution would help us to determine the time at which one species would become extinct. (2) Several diseases are caused by spread of an infection. Suppose that the susceptible population of a town is p. One person gets the infection. Because of contact another susceptible person is also infected. This process continues to cover the entire susceptible population. With some assumptions to simplify the mathematical considerations this situation can be framed into a mathematical model and a solution can be determined which would provide informations regarding the spread of the epidemic in the town. (3) If a dead body is brought for a medical examination at a particular time, the exact time of death can be determined by noting the temperature of the body at various time intervals, formulating it into a mathematical problem with available initial conditions and then solving it. (4) The determination of the amount of a radioactive material that disintegrates over a period of time is yet another mathematical formulation which yield the required result. (5) Several examples exist in which two nations have disputes on various issues. Each nation builds its own arms to defend the nation from attack. Naturally a spirit of race in building up arms persists between conflicting nations. A small grievance quite often creates a war-like
123
situation and adds to increasing the level of arms. These commonly experienced facts can be presented in a mathematical language and hence solved. It is a fact that such a model has been tested for some realistic situations that had prevailed in the First and Second World War between conflicting nations. From the above examples it is found that the mathematical formulation to all situations turn out to be differential equations. Thus the latent significance of differential equations in studying physical phenomena becomes apparent. This branch of Mathematics called ‘Differential Equations’ is like a bridge linking Mathematics and Science with its applications. Hence it is rightly considered as the language of Sciences. Galileo once conjectured that the velocity of a body falling from rest is proportional to the distance fallen. Later he decided that it is proportional to the time instead. Each of these statements can be formulated as an equation involving the rate of change of an unknown function and is therefore an ds example of what Mathematicians call a Differential Equation. Thus dt = kt is a differential equation which gives velocity of a falling body from a distance s proportional to the time t. Definition: An equation involving one dependent variable and its derivatives with respect to one or more independent variables is called a Differential Equation. dy If y = f(x) is a given function, then its derivative dx can be interpreted as the rate of change of y with respect to x. In any natural process the variables involved and their rates of change are connected with one another by means of the basic scientific principles that govern the process. When this expression is written in mathematical symbols, the result is often a differential equation. Thus a differential equation is an equation in which differential coefficients occur. Its importance can further be realised from the fact that every natural phenomena is governed by differential equations. Differential equation are of two types. (i) Ordinary and (ii) Partial. In this chapter we concentrate only on Ordinary differential equations. Definition : An ordinary differential equation is a differential equation in which a single independent variable enters either explicitly or implicitly.
124
dy d2y dy For instance (i) dx = x + 5 (ii) (y′)2 + (y′)3 + 3y = x2 (iii) 2 − 4 dx + 3y = 0 dx are all ordinary differential equations.
8.2 Order and degree of a differential equation : Definition : The order of a differential equation is the order of the highest order derivative occurring in it. The degree of the differential equation is the degree of the highest order derivative which occurs in it, after the differential equation has been made free from radicals and fractions as far as the derivatives are concerned. The degree of a differential equation does not require variables r, s, t … to be free from radicals and fractions. Example 8.1: Find the order and degree of the following differential equations: 3
d3y d2y dy5 + + +y=7 (i) dx3 dx2 dx
dy dx (ii) y = 4 dx + 3x dy
3
d2y dy 2 4 (iii) 2 = 4 + dx dx
(iv) (1 + y′)2 = y′2
Solution : (i) The order of the highest derivative in this equation is 3. The degree of the highest order is 1. ∴ (order, degree) = (3, 1) dx 1 dy dy (ii) y = 4 dx + 3x dy ⇒ y = 4dx + 3x dy dx dy Making the above equation free from fractions involving dx we get dy dy 2 y . dx = 4dx + 3x Highest order = 1 Degree of Highest order = 2 (order, degree) = (1, 2) 3
d2y dy 2 4 (iii) 2 = 4 + dx dx To eliminate the radical in the above equation, raising to the power 4 on 4
3
dy 2 d2y both sides, we get 2 = 4 + dx . Clearly (order, degree) = (2, 4). dx
125
(iv) (1 + y′)2 = y′2 ⇒ 1 + y′2 + 2y′ = y′2 from which it follows that dy 2 dx + 1 = 0 ∴ (order, degree) = (1, 1).
8.3 Formation of differential equations :
-2 x
+8
Let f (x, y, c1) = 0 be an equation containing x, y and one arbitrary constant c1. If c1 is eliminated by differentiating f (x, y, c1) = 0 with respect to the dy independent variable once, we get a relation involving x, y and dx , which is evidently a differential equation of the first order. Similarly, if we have an equation f(x, y, c1, c2) = 0 containing two arbitrary constants c1 and c2, then by differentiating this twice, we get three equations (including f). If the two arbitrary constants c1 and c2 are eliminated from these equations, we get a differential equation of second order. In general if we have an equation f(x, y, c1, c2, …cn) = 0 containing n arbitrary constants c1, c2 … cn, then by differentiating n times we get (n + 1) equations in total. If the n arbitrary constants c1, c2, … cn are eliminated we get a differential equation of order n. Note : If there are relations involving these arbitrary constants then the order of the differential equation may reduce to less than n. Illustration : Let us find the differential equation of straight lines y = mx + c where both m and c are arbitrary constants. Since m and c are two arbitrary constants differentiating twice we get y dy dx = m y=
d2y =0 dx2
y= - 2x
-x
+ 4
126
x y=
Both the constants m and c are seen to be eliminated. Therefore the required differential equation is d2 y =0 dx2
Fig. 8.1
Note : In the above illustration we have taken both the constants m and c as arbitrary. Now the following two cases may arise. Case (i) : m is arbitrary and c is fixed. Since m is the only arbitrary constant in y = mx + c ; … (1) y
1/3)x y =(
C
y=
-(1
/3) x
+c
+c
x
y= -2 c
Fig. 8.2 Case (ii) : c is an arbitrary constant and m is a fixed constant.
x+
Differentiating once we get dy … (2) dx = m Eliminating m between (1) and (2) we get the required differential equation dy x dx − y + c = 0
y
Since c is the only arbitrary constant differentiating once we get dy dx = m. Clearly c is eliminated from the above equation. Therefore the required differential equation is dy dx = m.
x
Fig. 8.3 Example 8.2: Form the differential equation from the following equations. (i) y = e2x (A + Bx) 2
2
(iii) Ax + By = 1 Solution : (i)
(ii) y = ex (A cos 3x + B sin 3x) (iv) y2 = 4a(x − a)
y = e2x (A + Bx)
ye−2x = A+ Bx … (1) Since the above equation contains two arbitrary constants, differentiating twice, we get y′e−2x − 2y e−2x = B {y′′e−2x − 2y′ e−2x} − 2{y′e−2x − 2y e−2x} = 0 e−2x {y′′ − 4y′ + 4y} = 0 [‡ e−2x ≠ 0] y′′ − 4y′ + 4y = 0 is the required differential equation.
127
(ii)
y = ex (A cos 3x + B sin 3x) ye−x = A cos 3x + B sin 3x We have to differentiate twice to eliminate two arbitrary constants y′e−x − ye−x = − 3A sin 3x + 3 B cos 3x y′′ e−x − y′e−x − y′e − x + ye−x = − 9 (A cos 3x + B sin 3x) i.e., e−x (y′′ − 2y′ + y) = − 9ye−x (‡ e−x ≠ 0)
⇒ y′′ − 2y′ + 10y = 0
(iii) Ax2 + By2 = 1 Differentiating, 2Ax + 2Byy′ = 0 i.e., Ax + Byy′ = 0
… (1) … (2)
Differentiating again, A + B (yy′′ + y′2) = 0 Eliminating A and B between (1), (2) and (3) we get
… (3)
x x 1
2
=0 0
y2
−1
yy′
0
yy′′ + y′2
⇒ (yy′′ + y′2) x − yy′ = 0
(iv) y2 = 4a(x − a) Differentiating, 2yy′ = 4a Eliminating a between (1) and (2) we get yy′ y2 = 2yy′ x − 2
… (1) … (2)
⇒ (yy′)2 − 2xyy′ + y2 = 0 EXERCISE 8.1 (1) Find the order and degree of the following differential equations. dy 2 (ii) y′ + y2 = x (i) dx + y = x (iii)
y′′ + 3y′2 + y3 = 0
(v)
d2y dy d3y2 y + − dx + 3 = 0 dx2 dx
(iv)
d2y +x= dx2
(vi)
y′′ = (y − y′3)3
3
(vii) y′ + (y′′)2 = (x + y′′)2 dx dy 2 (ix) dx + x = dy + x2
dy y + dx 2
(viii) y′ + (y′′)2 = x(x + y′′)2 (x)
128
sinx (dx + dy) = cosx (dx − dy)
(2) Form the differential equations by eliminating arbitrary constants given in brackets against each (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix)
y2 = 4ax y = ax2 + bx + c xy = c2 x2 y2 + =1 a2 b2
{a} {a, b} {c} {a, b}
y = Ae2x + Be−5x y = (A + Bx)e3x y = e3x {C cos 2x + D sin 2x) y = emx y = Ae2x cos (3x + B)
{A, B} {A, B} {C, D} {m} {A, B}
a (3) Find the differential equation of the family of straight lines y = mx + m when (i) m is the parameter ; (ii) a is the parameter ; (iii) a, m both are parameters (4) Find the differential equation that will represent the family of all circles having centres on the x-axis and the radius is unity.
8.4 Differential equations of first order and first degree : In this section we consider a class of differential equations, the order and degree of each member of the class is equal to one. For example, x+y (i) yy′ + x = 0 (ii) y′ + xy = sinx (iii) y′ = (iv) x dy + y dx = 0 x−y
Solutions of first order and first degree equations: We shall consider only certain special types of equations of the first order and first degree. viz., (i) Variable separable (ii) Homogeneous (iii) Linear.
8.4.1 Variable separable : Variables of a differential equation are to be rearranged in the form f1(x) g2(y) dx + f2(x) g1(y) dy = 0 i.e., the equation can be written as f2(x)g1(y)dy = − f1(x) g2(y) dx g1(y) f1(x) ⇒ g (y) dy = − f (x) dx 2 2
129
⌠ g1(y) ⌠ f1(x) The solution is therefore given by g (y) dy = − f (x) dx + c ⌡
⌡
2
2
dy Example 8.3: Solve : dx = 1 + x + y + xy Solution : The given equation can be written in the form dy dx = (1 + x) + y(1 + x) dy ⇒ dx = (1 + x) (1 + y) dy ⇒ 1 + y = (1 + x)dx Integrating, we have x2 log (1 + y) = x + 2 + c, which is the required solution. Example 8.4: Solve 3ex tan y dx + (1 + ex) sec2y dy = 0 Solution : The given equation can be written in the form 3ex sec2y dx + tan y dy = 0 1 + ex Integrating, we have 3 log (1 + ex) + log tan y = log c 3
⇒ log [tan y (1 + ex) ] = log c 3
⇒ (1 + ex) tan y = c, which is the required solution. 1 Note : The arbitrary constant may be chosen like c, c , log c, ec etc depending upon the problem. 1
dy 1 − y22 Example 8.5: Solve dx + =0 1 − x2 Solution : The given equation can be written as 1
dy 1 − y22 = − ⇒ dx 1 − x2
dy 1−y
2
=−
Integrating, we have sin−1y + sin−1x = c ⇒ sin−1 [x ⇒ x
1 − y2 + y
1 − y2 + y
1 − x2] = c
1 − x2 = C is the required solution.
130
dx 1 − x2
y 1 − y2 dx + x dy = 0
Example 8.6: Solve : ex
Solution : The given equation can be written as −y dy xexdx = 1 − y2 Integrating, we have y x ⌠ ⌡ xe dx = − ⌠ 1 − y2 dy ⌡ 1 dt ⌠ex dx = 2 ⌠ where t = 1 − y2 so that −2y dy = dt ⇒ xex − ⌡ ⌡ t
1
1 t2 ⇒ xe − e = 2 1/2 + c x
x
⇒ xex − ex = x
x
t+c
2
⇒ xe − e −
1 − y = c which is the required solution. dy Example 8.7: Solve : (x + y)2 dx = a2 Solution : Put x + y = z. Differentiating with respect to x we get dy dz dy dz 1 + dx = dx i.e., dx = dx − 1 dz The given equation becomes z2 dx − 1 = a2 a2 z2 dz dz = dx ⇒ dx − 1 = 2 or 2 z z + a2 2
Integrating we have,
⌠ z 2 2 dz = ⌠ ⌡ dx ⌡z + a
2 2 2 a2 ⌠z +a −a ⌠ 2 2 dz = x + c ⇒ 1 − 2 2 dz = x + c ⌡ z +a ⌡ z +a
1 z ⇒ z − a2 . a tan−1 a = x + c x+y ⇒ x + y − a tan−1 a = x + c
(‡ z = x + y) x+y i.e., y − a tan−1 a = c, which is the required solution.
131
4
Example 8.8: Solve : x dy = (y + 4x5 ex )dx Solution : 4
xdy − y dx = 4x5 ex dx 4 xdy − ydx = 4x3 ex dx 2 x 4 xdy − ydx Integrating we have, ⌠ =⌠ ⌡4x3 ex dx x2 ⌡
y ⌠ et dt ⇒ ⌠ d x = ⌡ ⌡
where t = x4
y t x = e +c 4 y i.e., x = ex + c which is the required solution.
⇒
Example 8.9: Solve: (x2−y)dx + (y2 − x) dy = 0, if it passes through the origin. Solution : (x2 − y)dx + (y2 − x) dy = 0 x2dx + y2dy = xdy + ydx x2dx + y2 dy = d(xy) x3 y3 3 + 3 = xy + c Since it passes through the origin, c = 0 Integrating we have,
∴ the required solution is
x3 y3 3 3 3 + 3 = xy or x + y = 3xy
Example 8.10 : Find the cubic polynomial in x which attains its maximum value 4 and minimum value 0 at x = − 1 and 1 respectively. Solution : Let the cubic polynomial be y = f(x). Since it attains a maximum at x = −1 and a minimum at x = 1. dy dx = 0 at x = − 1 and 1 dy 2 dx = k (x + 1) (x − 1) = k(x − 1) Separating the variables we have dy = k(x2 − 1) dx
132
2 ⌠dy = k ⌠ ⌡(x − 1) dx ⌡
x3 y = k 3 − x + c
… (1)
when x = − 1, y = 4 and when x = 1, y = 0 Substituting these in equation (1) we have 2k + 3c = 12 ; − 2k + 3c = 0 On solving we have k = 3 and c = 2. Substituting these values in (1) we get the required cubic polynomial y = x3 − 3x + 2. Example 8.11 : The normal lines to a given curve at each point (x, y) on the curve pass through the point (2, 0). The curve passes through the point (2, 3). Formulate the differential equation representing the problem and hence find the equation of the curve. Solution : dx Slope of the normal at any point P(x, y) = − dy Slope of the normal AP =
y−0 x−2
dx ∴ − dy =
y ⇒ ydy = (2 − x)dx x−2
x2 y2 Integrating both sides, 2 = 2x − 2 + c Since the curve passes through (2, 3) 9 4 5 5 2 = 4 − 2 + c ⇒ c = 2 ; put c = 2 in (1),
… (1)
x2 5 y2 2 2 = 2x − 2 2 + 2 ⇒ y = 4x − x + 5 EXERCISE 8.2 Solve the following : (1) sec 2x dy − sin5x sec2ydx = 0
(2) cos2xdy + yetanxdx = 0
(3) (x2 − yx2)dy + (y2 + xy2)dx = 0
(4) yx2dx + e−xdy = 0 dy (6) dx = sin(x + y)
(5) (x2 + 5x + 7) dy + dy (7) (x + y)2 dx = 1
9 + 8y − y2 dx = 0
(8) ydx + xdy = e−xy dx if it cuts the y-axis.
133
8.4.2 Homogeneous equations : Definition : A differential equation of first order and first degree is said to be dy f1(x, y) dy y homogeneous if it can be put in the form dx = f x or dx = f2(x, y) Working rule for solving homogeneous equation : By definition the given equation can be put in the form dy y dx = f x y = νx To solve (1) put Differentiating (2) with respect to x gives dν dy dx = ν + x dx Using (2) and (3) in (1) we have dν dν ν + x dx = f(ν) or x dx = f(ν) − ν Seperating the variables x and ν we have dν dx ⌠ dν x = f(ν) − ν ⇒ log x + c = ⌡ f(ν) − ν y where c is an arbitrary constant. After integration, replace ν by x . dy y y Example 8.12: Solve : dx = x + tan x Solution : Put y = vx dν L.H.S. = ν + x dx ; R.H.S. = v + tan v dν dx cos ν dv ∴ ν + x dx = ν + tan ν or x = sinν Integrating, we have logx = log sin ν + log c ⇒ x = c sin ν y i.e., x = c sin x , Example 8.13: Solve : (2 xy − x) dy + ydx = 0 −y dy Solution : The given equation is dx = 2 xy − x Put y = vx
134
… (1) … (2) … (3)
dv −v v L.H.S. = v + x dx ; R.H.S. = = 2 v−1 1−2 v v dv ∴ v + x dx = 1−2 v 2v v dx dv 1 − 2 v ⇒ ⇒ x dx = dv = 2 x 1 −2 v v v 1 dx i.e., v−3/2 − 2. v dv = 2 x ⇒ − 2v−1/2 − 2 log v = 2 log x + 2 log c − v−1/2 = log (v x c) − x/y x/y x − or ye =c y = log(cy) ⇒ cy = e Note : This problem can also be done easily by taking x = vy Example 8.14: Solve : (x3 + 3xy2)dx + (y3 + 3x2y)dy = 0 Solution :
x3 + 3xy2 dy = − dx y3 + 3x2y Put y = νx dν x3 + 3xy2 1 + 3ν2 L.H.S. = ν + x dx ; R.H.S. = − 3 2 =− 3 y + 3x y ν + 3ν dν 1 + 3ν2 ∴ ν + x dx = − 3 ν + 3ν ν4 + 6ν2 + 1 dν ⇒ x dx = − ν3 + 3ν ⇒
4dx 4ν3 + 12ν = dν − x ν4 + 6ν2 + 1
Integrating, we have 4 log x = − log (ν4 + 6ν2 + 1) + log c log[x4(ν4 + 6ν2 + 1)] = log c i.e., x4 (ν4 + 6ν2 + 1) = c or y4 + 6x2y2 + x4 = c Note (i) : This problem can also be done by using variable separable method.
135
Note (ii) : Sometimes it becomes easier in solving problems of the type dx f1(x/y) dy = f2(x/y) . The following example explains this case. Example 8.15: Solve : (1 + ex/y)dx + ex/y(1 − x/y) dy = 0 given that y = 1, where x = 0 Solution : The given equation can be written as dx (x / y − 1)ex/y dy = 1 + ex/y Put x = νy
… (1)
dν (v − 1)ev L.H.S. = ν + y dy ; R.H.S. = 1 + ev dν (ν − 1)eν ∴ ν + y dy = 1 + eν dν (eν + ν) or y dy = − 1 + eν ⇒ Integrating we have,
dy (eν + 1) = − dν y eν + ν
log y = − log (eν + ν) + log c
or y(eν + ν) = c ⇒ yex/y + x = c Now y = 1 when x = 0 ⇒ 1e0 + 0 = c ⇒ c = 1 ∴ yex/y + x = 1 Example 8.16: Solve : xdy − ydx = x2 + y2 dx Solution : From the given equation we have dy y + dx =
x2 + y2 x
… (1)
Put y = νx v + 1 + v2 dν L.H.S. = ν + x dx ; R.H.S. = 1 dν ∴ ν + x dx = ν +
136
1 + ν2
or
dx x =
dν 1 + ν2
Integrating, we have,log x + logc = log [v + i.e., xc = ν +
v2 + 1 ]
ν2 + 1 ⇒ x2c = y +
(y2 + x2)
EXERCISE 8.3 Solve the following : dy y y2 (1) dx + x = 2 x
dy y(x − 2y) (2) dx = x(x − 3y)
(3) (x2 + y2) dy = xy dx
dy (4) x2 dx = y2 + 2xy given that y = 1, when x = 1. (5) (x2 + y2) dx + 3xy dy = 0 (6) Find the equation of the curve passing through (1, 0) and which has slope y 1 + x at (x, y).
8.4.3 Linear Differential Equation : Definition : A first order differential equation is said to be linear in y if the power of the dy terms dx and y are unity. dy dy For example dx + xy = ex is linear in y, since the power of dx is one and dy also the power of y is one. If a term occurs in the form y dx or y2, then it is not linear, as the degree of each term is two. A differential equation of order one satisfying the above condition can dy always be put in the form dx + Py = Q, where P and Q are function of x only. Similarly a first order linear differential equation in x will be of the form dx dy + Px = Q where P and Q are functions of y only. The solution of the equation which is linear in y is given as ye∫ Pdx= ∫Qe∫ Pdx dx + c where e∫ Pdx is known as an integrating factor and it is denoted by I.F.
137
Similarly if an equation is linear in x then the solution of such an equation becomes x e∫ Pdy = ∫Q e∫ Pdy dy + c
(where e∫ Pdy is I.F.)
We frequently use the following properties of logarithmic and exponential functions : 1 (i) elog A = A (ii) em log A = Am (iii) e− m log A = m A dy Example 8.17 : Solve : dx + y cot x = 2 cos x dy Solution : The given equation is of the form dx + Py = Q. This is linear in y. Here P = cotx and Q = 2 cos x I.F. = e∫ Pdx= e∫ cot x dx = elog sin x = sin x ∴ The required solution is y (I.F.) = ⌠ ⌡(Q (I.F.)) dx + c ⇒ y(sinx) = ⌠ ⌡2 cosx sin x dx + c ⇒ y sin x = ⌠ ⌡sin 2x dx + c cos 2x 2 +c ⇒ 2y sin x + cos 2x = c dy Example 8.18 : Solve : (1 − x2) dx + 2xy = x (1 − x2) x 2x dy . This is linear in y Solution: The given equation is dx + 2 y = 1 − x (1 − x2) ⇒ y sin x = −
2x 2 Here ∫ Pdx = ⌠ 1 − x2 dx = − log (1 − x ) ⌡ 1 I.F. = e∫ Pdx = 1 − x2 The required solution is y.
1 =⌠ 1 − x2 ⌡ ∴
x 2
(1 − x )
×
1 dx. Put 1 − x2 = t ⇒ −2xdx = dt 1 − x2
y − 1 −3/2 ⌠ dt + c 2 = 2 ⌡t 1−x
138
y = t−1/2 + c 1 − x2 y 1 ⇒ +c 2 = 1−x 1 − x2
⇒
Example 8.19 : Solve : (1 + y2)dx = (tan−1y − x)dy dx Solution : The given equation can be written as dy +
x tan−1y = . 1 + y2 1 + y2
This is linear in x. Therefore we have ∫Pdy = ⌠
1 −1 2 dy = tan y 1 + y ⌡
−1 I.F. = e∫ Pdy = etan y The required solution is
xe
tan−1y
−1 tan−1y =⌠ dy + c ⌡e tan y 1 + y2
put tan−1y = t dy ∴ 1 + y2 = dt
−1 t ⇒ xe tan y = ⌠ ⌡ e . t dt + c −1 ⇒ xe tan y = tet − et + c
−1 −1 ⇒ xe tan y = e tan y (tan−1y − 1) + c dy Example 8.20 : Solve : (x + 1) dx − y = ex(x + 1)2 dy y Solution : The given equation can be written as dx − x + 1 = ex(x + 1) 1 This is linear in y. Here ∫Pdx = − ⌠ x + 1 dx = − log (x + 1) ⌡ 1 So I.F. = e ∫ Pdx = e−log(x + 1) = x + 1 ∴ The required solution is 1 1 y . x + 1 = ∫ex (x + 1) x + 1 dx + c = ∫ex dx + c y i.e., x + 1 = ex + c
139
dy Example 8.21 : Solve : dx + 2y tanx = sinx Solution : This is linear in y. Here ∫ Pdx = ⌠ ⌡2 tanx dx = 2 log secx 2 I.F. = e ∫ Pdx = elog sec x = sec2x
The required solution is y sec2x = ∫ sec2x . sinx dx = ∫ tanx sec x dx 2 ⇒ y sec x = sec x + c or y = cos x + c cos2x EXERCISE 8.4 Solve the following : dy (1) dx + y = x
dy 4x 1 (2) dx + 2 y = 2 x +1 (x + 1)2
dx x tan−1y (3) dy + = 1 + y2 1 + y2 dy y (5) dx + x = sin(x2)
dy (4) (1 + x2) dx + 2xy = cosx
dy (6) dx + xy = x dy (8) (y − x) dx = a2 (7) dx + xdy = e−y sec2y dy (9) Show that the equation of the curve whose slope at any point is equal to y + 2x and which passes through the origin is y = 2(ex − x − 1)
8.5 Second order linear differential equations with constant coefficients : A general second order non-homogeneous linear differential equation with constant coefficients is of the form … (1), a0y′′ + a1y′ + a2y = X where a0, a1, a2 are constants a0 ≠ 0, and X is a function of x. The equation a0y′′ + a1y′ + a2y = 0, a0 ≠ 0 … (2) is known as a homogeneous linear second order differential equation with constant coefficients, To solve (1), first we solve (2). To do this we proceed as follows : Consider the function y = epx, p is a constant. Now y′ = pepx and y′′ = p2epx
140
Note that the derivatives look similar to the function y = epx itself and if L(y) = a0y′′ + a1y′ + a2y then L(y) = L(epx) = (a0p2epx + a1pepx + a2 epx) = (a0p2 + a1p + a2)epx Hence if L(y) = 0 then it follows that (a0p2 + a1p + a2)epx = 0. Since epx ≠ 0 we get that a0p2 + a1p +a2 = 0 … (3) Note that epx satisfies the equation L(y) = a0y′′ + a1y′ + a2y = 0 then p must satisfy a0p2 + a1p + a2 = 0. Moreover if the various derivatives of a function look similar in form to the function itself then epx will be an ideal candidate to solve a0y′′ + a1y′ + a2y = 0 . Hereafter we will consider only those set of differential equations which admits epx as one of the solutions. Hence we have the following : Theorem : If λ is a root of a0p2 + a1p +a2 = 0, then eλx is a solution of a0y′′ + a1y′ + a2y = 0 8.5.1 Definition : The equation a0p2 + a1p + a2 = 0 is called the characteristic equation of (2). In general the characteristic equation has two roots say λ1 and λ2. Then the following three cases do arise. Case (i) : λ1 and λ2 are real and distinct. λ x λ x In this case, by the above theorem e 1 and e 2 are solutions of (2), and the λ x λ x linear combination y = c1 e 1 + c2e 2 is also a solution of (2). λ x λ x λ x λ x λ x λ x For L(y) = a0(c1e 1 + c2e 2 )′′+ a1(c1e 1 + c2e 2 )′ + a2(c1e 1 + c2e 2 ) λ x λ x = c1(a0λ12 + a1λ1 + a2)e 1 + c2(a0λ22 + a1 λ2 + a2)e 2 = c1 . 0 + c2 . 0 = 0. λ x λ x and the solution c1e 1 + c2e 2 is known as the complementary function. Case (ii) : λ1 and λ2 are complex λ1 = a + ib and λ2 = a − ib
In this case as the two roots λ1 and λ2 are complex from theory of equations λ x e 1 = e(a + ib)x = eax . eibx= eax (cos bx + i sin bx) and
141
λ x e 2 = eax (cos bx − i sin bx) Hence the solution λ x λ x y = c1e 1 + c2e 2 = eax [(c1 + c2) cos bx + i(c1 − c2) sinbx]
= eax [A cos bx + B sin bx] where A = c1 + c2 and B = (c1 − c2)i and the complementary function is eax [A cos bx + B sin bx]. Case (iii) :The roots are real and equal λ1 = λ2 (say) Clearly eλ1x is one of the solutions of (2). By using the double root property, we will obtain xeλ1x as the other solution of (2). Now the linear combination c1eλ1x + c2xeλ1x becomes the solution. i.e., y = (c1 + c2x)eλ1x is the solution or C.F. The above discussion is summarised as follows : Given a0y′′ + a1y′ + a2y = 0 Determine its characteristic equation a0p2 + a1p + a2 = 0 … (3). Let λ1, λ2 be the two roots of (3), then the solution of (2) is λ x
y=
λ x
Ae 1 + Be 2 if λ1 and λ2 are real and distinct eax (A cos bx + B sin bx) if λ1 = a + ib and λ2 = a − ib λ x (A + Bx)e 1 if λ1 = λ2 (real)
A and B are arbitrary constants. General solution : The general solution of a linear equation of second order with constant co-efficient consists of two parts namely the complementary function and the particular integral. Working rule : To obtain the complementary function (C.F.) we solve the equation d2y dy 2 + a1 dx + a2y = 0 and obtain a solution y = u (say). Then the general dx solution is given by y = u + ν where ν is called the particular integral of (1). The function u, the complementary function is associated with the homogeneous equation and v, the particular integral is associated with the term X. If X = 0 then the C.F. becomes the general solution of the equation. a0
142
Note : In this section we use the differential operators d2 dy d D ≡ dx and D2 ≡ 2 ; Dy = dx dx
; D2y =
d2y dx2
8.5.2 Method for finding Particular Integral : (a) Suppose X is of the form eαx, α a constant D(eαx) = αeαx ; D2 (eαx) = α2 eαx … Dn(eαx) = αneαx , then f(D) eαx = f(α) eαx … (1) 1 Note that f(D) is the inverse operator to f(D). 1 Operating both sides of (1) by f(D) we have, 1 1 f(D) . f(D) eαx = f(D) f(α)eαx 1 1 ⇒ eαx = f(D) f(α)eαx (‡ f(D) . f(D) = I) 1 1 αx e = f(D) eαx then f(α) 1 αx 1 e represented symbolically. …(2) Thus the P.I. is given by f(D)eαx= f(α) (2) holds when f(α) ≠ 0. If f(α) = 0 then D = α is a root of the characteristic equation for the differential equation f(D) = 0 ⇒ D − α is a factor of f(D). Let f(D) = (D − α) θ(D), where θ(α) ≠ 0 then 1 1 αx . αx f(D) e = (D − α) θ(D) e 1 . 1 = eαx D − α θ(α) 1 1 eαx … (3) = θ(α) D − α 1 Put eαx = y ⇒ (D − α)y = eαx (D − α)
⌠eαx. e− ∫ α dx. dx then ye− ∫ α dx = ⌡ ⌠eαx e−αx dx ⇒ y = eαxx i.e., ye−αx = ⌡
143
Substituting in (3) we have 1 1 αx αx f(D) e = θ(α) xe If further, θ(α) = 0, then D = α is a repeated root for f(D) = 0. x2 1 Then f(D) eαx = 2 eαx Example 8.22 : Solve : (D2 + 5D + 6)y = 0 or y″ + 5y′ + 6y = 0 Solution : To find the C.F. solve the characteristic equation p2 + 5p + 6 = 0 ⇒ (p + 2) (p + 3) = 0 ⇒ p = − 2 and p = − 3 −2x The C.F. is Ae + Be−3x. Hence the general solution is y = Ae−2x + Be−3x where A and B are arbitrary constants. Example 8.23 : Solve : (D2 + 6D + 9)y = 0 Solution : The characteristic equation is p2 + 6p + 9 = 0 i.e., (p + 3)2 = 0 ⇒ p = − 3, − 3 The C.F. is (Ax + B)e− 3x Hence the general solution is y = (Ax + B)e−3x where A and B are arbitrary constants. Example 8.24 : Solve : (D2 + D + 1)y = 0 Solution : The characteristic equation is p2 + p + 1 = 0 −1±
−1 3 = 2 ±i 2 3 3 Hence the general solution is y = e−x/2 A cos 2 x + B sin 2 x where A and B are arbitrary constant. ∴ p=
1−4
2
Example 8.25 : Solve : (D2 − 13D + 12)y = e−2x Solution : The characteristic equation is p2 − 13p + 12 = 0 ⇒ (p − 12) (p − 1) = 0 ⇒ p = 12 and 1 The C.F. is Ae12x + Bex Particular integral
P.I. =
1 e−2x D − 13D + 12 2
144
1 1 e−2x = 4 + 26 + 12 e−2x (− 2) − 13 (− 2) + 12 1 = 42 e−2x 1 Hence the general solution is y = CF + PI ⇒ y = Ae12x + Bex + 42 e−2x =
2
Example 8.26 : Solve : (D2 + 6D + 8)y = e−2x Solution : The characteristic equation is p2 + 6p + 8 = 0 ⇒ (p + 4) (p + 2) = 0 ⇒ p = − 4 and − 2 The C.F. is Ae− 4x + Be−2x Particular integral
P.I. =
=
1 1 e−2x e−2x = (D + 4) (D + 2) D + 6D + 8 Since f(D) = (D + 2) θ(D)) 2
1 xe−2x 1 xe−2x =2 θ (−2)
1 xe−2x Hence the general solution is y = Ae− 4x + Be−2x + 2 Example 8.27 : Solve : (D2 − 6D + 9)y = e3x Solution : The characteristic equation is p2 − 6p + 9 = 0 i.e., (p − 3)2 = 0 ⇒ p = 3, 3 The C.F. is (Ax + B)e3x Particular integral
P.I. = =
1 e3x D − 6D + 9 2
1 e3x x2 e3x = 2 2 (D − 3)
x2 e3x Hence the general solution is y = (Ax + B)e3x + 2 1
2
Example 8.28 : Solve : (2D + 5D + 2)y = e
−2x
Solution : The characteristic equation is 2p2 + 5p + 2 = 0 ∴ p=
−5±
25 − 16 − 5 ± 3 = 4 4
145
1 ⇒ p = − 2 and − 2 1
−2x + Be−2x The C.F. is Ae 1
Particular integral
1 P.I. = 2 2D + 5D + 2
− x e 2
1
− x e 2 1 = 1 2D + 2 (D + 2)
1
1 − x xe 2 1 x e− 2 x 1 =3 = 1 θ − 2 . 2 1
Hence the general solution is y = Ae
−2x
1
+ Be
−2x
1 −2x +3xe
Caution : In the above problem we see that while calculating the particular integral the coefficient of D expressed as factors is made unity. (b) When X is of the form sin ax or cos ax. Working rule : Formula 1: Express f(D) as function of D2, say φ(D2) and then replace D2 by − a2. If φ(− a2) ≠ 0. Then we use the following result. 1 1 1 P.I. = f(D) cos ax = cos ax 2 cos ax = φ(D ) φ(− a2) 1 1 1 For example PI = 2 cos 2x = cos 2x = − 3 cos 2x 2 D +1 −2 +1 Formula 2 : Sometimes we cannot form φ(D2). Then we shall try to get φ(D, D2), that is, a function of D and D2. In such cases we proceed as follows : 1 cos3x For example : P.I. = 2 D − 2D + 1 1 cos3x = Replace D2 by − 32 2 − 3 − 2D + 1 − 1 cos3x = 2(D + 4) − 1 D − 4 cos3x = 2 Multiply and divide by D − 4 D2 − 42
146
−1 1 = 2 (D − 4) cos3x 2 − 3 − 42 1 = 50 (D − 4) cos 3x 1 1 = 50 [D cos 3x − 4 cos 3x] = 50 [− 3 sin 3x − 4 cos 3x] Formula 3 : If φ(− a2) = 0 then we proceed as shown in the following example: 1 1 cosax cosax = 2 Example : P.I. = D + a2 φ(D2) 1 cosax = (D + ia) (D − ia) eiax 1 1 xeiax = R.P. = R.P. (D + ia) (D − ia) θ(ia)
xeiax = Real part of 2ia as θ (ia) = 2ia −x = 2a [Real part of i [cos ax + i sin ax]] −x x sin ax = 2a (− sin ax) = 2a Note : If X = sin ax 1 sin ax Formula 1 : φ(− a2) Formula 2 : Same as cos ax method eiax − x 1 1 Formula 3 : 2 cos ax = 2 sin ax = I.P. (D + ia) (D − ia) 2a D +a Example 8.29 : Solve : (D2 − 4)y = sin 2x Solution : The characteristic equation is p2 − 4 = 0 ⇒ p = ± 2 C.F. = Ae2x + Be−2x ; 1 1 1 P.I. = 2 (sin 2x) = (sin 2x) = − 8 sin 2x −4−4 D −4 1 Hence the general solution is y = C.F. + P.I. ⇒ y = Ae2x + Be− 2x − 8 sin 2x Example 8.30 : Solve : (D2 + 4D + 13)y = cos 3x Solution : The characteristic equation is p2 + 4p + 13 = 0
147
p=
−4±
16 − 52 − 4 ± − 36 − 4 ± i6 = = = − 2 ± i3 2 2 2
C.F. = e−2x (A cos 3x + B sin 3x) P.I. =
1 (cos 3x) D + 4D + 13 2
=
1 1 (cos 3x) = 4D + 4 (cos 3x) − 3 + 4D + 13
=
(4D − 4) 4D − 4 (cos 3x) = (cos 3x) (4D + 4) (4D − 4) 16D2 − 16
=
4D − 4 1 (cos 3x) = 40 (3 sin 3x + cos 3x) −160
2
The general solution is y = C.F. + P.I. 1 y = e−2x (A cos 3x + B sin 3x) + 40 (3 sin 3x + cos 3x) Example 8.31 : Solve (D2 + 9)y = sin 3x Solution : The characteristic equation is p2 + 9 = 0 ⇒ p = ± 3i C.F. = (A cos 3x + B sin 3x) P.I. =
1 sin 3x D2 + 9
−x = 6 cos 3x
1 −x since 2 2 sin ax = 2a cos ax D +a
Hence the solution is y = C.F. + P.I. i.e., y = (A cos 3x + B sin 3x) −
x cos 3x 6
(c) When X is of the form x and x2 Working rule : Take the P.I. as c0 + c1x if f(x) = x and c0 + c1x + c2x2 if f(x) = x2. Since P.I. is also a solution of (aD2 + bD + c)y = f(x), take y = c0 + c1x or y = c0 + c1x + c2x2 according as f(x) = x or x2. By substituting y value and comparing the like terms, one can find c0, c1 and c2.
148
Example 8.32 : Solve : (D2 − 3D + 2)y = x Solution : The characteristic equation is p2 − 3p + 2 = 0 ⇒ (p − 1) (p − 2) = 0 p = 1, 2 The C.F. is (Aex + Be2x) Let P.I. = c0 + c1x
∴ c0 + c1x is also a solution. ∴ (D2 − 3D + 2) (c0 + c1x) = x i.e., (− 3c1 + 2 c0) + 2c1x = x 1
⇒ 2c1 = 1 ∴ c1 = 2 3
(− 3c1 + 2 c0) = 0 ⇒ c0 = 4 x 3 ∴ P.I. = 2 + 4 Hence the general solution is y = C.F. + P.I. x 3 y = Aex + Be2x + 2 + 4 Example 8.33 : Solve : (D2 − 4D + 1)y = x2 Solution : The characteristic equation is p2 − 4p + 1 = 0 ⇒ p=
4±
C.F. = Ae
16 − 4 4 ± 2 3 = =2± 3 2 2
(2 +
3)x
(2 −
+ Be
3 )x
Let P.I. = c0 + c1x + c2x2 But P.I. is also a solution.
∴ (D2 − 4D + 1) (c0 + c1x + c2x2 ) = x2 i.e., (2c2 − 4c1 + c0) + (− 8c2 + c1)x + c2x2 = x2 c2 = 1 − 8c2 + c1 = 0 ⇒ c1 = 8
2c2 − 4c1 + c0 = 0 ⇒ c0 = 30 P.I. = x2 + 8x + 30
149
Hence the general solution is y = C.F. + P.I.
(2 +
y = Ae
3 )x
+ Be
(2 −
3)x
+ (x2 + 8x + 30)
EXERCISE 8.5 Solve the following differential equations : (2) (D2 − 4D + 13)y = e−3x
(1) (D2 + 7D + 12)y = e2x (3) (D2 + 14D + 49)y = e−7x + 4
(4) (D2 − 13D + 12)y = e−2x + 5ex π (5) (D2 + 1) y = 0 when x = 0, y = 2 and when x = 2, y = − 2 (6)
d2y dy 3x 2 − 3 dx + 2y = 2e when x = log2, y = 0 and when x = 0, y = 0 dx
(7) (D2 + 3D − 4) y = x2
(8) (D2 − 2D − 3)y = sinx cosx
2
(10) (D2 − 6D + 9) y = x + e2x
(9) D y = − 9 sin 3x (11) (D2 − 1)y = cos2x − 2 sin 2x
(12) (D2 + 5)y = cos2x
(13) (D2 + 2D + 3)y = sin 2x
(14) (3D2 + 4D + 1)y = 3e−x/3
8.6 Applications : In this section we solve problems on differential equations which have direct impact on real life situation. Solving of these types of problems involve (i) Construction of the mathematical model describing the given situation (ii) Seeking solution for the model formulated in (i) using the methods discussed earlier. Illustration : Let A be any population at time t. The rate of change of population is directly proportional to initial population i.e., dA dA dt α A i.e., dt = kA where k is called the constant of proportionality (1) If k > 0, we say that A grows exponentially with growth constant k (growth problem). (2) If k < 0 we say that A decreases exponentially with decreasing constant k (decay problem). In all the practical problems we apply the principle that the rate of change of population is directly proportional to the initial population
150
dA dA i.e., dt α A or dt = kA (Here k may be positive or negative depends on the problem). This linear equation can be solved in three ways i.e., (i) variable separable (ii) linear (using I.F.) (iii) by using characteristic equation with single root k. In all the ways we get the solution as A = cekt where c is the arbitrary constant and k is the constant of proportionality. In general we have to find out c as well as k from the given dA data. Sometimes the value of k may be given directly as in 8.35. dt is directly given in 8.38. dA Solution : dt = kA dA (i) ⇒ log A = kt + log c A = kdt ⇒ (ii)
A = ekt + log c ⇒ A = cekt
dA dt − kA = 0 is linear in A I.F. = e−kt −kt −kt Ae−kt = ⌠ ⌡e O dt + c ⇒ Ae = c
A = cekt (iii) (D − k)A = 0 Chr. equation is p − k = 0 ⇒ p = k The C.F. is cekt But there is no P.I. ∴ A = cekt (iv) In the case of Newton’s law of cooling (i.e., the rate of change of temperature is proportional to the difference in temperatures) we get the equation as dT dt = k(T − S) [T− cooling object temperature, S − surrounding temperature] dT = kdt ⇒ log (T − S) = kt + log c ⇒ T − S = cekt T−S ⇒ T = S + cekt
151
Example 8.34 : In a certain chemical reaction the rate of conversion of a substance at time t is proportional to the quantity of the substance still untransformed at that instant. At the end of one hour, 60 grams remain and at the end of 4 hours 21 grams. How many grams of the substance was there initially? Solution : Let A be the substance at time t
dA dA kt dt α A ⇒ dt = kA ⇒ A = ce When t = 1, A = 60 ⇒ cek = 60 When t = 4, A = 21 ⇒ ce 4 4k
(1) ⇒ c e
4k
= 21
4
… (1) … (2) … (3)
= 60
4
(3) 60 3 (2) ⇒ c = 21 ⇒ c = 85.15 (by using log) Initially i.e., when t = 0, A = c = 85.15 gms (app.) Hence initially there was 85.15 gms (approximately) of the substance. Example 8.35 : A bank pays interest by continuous compounding, that is by treating the interest rate as the instantaneous rate of change of principal. Suppose in an account interest accrues at 8% per year compounded continuously. Calculate the percentage increase in such an account over one year. [Take e.08 ≈ 1.0833] Solution : Let A be the principal at time t dA dA dA dt α A ⇒ dt = kA ⇒ dt = 0.08 A, since k = 0.08 ⇒ A(t) = ce0.08t A(1) − A(0) × 100 Percentage increase in 1 year = A(0)
c. e0.08 A(1) = A(0) − 1 × 100 = c − 1 × 100 = 8.33%
Hence percentage increase is 8.33% Example 8.36 : The temperature T of a cooling object drops at a rate proportional to the difference T − S, where S is constant temperature of surrounding medium. If initially T = 150°C, find the temperature of the cooling object at any time t.
152
Solution : Let T be the temperature of the cooling object at any time t dT dT kt dt α (T− S) ⇒ dt = k (T − S) ⇒ T − S = ce , where k is negative ⇒ T = S + cekt When t = 0, T = 150 ⇒ 150 = S + c ⇒ c = 150 − S ∴ The temperature of the cooling object at any time is T = S + (150 − S)ekt Note : Since k is negative, as t increases T decreases. It is a decay problem. Instead of k one may take − k where k > 0. Then the answer is T = S + (150 − S)e− kt . Again, as t increases T decreases. Example 8.37 : For a postmortem report, a doctor requires to know approximately the time of death of the deceased. He records the first temperature at 10.00 a.m. to be 93.4°F. After 2 hours he finds the temperature to be 91.4°F. If the room temperature (which is constant) is 72°F, estimate the time of death. (Assume normal temperature of a human body to be 98.6°F). log 19.4 = − 0.0426 × 2.303 and log 26.6 = 0.0945 × 2.303 e 21.4 e 21.4 Solution : Let T be the temperature of the body at any time t dT By Newton’s law of cooling dt α (T − 72) since S = 72°F dT kt dt = k (T − 72) ⇒ T− 72 = ce or T = 72 + cekt At t = 0, T = 93.4 ⇒ c = 21.4 [ First recorded time 10 a.m. is t = 0] ∴ T = 72 + 21.4ekt 19.4 1 19.4 When t = 120, T = 91.4 ⇒ e120k = 21.4 ⇒ k = 120 loge21.4 1 = 120 (− 0.0426 × 2.303) Let t1 be the elapsed time after the death. kt1
When t = t1 ; T = 98.6 ⇒ 98.6 = 72 + 21.4 e
153
1 26.6 − 120 × 0.0945 × 2.303 ⇒ t1 = k loge 21.4 = = − 266 min 0.0426 × 2.303 [For better approximation the hours converted into minutes] i.e., 4 hours 26 minutes before the first recorded temperature. The approximate time of death is 10.00 hrs − 4 hours 26 minutes. ∴ Approximate time of death is 5.34 A.M. dT Note : Since it is a decay problem, we can even take dt = − k(T − 72) where k > 0. The final answer will be the same. Example 8.38 : A drug is excreted in a patients urine. The urine is monitored continuously using a catheter. A patient is administered 10 mg of drug at time t = 0, which is excreted at a Rate of − 3t1/2 mg/h. (i) What is the general equation for the amount of drug in the patient at time t > 0? (ii) When will the patient be drug free? Solution : (i) Let A be the quantum of drug at any time t 1
The drug is excreted at a rate of − 3t2 1
3
dA i.e., dt = − 3t2 ⇒ A = − 2t2 + c When t = 0, A = 10 ⇒ c = 10 3
At any time t
A = 10 − 2t2 3
(ii) For drug free, A = 0 ⇒ 5 = t2 ⇒ t3 = 25 ⇒ t = 2.9 hours. Hence the patient will be drug free in 2.9 hours or 2 hours 54 min. Example 8.39 : The number of bacteria in a yeast culture grows at a rate which is proportional to the number present. If the population of a colony of yeast bacteria triples in 1 hour. Show that the number of bacteria at the end of five hours will be 35 times of the population at initial time. Solution : Let A be the number of bacteria at any time t dA dA kt dt α A ⇒ dt = kA ⇒ A = ce
154
Initially, i.e., when t = 0, assume that A = A0 ∴ A0 = ce° = c ∴ A = A0ekt when t = 1, A = 3A0 ⇒ 3A0 = A0ek ⇒ ek = 3 When t = 5, A = A0e5k = A0(ek)5 = 35. A0 ∴ The number of bacteria at the end of 5 hours will be 35 times of the number of bacteria at initial time EXERCISE 8.6 (1) Radium disappears at a rate proportional to the amount present. If 5% of the original amount disappears in 50 years, how much will remain at the end of 100 years. [Take A0 as the initial amount]. (2) The sum of Rs. 1000 is compounded continuously, the nominal rate of interest being four percent per annum. In how many years will the amount be twice the original principal? (loge2 = 0.6931). (3) A cup of coffee at temperature 100°C is placed in a room whose temperature is 15°C and it cools to 60°C in 5 minutes. Find its temperature after a further interval of 5 minutes. (4) The rate at which the population of a city increases at any time is proportional to the population at that time. If there were 1,30,000 people in the city in 1960 and 1,60,000 in 1990 what population may be 16 anticipated in 2020. log e 13 = .2070 ; e.42 = 1.52 (5) A radioactive substance disintegrates at a rate proportional to its mass. When its mass is 10 mgm, the rate of disintegration is 0.051 mgm per day. How long will it take for the mass to be reduced from 10 mgm to 5 mgm. [loge2 = 0.6931]
155
9. DISCRETE MATHEMATICS Discrete Mathematics deals with several selected topics in Mathematics that are essential to the study of many Computer Science areas. Since it is very difficult to cover all the topics, only two topics, namely “Mathematical Logic”, and “Groups” have been introduced. These topics will be very much helpful to the students in certain practical applications related to Computer Science.
9.1. Mathematical Logic : Introduction : Logic deals with all types of reasonings. These reasonings may be legal arguments or mathematical proofs or conclusions in a scientific theory. Aristotle (384 – 322 BC) wrote the first treatise on logic. Gottfried Leibnitz framed the idea of using symbols in logic and this idea was realised in the nineteenth century by George Boole and Augustus De’Morgan. Logic is widely used in many branches of sciences and social sciences. It is the theoretical basis for many areas of Computer Science such as digital logic circuit design, automata theory and artificial intelligence. We express our thoughts through words. Since words have many associations in every day life, there are chances of ambiguities to appear. In order to avoid this, we use symbols which have been clearly defined. Symbols are abstract and neutral. Also they are easy to write and manipulate. This is because the mathematical logic which we shall study is also called symbolic logic.
9.1.1 Logical statement or Proposition : A statement or a proposition is a sentence which is either true or false but not both. A sentence which is both true and false simultaneously is not a statement, rather it is a paradox. Example 1 : (a) Consider the following sentences : (i) Chennai is the capital of Tamilnadu. (ii) The earth is a planet. (iii) Rose is a flower. Each of these sentences is true and so each of them is a statement. (b) Consider the following sentences : (iv) Every triangle is an isosceles triangle.
156
(v) Three plus four is eight (vi) The sun is a planet. Each of these sentences is false and so each of them is a statement. Example 2 : Each of the sentences (vii) Switch on the light. (viii) Where are you going? (ix) May God bless you with success. (x) How beautiful Taj Mahal is! cannot be assigned true or false and so none of them is a statement. In fact, (vii) is a command, (viii) is a question (ix) is an optative and (x) is exclamatory. Truth value of a statement : The truth or falsity of a statement is called its truth value. If a statement is true, we say that its truth value is TRUE or T and if it is false, we say that its truth value is FALSE or F. All the statements in Example 1(a) have the truth value T, while all the statements in Example 1 (b) have the truth value F. Simple statements : A statement is said to be simple if it cannot be broken into two or more statements. All the statements in (a) and (b) of Example 1 are simple statements. Compound statements : If a statement is the combination of two or more simple statements, then it is said to be a compound statement. Example : It is raining and it is cold. This is a compound statement and it is a combination of two simple statements “It is raining”, “It is cold”. Simple statements which on combining form compound statements are called sub-statements or component statements of the compound statement. The fundamental property of a compound statement is that its truth value is completely determined by the truth values of its sub-statements together with the way in which they are combined to form the compound statement. Basic logical connectives The words which combine simple statements to form compound statements are called connectives. We use the connectives ‘and’, ‘or’, etc., to form new statements by combining two or more statements. But the use of these connectives in English language is not always precise and unambiguous. Hence it is necessary to define a set of connectives with definite meanings in the
157
language of logic, called object language. Three basic connectives are conjunction which corresponds to the English word ‘and’, ‘disjunction’ which corresponds to the word ‘or’ ‘negation’ which corresponds to the word ‘not’. We use the symbol “∧” to denote conjunction, “∨” to denote disjunction and “ ~ ” to denote negation. Conjunction : If two simple statements p and q are connected by the word ‘and’, then the resulting compound statement ‘p and q’ is called the conjunction of p and q and is written in the symbolic form as ‘p ∧ q’. Example 1 : Form the conjunction of the following simple statements p : Ram is intelligent. q : Ravi is handsome. p ∧ q : Ram intelligent and Ravi is handsome. Example 2 : Convert the following statement into symbolic form : ‘Usha and Mala are going to school’. the given statement can be rewritten as : ‘Usha is going to school’, and ‘Mala is going to school’. Let p : Usha is going to school. q : Mala is going to school. The given statement in symbolic form is p ∧ q. Rule : (A1) The statement p ∧ q has the truth value T whenever both p and q have the truth value T. (A2) The statement p ∧ q has the truth value F whenever either p or q or both have the truth value F. Example : Write the truth value of each of the following statements : (i)
Ooty is in Tamilnadu and 3 + 4 = 8
(ii)
Ooty is in Tamilnadu and 3 + 4 = 7
(iii) Ooty is in Kerala and 3 + 4 = 7 (iv) Ooty is in Kerala and 3 + 4 = 8 In (i) the truth value of the statement 3 + 4 = 8 is F. ∴ By (A2) (i) has the truth value F.
158
In (ii) both the sub-statements have truth value T and hence by (A1). (ii) has truth value T. The truth values of (iii) and (iv) are F. Disjunction : If two simple statements p and q are connected by the word ‘or’, then the resulting compound statement ‘p or q’ is called the disjunction of p and q and is written in symbolic form as p ∨ q. Example : Form the disjunction of the following simple statements : p : John is playing cricket. q : There are thirty students in the class room. p ∨ q : John is playing cricket or there are thirty students in the class room. Example : Convert the following statement into symbolic form. “5 is a positive integer or a square is a rectangle”. Let p : 5 is a positive integer. q : A square is a rectangle. The given statement in symbolic form is p w q. Rule : (A3) The statement p ∨ q has the truth value F whenever both p and q have the truth value F. (A4) The statement p ∨ q has the truth value T whenever either p or q or both have the truth value T. Example : (i) Chennai is in India or 2 is an integer. (ii) Chennai is in India or 2 is an irrational number. (iii) Chennai is in China or 2 is an integer. (iv) Chennai is in China or 2 is an irrational number. By (A4), we see that the truth values of (i), (ii) and (iv) are T and by (A3), the truth value of (iii) is F. Negation : The negation of a statement is generally formed by introducing the word ‘not’ at some proper place in the statement or by prefixing the statement with ‘It is not the case that’ or ‘It is false that’. If p denotes a statement, then the negation of p is written as ∼p or p. We use the symbol ∼p to denote the negation of p.
159
Rule :
(A5) If the truth value of p is T then the truth value of ∼p is F. Also, if the truth value of p is F, then the truth value of ∼p is T.
Example : p : All men are wise. ∼p : Not all men are wise. (or) ∼p : It is not the case that all men are wise (or) ∼p : It is false that all men are wise. Note : Negation is called a connective although it does not combine two or more statements. It only modifies a statement. EXERCISE 9.1 Find out which of the following sentences are statements and which are not? Justify your answer. (1) All natural numbers are integers. (2) A square has five sides. (3) The sky is blue. (4) How are you? (5) 7 + 2 < 10. (6) The set of rational numbers is finite. (7) How beautiful you are! (8) Wish you all success. (9) Give me a cup of tea. (10) 2 is the only even prime. Write down the truth value (T or F) of the following statements : (11) All the sides of a rhombus are equal in length. (12) 1 + 8 is an irrational number. (13) Milk is white. (14) The number 30 has four prime factors. (15) Paris is in France. (16) Sin x is an even function. (17) Every square matrix is non-singular. (18) Jupiter is a planet. (19) The product of a complex number and its conjugate is purely imaginary. (20) Isosceles triangles are equilateral.
160
(21) Form the conjunction and the disjunction of (i) p : Anand reads newspaper, q : Anand plays cricket. (ii) p : I like tea. q : I like ice-cream. (22) Let p be “Kamala is going to school” and q be “There are twenty students in the class “. Give a simple verbal sentence which describes each of the following statements : (i) p ∨ q (ii) p ∧ q (iii) ∼ p (iv) ∼ q (v) ∼p ∨ q (23) Translate each of the following compound statements into symbolic form : (i) Rose is red and parrot is a bird. (ii) Suresh reads ‘Indian Express’ or ‘The Hindu’. (iii) It is false that the mangoes are sweet. (iv) 3 + 2 = 5 and Ganges is a river. (v) It is false that sky is not blue. (24) If p stands for the statement “Sita likes reading” and q for the statement “Sita likes playing’ what does ∼p ∧ ∼ q stand for? (25) Write negation of the each of the following statements : (i) 5 is an irrational number. (ii) Mani is sincere and hardworking. (iii) This picture is good or beautiful.
9.1.2 Truth tables : A table that shows the relation between the truth values of a compound statement and the truth values of its sub-statements is called the truth table. A truth table consists of rows and columns. The initial columns are filled with the possible truth values of the sub-statements and the last column is filled with the truth values of the compound statement on the basis of the truth values of the sub-statements written in the initial columns. If the compound statement is made up of n sub-statements, then its truth table will contain 2n rows. Example 9.1 : Construct the truth table for ∼p Solution: The statement ∼p consists of only one simple statement p. Therefore, its truth table will contain 21(= 2) rows. Also we know that if p has the truth value T then ∼p has the truth value F and if p has the truth value F, then ∼p has the truth value T. Thus the truth table for ∼p is as given below : Truth table for ∼ p p ∼p T F F T
161
Example 9.2 : Construct the truth table for p ∨ (∼p). Solution: The compound statement p∨ (∼p) consists of only one single statement p. Therefore its truth table will contain 21(= 2) rows. In the first column, enter all possible truth values of p. In the second column, enter the truth values of ∼p based on the corresponding truth values of p. Finally, in the last column, enter the truth values of p ∨ (∼p), using (A4). Truth table for p∨(∼p) p ∼p p∨(∼p) T F T F T T Example 9.3 : Construct the truth table for p ∧ q. Solution: The compound statement p ∧ q consists of two simple statements p and q. Therefore, there must be 22(= 4) rows in the truth table of p ∧ q. Now enter all possible truth values of statements p and q namely TT, TF, FT and FF in the first two columns of the truth table. Using (A1) and (A2), enter the truth values of p ∧ q in the final column based on the corresponding truth values of p and q in the first two columns. Truth table for p ∧ q p q p∧q T T T T F F F T F F F F Note : Similarly, by using (A3) and (A4) we can construct the truth table for p ∨ q, as given below : Truth table for p ∨ q p q p∨q T T T T F T F T T F F F
162
Example 9.4 : Construct the truth table for the following statements :
(i) ((∼p) ∨ (∼ q)) (iii) (p ∨ q) ∧ (∼ q) Solution: (i) p T T F F (ii) p T T F F (iii) p T T F F
(ii) ∼ ((∼ p) ∧ q) (iv) ∼ ((∼ p) ∧ (∼ q))
Truth table for ((∼p) ∨ (∼ q)) q ∼p ∼q ((∼p) ∨ (∼ q)) T F F F F F T T T T F T F T T T Truth table for ∼ ((∼ p) ∧ q) q ∼p (∼ p) ∧ q ∼ ((∼ p) ∧ q) T F F T F F F T T T T F F T F T Truth table for (p ∨ q) ∧ (∼ q) q p∨q ∼q (p ∨ q) ∧ (− q) T T F F F T T T T T F F F F T F Truth table for ∼ ((∼ p) ∧ (∼ q))
(iv) p q ∼p ∼q (∼ p) ∧ (∼ q) ∼ ((∼ p) ∧ (∼ q)) T T F F F T T F F T F T F T T F F T F F T T T F Example 9.5 : Construct the truth table for (p ∧ q) ∨ (∼ r) Solution: The compound statement (p ∧ q) ∨ (∼ r) consists of three simple statements p, q and r. Therefore, there must be 23(= 8) rows in the truth table of (p ∧ q) ∨ (∼r). The truth value of p remains at the same value of T or F for each of four consecutive assignments of logical values. The truth value of q remains at T or F for two assignments and that of r remains at T or F for one assignment. 163
p q r p∧q ∼r (p ∧ q) ∨ (∼r) T T T T F T T T F T T T T F T F F F T F F F T T F T T F F F F T F F T T F F T F F F F F F F T T Example 9.6 : Construct the truth table for (p ∨ q) ∧ r Solution: p q r p∨q (p ∨ q) ∧ r T T T T T T T F T F T F T T T T F F T F F T T T T F T F T F F F T F F F F F F F EXERCISE 9.2 Construct the truth tables for the following statements : (1) p ∨ (∼ q) (2) (∼ p) ∧ (∼ q) (3) ∼ (p ∨ q) (4) (p ∨ q) ∨ (∼ p) (5) (p ∧ q) ∨ (∼ q) (6) ∼ (p ∨ (∼ q)) (7) (p ∧ q) ∨ [∼ (p ∧ q)] (8) (p ∧ q) ∨ (∼ q) (9) (p ∨ q) ∨ r (10) (p ∧ q) ∨ r Logical Equivalence : Two compound statements A and B are said to be logically equivalent or simply equivalent, if they have identical last columns in their truth tables. In this case we write A ≡ B. Example 9.7 : Show that ∼ (p ∨ q) ≡ (∼ p) ∧ (∼ q)
164
Solution: Truth table for ∼ (p ∨ q) p q p∨q ∼ (p ∨ q) T T T F T F T F F T T F F F F T Truth table for ((∼ p) ∧ (∼ q)) p q ∼p ∼q ((∼ p) ∧ (∼ q)) T T F F F T F F T F F T T F F F F T T T The last columns are identical. ∴ ∼ (p ∨ q) ≡ ((∼ p) ∧ (∼ q)) Negation of a negation : Negation of a negation of a statement is the statement itself. Equivalently we write ∼ (∼ p) ≡ p p ∼p ∼ (∼ p) T F T F T F In the truth table, the columns corresponding to p and ∼ (∼ p) are identical. Hence p and ∼ (∼ p) are logically equivalent. Example 9.8 : Verify ∼ (∼ p) ≡ p for the statement p : the sky is blue. Solution: p : The sky is blue : The sky is not blue ∼p ∼ (∼ p) : It is not the case that the sky is not blue or It is false that the sky is not blue or The sky is blue Conditional and bi-conditional statements : In Mathematics, we frequently come across statements of the form “If p then q”. Such statements are called conditional statements or implications. They are denoted by p → q, read as ‘p implies q’. The conditional p → q is false only if p is true and q is false. Accordingly, if p is false then p → q is true regardless of the truth value of q.
165
Truth table for p → q p q p→q T T T T F F F T T F F T If p and q are two statements, then the compound statement p → q and q → p is called a bi-conditional statement and is denoted by p ↔ q, read as p if and only if q. p ↔ q has the truth value T whenever p and q have the same truth values; otherwise it is F. Truth table for p ↔ q p q p↔q T T T T F F F T F F F T
9.1.3 Tautologies : A statement is said to be a tautology if the last column of its truth table contains only T, i.e., it is true for all logical possibilities. A statement is said to be a contradiction if the last column of its truth table contains only F, i.e., it is false for all logical possibilities. Example 9.9 : (i) p ∨ (∼ p) is a tautology. (ii) p ∧ (∼ p) is a contradiction Solution: (i) Truth table for p ∨ (∼ p) p ∼p p ∨ (∼ p) T F T F T T The last column contains only T. ∴ p ∨ (∼p) is a tautology. (ii) Truth table for p ∧ (∼ p) p ∼p p ∧ (∼ p) T F F F T F The last column contains only F. ∴ p ∧ (∼p) is a contradiction.
166
Example 9.10 : (i) Show that ((∼ p) ∨ (∼ q)) ∨ p is a tautology. (ii) Show that ((∼ q) ∧ p) ∧ q is a contradiction. Solution: (i) Truth table for ((∼ p) ∨ (∼ q)) ∨ p p q ∼ p ∼ q (∼ p) ∨ (∼ q) ((∼ p) ∨ (∼ q))∨ p T T F F F T T F F T T T F T T F T T F F T T T T The last column contains only T. ∴ ((∼ p) ∨ (∼ q)) ∨ p is a tautology. (ii) Truth table for ((∼ q) ∧ p) ∧ q p q ∼q (∼ q) ∧ p ((∼ q) ∧ p) ∧ q T T F F F T F T T F F T F F F F F T F F The last column contains only F. ∴ ((∼ q) ∧ p) ∧ q is a contradiction. Example 9.11 : Use the truth table to determine whether the statement ((∼ p) ∨ q) ∨ (p ∧ (∼ q)) is a tautology. Solution: Truth table for ((∼ p) ∨ q) ∨ (p ∧ (∼ q)) p q ∼ p ∼ q (∼ p) ∨ q p ∧ (∼ q) ((∼ p) ∨ q) ∨ (p ∧ (∼ q) T T F F T F T T F F T F T T F T T F T F T F F T T T F T The last column contains only T. ∴ The given statement is a tautology. EXERCISE 9.3 (1) Use the truth table to establish which of the following statements are tautologies and which are contradictions. (i) ((∼ p) ∧ q) ∧ p (ii) (p ∨ q) ∨ (∼ (p ∨ q)) (iii) (p ∧ (∼ q)) ∨ ((∼ p) ∨ q) (iv) q ∨ (p ∨ (∼ q)) (v) (p ∧ (∼ p)) ∧ ((∼ q) ∧ p)
167
(2) (3) (4) (5) (6) (7)
Show that p → q ≡ (∼ p) ∨ q Show that p ↔ q ≡ (p → q) ∧ (q → p) Show that p ↔ q ≡ ((∼ p) ∨ q) ∧ ((∼ q) ∨ p) Show that ∼(p ∧ q) ≡ ((∼ p) ∨ (∼ q)) Show that p → q and q → p are not equivalent. Show that (p ∧ q) → (p ∨ q) is a tautology.
9.2 Groups : 9.2.1 Binary Operation : We know that the addition of any two natural numbers is a natural number, the product of any two natural numbers is also a natural number. Each of these operations associates with the two given numbers, a third number, their sum in the case of addition, and their product in the case of multiplication. In this section we are going to deal with the notion of a binary operation or a binary composition on a set which is nothing but a generalisation of the usual addition and usual multiplication on the number systems. Definition : A binary operation * on a non-empty set S is a rule, which associates to each ordered pair (a, b) of elements a, b in S an element a * b in S. Thus a binary operation * on S is just a map, * : S × S → S by (a, b) → a * b. Where we denote by a * b, the image of (a, b) in S under *. From the definition we see that, if * is a binary operation on S then a, b ∈ S ⇒ a * b ∈ S. In this case, we also say that S is closed under *. This property is known as the “closure axiom” or “closure property”. List of symbols used in this chapter : N - The set of all natural numbers. Z - The set of all integers. W - The set of all non-negative integers (whole numbers). E - The set of all even integers. O - The set of all odd integers. Q - The set of all rational numbers. R - The set of all real numbers. C - The set of all complex numbers. Q − {0} - The set of all non-zero rational numbers. R – {0} - The set of all non-zero real numbers. C - }0} - The set of all non-zero complex numbers. - for every ∀ ∃ - there exists ∋ - such that 168
Illustrative examples : The usual addition + is a binary operation on N. Since a, b ∈ N ⇒ a + b ∈ N. i.e., N is closed under +. But the usual subtraction is not binary on N. Since 2, 5 ∈ N, but 2 − 5 = − 3 ∉ N. ∴ N is not closed under subtraction. At the same time, we see that − is a binary operation on Z. From this we see that, an operation becoming binary or not binary depends on the set. The following table gives which number systems are closed under the usual algebraic operations, namely addition, subtraction, multiplication and division denoted by +, −, . , ÷ respectively. Number Systems
N
Z
Q
R
C
Q − {0}
R − {0}
C − {0}
+
binary
binary
binary
binary
binary
not binary
not binary
not binary
−
not binary
binary
binary
binary
binary
not binary
not binary
not binary
.
binary
binary
binary
binary
binary
binary
binary
binary
÷
not binary
not binary
not binary
not binary
not binary
binary
binary
binary
Operations
Apart from the usual algebraic operations, some new operations on the number systems can also be defined. For example, consider the operation * on N defined by a * b = ab. It is clear that * is binary on N, ‡ a, b ∈ N ⇒ a * b = ab ∈ N. Some more facts about binary operations : (1) Let the set S be R or any subset of real number system. Define * as
(i)
a * b = minimum of {a, b}
(ii)
a * b = maximum of {a, b}
(iii) a * b = a (iv) a * b = b All the above operations (*) are binary operations on the corresponding sets.
169
(2) (N, *) * is defined as a * b = ab + 5. Since ab and 5 are natural numbers, ab + 5 is also a natural number. ∴ * is a binary operation on N. On the other hand, the operation * defined by a * b = ab − 5 is not binary on N because 2 * 1 = (2)(1) − 5 = − 3 ∉ N. (3) (Z, *), where * is defined by, a * b = ab, is not a binary operation on z. Since take a = 2, b = − 1 1 ab = 2−1 = 2 ∉ Z Note that * is also not a binary operaton on R − {0} 1 ab = (− 1)1/2 ∉ R − {0} because take a = − 1, b = 2 (4) (R, *) Define a * b = a + b + ab Clearly * is a binary operation on R since a + b and ab are real numbers and their sum is also a real number. (5) (O, +) Addition is not a binary operation on the set of odd integers, since addition of two odd integers is not odd. (6) (O, .) Multiplication is a binary operation on the set of odd integers. Since product of two odd integers is an odd integer. (7) Matrix addition is a binary operation on the set of m × n matrices. Since sum of two m × n matrices is again an m × n matrix. (8) Matrix addition is not a binary operation on the set of n × n singular matrices as well as on the set of n × n non-singular matrices. Because, sum of two non-singular matrices need not be non-singular and sum of two singular matrices need not be singular. (9) Matrix multiplication is a binary operation on the set of singular matrices as well as on the set of non-singular matrices. (10) Cross product is a binary operation on the set of vectors, but dot product is not a binary operation on the set of vectors. Multiplication table for a binary operation Any binary operation * on a finite set S = {a1, a2 ... an} can be described by means of multiplication table. This table consists of ‘n’ rows and ‘n’ columns. Place each element of S at the head of one row and one column, usually taking them in the same order for columns as for rows. The operator * is placed at the left hand top corner. The n × n = n2 spaces can be filled by writing ai * aj in the space common to the ith row and the jth column of the table.
170
a2 aj a1 .................................. ... * a1 . . . . ai ai * a j . . . This table is also known as Cayley’s table or composition table. In the next section we will see that these composition tables are very much helpful in exhibiting finite groups.
9.2.2 Groups : Given any non-empty set S, the possibility of combining two of its elements to get yet another element of S endows S with an algebraic structure. A non-empty set S together with a binary operation * is called an algebraic structure. Group is the simplest of all algebraic structures. It is the one operational algebraic system. The study of groups was started in the nineteenth century in connection with the solution of equations. The concept of group arises not only in Mathematics but also in other fields like Physics, Chemistry and Biology. Definition : A non-empty set G, together with an operation * i.e., (G, *) is said to be a group if it satisfies the following axioms (1) Closure axiom : a, b ∈ G ⇒ a * b ∈ G (2) Associative axiom : ∀a, b, c ∈ G, (a * b) * c = a * (b * c) (3) Identity axiom : There exists an element e ∈ G such that a * e = e * a = a, ∀a ∈ G. (4) Inverse axiom
: ∀a ∈ G there exists an element a−1∈G such that a−1 * a = a * a−1 = e.
e is called the identity element of G and a−1 is called the inverse of a in G. Definition (Commutative property) : A binary operation * on a set S is said to be commutative if a * b = b * a ∀ a, b ∈ S
171
Definition : If a group satisfies the commutative property then it is called an abelian group or a commutative group, otherwise it is called a non-abelian group. Note (1) : If the operation * is a binary operation, the closure axiom will be satisfied automatically. Note (2) : We shall often use the same symbol G to denote the group and the underlying set. Order of a group : The order of a group is defined as the number of distinct elements in the underlying set. If the number of elements is finite then the group is called a finite group and if the number of elements is infinite then the group is called an infinite group. The order of a group G is denoted by o(G). Definition : A non-empty set S with an operation * i.e., (S, *) is said to be a semi-group if it satisfies the following axioms. (1) Closure axiom : a, b ∈ S ⇒ a * b ∈ S (2) Associative axiom : (a * b) * c = a * (b * c), ∀ a, b, c ∈ S. Definition : A non-empty set M with an operation * i.e., (M, *) is said to be a monoid if it satisfies the following axioms : (1) Closure axiom : a, b ∈ M ⇒ a * b ∈ M (2) Associative axiom : (a * b) * c = a * (b * c) ∀a, b, c ∈ M (3) Identity axiom : There exists an element e ∈ M such that a * e = e * a = a, ∀a ∈ M. (N, +) is a semi-group but it is not a monoid, because the identity element O ∉ N. (N, *) where * is defined by a * b = ab is not a semi-group, because, consider (2 * 3) * 4 = 23 * 4 = 84 = 212 and 2 * (3 * 4) = 2 * 34 = 2 * 81 = 281 ∴ (2 * 3) * 4 ≠ 2 * (3 * 4) i.e., associative axiom is not satisfied. (Z, .) is a monoid. But it is not a group, because, the inverse axiom is not 1 satisfied. (5 ∈ Z, but 5 ∉ Z). (Z, +) and (Z, .) are semi-groups as well as monoids. From the definitions, it is clear that every group is a monoid. 172
Example 9.12 : Prove that (Z, +) is an infinite abelian group. Solution: (i) Closure axiom : We know that sum of two integers is again an integer. (ii) Associative axiom : Addition is always associative in Z i.e., ∀a, b, c ∈ Z, (a + b) + c = a + (b + c) (iii) Identity axiom : The identity element O ∈ Z and it satisfies O + a = a + O = a, ∀ a ∈ Z Identity axiom is true. (iv) Inverse axiom : For every a ∈ Z, ∃ an element − a ∈ Z such that − a + a = a + (− a) = 0 ∴ Inverse axiom is true. ∴ (Z, +) is a group. (v) ∀ a, b ∈ Z, a + b = b + a ∴ addition is commutative. ∴ (Z, +) is an abelian group. (vi) Since Z is an infinite set (Z, +) is infinite abelian group. Example 9.13 : Show that (R − {0}, .) is an infinite abelian group. Here ‘.’ denotes usual multiplication. Solution: (i) Closure axiom : Since product of two non-zero real numbers is again a non-zero a real number. i.e., ∀ a, b ∈ R, a . b ∈ R. (ii) Associative axiom : Multiplication is always associative in R− {0} i.e., a . (b . c) = (a . b) . c ∀ a, b, c ∈ R − {0} ∴ associative axiom is true. (iii) Identity axiom : The identity element is 1 ∈ R − {0} under multiplication and 1 . a = a . 1 = a, ∀ a ∈ R − {0} ∴ Identity axiom is true. 1 (iv) Inverse axiom : ∀ a ∈ R − {0}, a ∈ R − {0} such that 1 1 a . a = a . a = 1 (identity element). ∴ Inverse axiom is true. ∴ (R − {0}, .) is a group. (v) ∀ a, b ∈ R − {0}, a . b = b . a ∴ Commutative property is true. ∴ (R − {0}, .) is an abelian group.
173
(vi) Further R − {0} is an infinite set, (R − {0}, .) is an infinite abelian group. Example 9.14 : Show that the cube roots of unity forms a finite abelian group under multiplication. Solution: Let G = {1, ω, ω2}. The Cayley’s table is From the table, we see that, . 1 ω ω2 (i) all the entries in the table are members of G. 1 1 ω ω2 So, the closure property is true. (ii) multiplication is always associative. ω ω ω2 1 (iii) the identity element is 1 and it satisfies the ω identity axiom. ω2 ω 2 1 (iv) The inverse of 1 is 1 The inverse of ω is ω2 the inverse of ω2 is ω and it satisfies the inverse axiom also. ∴ (G, .) is a group. (v) the commutative property is also true. ∴ (G, .) is an abelian group. (vi) Since G is a finite set, (G, .) is a finite abelian group. Example 9.15 : Prove that the set of all 4th roots of unity forms an abelian group under multiplication. Solution: We know that the fourth roots of unity are 1, i, − 1, − i. Let G = {1, i, − 1, − i}. The Caylely’s table is . 1 −1 i − i From the table, 1 1 −1 i − i (i) the closure axiom is true. (ii) multiplication is always associative in C and i −1 −1 1 −i hence in G. i i −i −1 1 (iii) the identity element is 1 ∈ G and it satisfies i 1 −1 −i −i the identity axiom. (iv) the inverse of 1 is 1 ; i is − i ; − 1 is − 1 ; and − i is i. Further it satisfies the inverse axiom. hence (G, .) is a group. (v) From the table, the commutative property is also true. ∴ (G, .) is an abelian group. Example 9.16 : Prove that (C, +) is an infinite abelian group. Solution: (i) Closure axiom : Sum of two complex numbers is always a complex number.
174
i.e., z1, z2 ∈ C ⇒ z1 + z2 ∈ C Closure axiom is true. (ii) Associative axiom : Addition is always associative in C i.e., (z1 + z2) + z3 = z1 + (z2 + z3) ∀ z1, z2, z3 ∈ C ∴ Associative axiom is true. (iii) Identity axiom : The identity element o = o + io ∈ C and o + z = z + o = z ∀ z ∈ C ∴ Identity axiom is true. (iv) Inverse axiom : For every z ∈ C there exists a unique − z ∈ C such that z + (− z) = − z + z = 0. Inverse is true. ∴ (C, +) is a group. (v) Commutative property : ∀ z1, z2 ∈ C , z1 + z2 = z2 + z1 ∴ the commutative property is true. Hence (C , +) is an abelian group. Since C is an infinite set (C, +) is an infinite abelian group. Example 9.17 : Show that the set of all non-zero complex numbers is an abelian group under the usual multiplication of complex numbers. Solution: (i) Closure axiom : Let G = C − {0} Product of two non-zero complex numbers is again a non-zero complex number. ∴ Closure axiom is true. (ii) Associative axiom : Multiplication is always associative. ∴ Associative property is true. (iii) Identity axiom : 1 = 1 + io ∈ G, 1 is the identity element and 1.z = z . 1 = z ∀ z ∈ G. ∴ Identity axiom is true. (iv) Inverse axiom : Let z = x + iy ∈ G. Here z ≠ 0 ⇒ x and y are not both zero. ∴ x2 + y2 ≠ 0 1 1 x − iy x − iy x −y z = x + iy = (x + iy) (x − iy) = x2 + y2 = x2 + y2 + i x2 + y2 ∈ G 1 1 1 Further z . z = z . z = 1 ∴ z has the inverse z ∈ G. Thus inverse axiom is satisfied. ∴ (G, .) is a group.
175
(v) Commutative property : z1 z2 = (a + ib) (c + id) = (ac − bd) + i (ad + bc) = (ca − db) + i (da + cb) = z2 z1 ∴ It satisfies the commutative property. ∴ G is an abelian group under the usual multiplication of complex numbers. Note : Here the number 0 is removed, because 0 has no inverse under multiplication. We can also show that Q − {0}, R − {0} are abelian groups under multiplication. But Z − {0} is not a group under multiplication. 1 Q 7 ∈ Z − {0} while its inverse 7 ∉ Z − {0} Note : While verifying the axioms, follow the order given in the definition. If one axiom fails, stop the process at that stage. There is no use in continuing further. The following table shows which number systems are satisfying the axioms of a group in the order for a particular operation. *
N
E
Z
Q
R
C
Q − {0}
R − {0}
C − {0}
+
Semi
group
group
group
group
group
not closed
not closed
not closed
semi-group
monoid
monoid
monoid
monoid
group
group
group
not closed
not closed
not closed
not
not
group . − ÷
monid not
not
not
not
not
not
closed
associative
associative
associative
associative
associative
not
not closed
not closed
not closed
not closed
not closed
closed
not associative
associative associative
Example 9.18 : Show that (Z, *) is an infinite abelian group where * is defined as a * b = a + b + 2. Solution: (i) Closure axiom : Since a, b and 2 are integers a + b + 2 is also an integer. ∴ a * b ∈ z ∀ a, b ∈ z Thus closure axiom is true. (ii) Associative axiom : Let a, b, c ∈ G (a * b) * c = (a + b + 2) * c = (a + b + 2) + c + 2 = a + b + c + 4 a * (b * c) = a * (b + c + 2) = a + (b + c + 2) + 2 = a + b + c + 4 ⇒ (a * b) * c = a * (b * c) Thus associative axiom is true. 176
(iii) Identity axiom : Let e be the identity element. By the definition of e, a * e = a By the definition of *, a * e = a + e + 2 ⇒ a+e+2=a ⇒e=−2 − 2 ∈ Z. Thus identity axiom is true. (iv) Inverse axiom : Let a ∈ G and a−1 be the inverse element of a By the definition of
a−1, a * a−1 = e = − 2
By the definition of
*, a * a−1 = a + a−1 + 2 ⇒ a + a−1 + 2 = − 2
⇒ a−1 = − a − 4 Clearly − a − 4 ∈ Z. ∴ Inverse axiom is true. ∴ (Z, *) is a group. (v) Commutative property : Let a, b ∈ G a * b = a + b + 2 = b + a + 2 = b * a ∴ * is commutative. ∴ (Z, *) is an abelian group. further, Z is an infinite set. The group is an infinite abelian group. Example 9.19 : Show that the set of all 2 × 2 non-singular matrices forms a non-abelian infinite group under matrix multiplication, (where the entries belong to R). Solution: Let G be the set of all 2 × 2 non-singular matrices, where the entries belong to R. (i) Closure axiom : Since product of two non-singular matrices is again non-singular and the order is 2 × 2, the closure axiom is satisfied. i.e., A, B ∈ G ⇒ AB ∈ G. (ii) Associative axiom : Matrix multiplication is always associative and hence associative axiom is true. i.e., A (BC) = (AB) C ∀ A, B, C ∈ G. 1 0 (iii) Identity axiom : The identity element is I2 = ∈ G and it satisfies 0 1 the identity property.
177
(iv) Inverse axiom : the inverse of A ∈ G, exists i.e. A−1 exists and is of order 2 × 2 and AA−1 = A−1A = I. Thus the inverse axiom is satisfied. Hence the set of all 2 × 2 non-singular matrices forms a group under matrix multiplication. Further, matrix multiplication is non-commutative (in general) and the set contain infinitely many elements. The group is an infinite non-abelian group. Example 9.20 : Show that the set of four matrices
1 0 − 1 0 1 0 − 1 0 , , , form an abelian group, under 0 1 0 1 0 − 1 0 − 1 multiplication of matrices. Solution: − 1 0 − 1 0 1 0 1 0 Let I = , B = , C = and let , A = 0 1 0 1 0 − 1 0 − 1 G = {I, A, B, C} By computing the products of these matrices, taken in pairs, we can form the multiplication table as given below : I A B C . I I A B C A A I C B B B C I A C C B A I (i) All the entries in the multiplication tables are members of G. So, G is closed under . ∴ Closure axiom is true. (ii) Matrix multiplication is always associative (iii) Since the row headed by I coincides with the top row and the column headed by I coincides with the extreme left column, I is the identity element in G. (iv)
I . I = I ⇒ I is the inverse of I A . A = I ⇒ A is the inverse of A B . B = I ⇒ B is the inverse of B C . C = I ⇒ C is the inverse of C
From the table it is clear that . is commutative. ∴ G is an abelian group under matrix multiplication.
178
x x Example 9.21 : Show that the set G of all matrices of the form , where x x x ∈ R − {0}, is a group under matrix multiplication. Solution: x x Let G = / x ∈ R − {0} we shall show that G is a group under x x matrix multiplication. (i) Closure axiom : x x y y A= ∈ G, B = ∈G x x y y 2xy 2xy AB= ∈ G , ( ‡ x ≠ 0, y ≠ 0 ⇒ 2xy ≠ 0) 2xy 2xy i.e., G is closed under matrix multiplication. (ii) Matrix multiplication is always associative. e e (iii) Let E= ∈ G be such that AE = A for every A ∈ G. e e x x e e x x AE = A ⇒ = x x e e x x 1 2xe 2xe x x ⇒ = ⇒ 2xe = x ⇒ e = 2 (‡ x ≠ 0) 2xe 2xe x x 1/2 1/2 ∈ G is such that AE = A, for every A ∈ G 1/2 1/2
Thus E =
We can similarly show that EA = A for every A ∈ G. ∴ E is the identity element in G and hence identity axiom is true. y y (iv) Suppose A−1 = ∈ G is such that A−1A = E y y 1 1 2xy 2xy 1/2 1/2 Then we have = ⇒ 2xy = 2 ⇒ y = 4x 2xy 2xy 1/2 1/2 1/4 x 1/4 x ∴ A−1 = ∈ G is such that A−1A = E 1/4 x 1/4 x Similarly we can show that A A−1 = E. ∴ A−1 is the inverse of A. ∴ G is a group under matrix multiplication.
179
Note : The above group is abelian since AB = BA. But in general matrix multiplication is not commutative. Example 9.22 : Show that the set G = {a + b 2 / a, b ∈ Q} is an infinite abelian group with respect to addition. Solution: (i) Closure axiom : Let x, y ∈ G. Then x = a + b 2, y = c + d 2 ; a, b, c, d ∈ Q. x + y = (a + b 2) + (c + d 2) = (a + c) + (b + d) 2 ∈ G, since (a + c) and (b + d) are rational numbers. ∴ G is closed with respect to addition. (ii) Associative axiom : Since the elements of G are all real numbers, addition is associative. (iii) Identity axiom : There exists 0 = 0 + 0 2 ∈ G such that for all x = a + b 2 ∈ G, x + 0 = (a + b 2) + (0 + 0 2) = a+b 2=x Similarly, we have 0 + x = x. ∴ 0 is the identity element of G and satisfies the identity axiom. (iv) Inverse axiom : For each x = a + b 2 ∈ G, there exists − x = (− a) + (− b) 2 ∈ G such that x + (− x) = (a + b 2) + ((− a) + (− b) 2) = (a + (− a)) + (b + (− b)) 2 = 0 Similarly we have (− x) + x = 0 ∴ (− a) + (− b) 2 is the inverse of a + b 2 and satisfies the inverse axiom. ∴ G is a group under addition. (v) Commutative axiom : x + y = (a + c) + (b + d) 2 = (c + a) + (d + b) 2 = (c + d 2) + (a + b 2) = y + x, for all x, y ∈ G. ∴ The commutative property is true. ∴ (G, +) is an abelian group. Since G is infinite, we see that (G, +) is an infinite abelian group. Example 9.23 : Let G be the set of all rational numbers except 1 and * be defined on G by a * b = a + b − ab for all a, b ∈ G. Show that (G, *) is an infinite abelian group. Solution: Let G = Q − {1} Let a, b ∈ G. Then a and b are rational numbers and a ≠ 1, b ≠ 1.
180
(i) Closure axiom : Clearly a * b = a + b − ab is a rational number. But to prove a * b ∈ G, we have to prove that a * b ≠ 1. On the contrary, assume that a * b = 1 then a + b − ab = 1 ⇒ b − ab = 1 − a ⇒ b(1 − a) = 1 − a ⇒ b = 1 (‡ a ≠ 1, 1− a ≠ 0) This is impossible, because b ≠ 1. ∴ Our assumption is wrong. ∴ a * b ≠ 1 and hence a * b ∈ G. ∴ Closure axiom is true. (ii) Associative axiom : a * (b * c) = a * (b + c − bc) = a + (b + c − bc) − a (b + c − bc) = a + b + c − bc − ab − ac + abc (a * b) * c = (a + b − ab) * c = (a + b − ab) + c − (a + b − ab) c = a + b + c − ab − ac − bc + abc ∴ a * (b * c) = (a * b) * c ∀ a, b, c ∈ G ∴ Associative axiom is true. (iii) Identity axiom : Let e be the identity element. By definition of e, a * e = a By definition of *, a * e = a + e − ae ⇒ a + e − ae = a ⇒ e(1 − a) = 0 ⇒ e = 0 since a ≠ 1 e = 0∈G ∴ Identity axiom is satisfied. (iv) Inverse axiom : Let a−1 be the inverse of a ∈ G. By the definition of inverse, a * a−1 = e = 0 By the definition of *, a * a−1 = a + a−1 − aa − 1 ⇒ a + a−1 − aa−1 = 0 ⇒ a−1 (1 − a) = − a a ⇒ a−1 = ∈ G since a ≠ 1 a−1 ∴ Inverse axiom is satisfied. ∴ (G, *) is a group. 181
(v) Commutative axiom : For any a, b ∈ G,
a * b = a + b − ab = b + a − ba = b*a ∴ * is commutative in G and hence (G, *) is an abelian group. Since G is infinite, (G, *) is an infinite abelian group. Example 9.24 : Prove that the set of four functions f1, f2, f3, f4 on the set of nonzero complex numbers C − {0} defined by 1 1 f1(z) = z, f2(z) = − z, f3(z) = z and f4(z) = − z ∀ z ∈ C − {0} forms an abelian group with respect to the composition of functions. Let G = {f1, f2, f3, f4} Solution: (f1° f1) (z) = f1(f1(z)) = f1(z) ∴ f1°f1 = f1 f2° f1 = f2 , f3°f1 = f3, f4°f1 = f4 Again
(f2°f2) (z) = f2(f2(z)) = f2(− z) = − (− z) = z = f1(z)
∴ f2°f2 = f1 Similarly f2°f3 = f4, f2°f4 = f3 1 (f3°f2) (z) = f3 (f2 (z)) = f3(− z) = − z = f4(z) ∴ f3°f2 = f4 Similarly
f3°f3 = f1, f3°f4 = f2 1 1 = = f (z) (f4°f2) (z) = f4(f2(z)) = f4(− z) = − −z z 3
∴ f4°f2 = f3 Similarly f4°f3 = f2, f4°f4 = f1 Using these results we have the composition table as given below : f2 f3 f4 f1 ° f1 f1 f2 f3 f4 f2
f2
f1
f4
f3
f3
f3
f4
f1
f2
f4
f4
f3
f2
f1
182
From the table (i) All the entries of the composition table are the elements of G . ∴ Closure axiom is true. (ii) Composition of functions is in general associative. (iii) Clearly f1 is the identity element of G and satisfies the identity axiom. (iv) From the table : Inverse of f1 is f1 ;
Inverse of f2 is f2
Inverse of f3 is f3 ;
Inverse of f4 is f4
Inverse axiom is satisfied. (G, o) is a group. (v) From the table the commutative property is also true. ∴ (G, o) is an abelian group.
9.2.3 Modulo Operation We shall now define new types of operations called “Addition modulo n” and “Multiplication modulo n”, where n is a positive integer. To define these operations we require the notion of “Division Algorithm”. Let a, b ∈ Z with b ≠ 0. Then we can divide a by b to get a quotient q and a non-negative remainder r which is smaller in size than b. i.e., a = qb + r, where 0 ≤ r < | b |. This is called “Division Algorithm”. For example, if a = 17, b = 5 then 17 = (3 × 5) + 2 Here q = 3 and r = 2 Addition modulo n (+ n) : Let a, b ∈ Z and n be a fixed positive integer. We define addition modulo n by a +n b = r ; 0 ≤ r < n where r is the least non-negative remainder when a + b is divided by n. For example, if a = 25, b = 8 and n = 7 then 25 +78 = 5 (‡ 25 + 8 = 33 = (4 × 7) + 5) Multiplication modulo n (. n) As given above a .n b = r ; 0 ≤ r < n, where r is the least non-negative remainder when ab is divided by n. For example, 2 .54 = 3 7 .98 = 2
183
Congruence modulo n : Let a, b ∈ Z and n be a fixed positive integer. We say that “a is congruent to b modulo n” ⇔ (a − b) is divisible by n Symbolically, a ≡ b (mod n) ⇔ (a − b) is divisible by n. 15 ≡ 3 (mod 4) is true because 15 − 3 is divisible by 4. 17 ≡ 4 (mod 3) is not true because 17 − 4 is not divisible by 3. Congruence classes modulo n : Let a ∈ Z and n be a fixed positive integer. Collect all numbers which are congruent to ‘a’ modulo n. This set will be denoted as [a] and is called the congruence class modulo n or residue class modulo n. Thus [a] = {x ∈ Z / x ≡ a (mod n)} = {x ∈ Z / (x − a) is divisible by n} = {x ∈ Z / (x − a) is a multiple of n} = {x ∈ Z / (x − a) = kn}, k ∈ Z = {x ∈ Z / x = a + kn}, k ∈ Z consider the congruence classes modulo 5. [a] = {x ∈ Z / x = a + kn} [0] = {x ∈ Z / x = 5k, k ∈ Z} = {... − 10, − 5, 0, 5, 10...} [1] = {x ∈ Z / x = 5k + 1, k ∈ Z} = {... − 9, − 4, 1, 6, 11, ...} [2] = {x ∈ Z / x = 5k + 2, k ∈ Z} = {... − 8, − 3, 2, 7, 12, ...} [3] = {x ∈ Z / x = 5k + 3, k ∈ Z} = {... − 7, − 2, 3, 8, 13, ...} [4] = {x ∈ Z / x = 5k + 4, k ∈ Z} = {... − 6, − 1, 4, 9, 14 ...} [5] = {x ∈ Z / x = 5k + 5, k ∈ Z} = {... − 5, 0, 5, 10, ...} = [0] Similarly [6] = [1] ; [7] = [2] ; etc. Note that, we have only 5 distinct classes whose union gives the entire Z. Thus the set of congruence classes corresponding to 5 is [0], [1], [2], [3], [4]} and it will be deonoted by Z5. { i.e., Z5 = {[0], [1], [2], [3], [4]} If we take the modulo 6, we have Z6 = {[0], [1] .... [5]}. Thus for any positive integer n, we have Zn = {[0], [1] ... [n − 1]} Here [n] = [0] and the union of these classes gives Z.
184
Operations on congruence classes : (1) Addition : Let [a], [b] ∈ Zn [a] + n [b] = [a + b] if a + b < n = [r] if a + b ≥ n Where r is the least non-negative remainder when a + b is divided by n. For example, In Z10 , [5] +10 [7] = [2] In Z8 , [3] +8 [5] = [0] (ii) Multiplication : [ab] if ab < n [a] .n [b] = if ab ≥ n [r] where r is the least non-negative remainder when ab is divided by n In Z5 [2] .5[2] = [4] [3] .5 [4] = 2 In Z7,
[3] .7 [3] = [2]
In Z8 ,
[5] .8 [3] = [7]
Example 9.25 : Show that (Zn, +n) forms group. Solution: Let Zn = {[0], [1], [2], ... [n − 1]} be the set of all congruence classes modulo n. and let [l], [m], ∈ Zn 0 ≤ l, m, < n (i) Closure axiom : By definition [l + m] if l + m < n [l] + n [m] = [r] if l + m ≥ n where l + m = q . n + r 0 ≤ r < n In both the cases, [l + m] ∈ Zn and [r] ∈ Zn ∴ Closure axiom is true. (ii) Addition modulo n is always associative in the set of congruence classes modulo n. (iii) The identity element [0] ∈ Zn and it satisfies the identity axiom. (iv) The inverse of [l] ∈ Zn is [n − l] Clearly [n − l] ∈ Zn and
185
[l] + n [n − l] = [0] [n − l] + n [l] = [0] ∴ The inverse axiom is also true. Hence (Zn, +n) is a group. Note : (Zn, +n) is a finite abelian group of order n. Example 9.26 : Show that (Z7 − {[0]}, .7) forms a group. Solution: Let G = [[1], [2], ... [6]] The Cayley’s table is .7 [1] [2] [1] [2] [3] [4] [5] [6]
[1] [2] [3] [4] [5] [6]
[2] [4] [6] [1] [3] [5]
[3]
[4]
[5]
[6]
[3] [6] [2] [5] [1] [4]
[4] [1] [5] [2] [6] [3]
[5] [3] [1] [6] [4] [2]
[6] [5] [4] [3] [2] [1]
From the table : (i) all the elements of the composition table are the elements of G. ∴ The closure axiom is true. (ii) multiplication modulo 7 is always associative. (iii) the identity element is [1] ∈ G and satisfies the identity axiom. (iv) the inverse of [1] is [1] ; [2] is [4] ; [3] is [5] ; [4] is [2] ; [5] is [3] and [6] is [6] and it satisfies the inverse axiom. ∴ the given set forms a group under multiplication modulo 7. In general, it can be shown that (Zp − {(0)}, . p) is a group for any prime p. But the proof is beyond the scope of this book. Note : Does the set of all non-zero congruence classes modulo n, a positive integer, form a group under multiplication modulo n, ? Example 9.27 : Show that the nth roots of unity form an abelian group of finite order with usual multiplication. Solution: We know that 1, ω, ω2...... ωn − 1 are the nth roots of unity, where 2π ω = cis n . Let G = {1, ω, ω2 ... ωn − 1} (i) Closure axiom : Let ω l, ωm ∈ G, 0 ≤ l, m ≤ (n − 1) To prove ωl ωm = ωl + m ∈ G 186
Case (i) l + m < n If l + m < n then clearly ωl + m ∈ G Case (ii) l + m ≥ n By division algoritham, l + m = (q . n) + r where 0 ≤ r < n, q is a positive integer. q
ωl + m = ωqn + r = (ωn) . ωr = (1)qωr = ωr ∈ G ‡ 0 ≤ r < n Closure property is true. (ii) Associative axiom : Multiplication is always associative in the set of complex numbers and hence in G ωl .(ωp.ωm) = ωl . ω(p + m) = ωl + (p + m) = ω( l + p) + m = (ωl + p) . ωm = (ωl . ωp) . ωm = ∀ ω l, ωm, ωp ∈ G (iii) Identity axiom : The identity element 1 ∈ G and it satisfies 1.ωl = ωl .1 = ωl ∀ ωl ∈ G (iv) Inverse axiom : For any ωl ∈ G, ωn − l ∈ G and ωl . ωn − l = ωn − l .ωl = ωn = 1 Thus inverse axiom is true. ∴ (G, .) is a group. (v) Commutative axiom : ωl . ωm = ωl + m = ωm + l = ωm . ωl ∀ ωl , ωm ∈ G ∴ (G, .) is an abelian group. Since G contains n elements, (G, .) is a finite abelian group of order n.
9.2.4 Order of an element : Let G be a group and a ∈ G. The order of ‘a’ is defined as the least positive integer n such that an = e, e is the identity element. If no such positive integer exists, then a is said to be of infinite order. The order of a is denoted by 0(a). Note : Here an = a * a * a ... *a (n times). If * is usual multiplication ‘.’ then an is a . a .a... (n times) i.e., an . If * is usual addition then an is a + a + a + ... + a (n times) i.e., na. Thus an is not “a to the power n”, it is a symbol to denote a * a * a ... * a (n times). Clearly an ∈ G, if a ∈ G . (By the repeated application of closure axiom). Theorem : For any group G, the identity element is the only element of order 1.
187
Proof : If a (≠ e) is another element of order 1 then by the definition of order of an element, we have (a)1 = e ⇒ a = e which is a contradiction. ∴ e is the only element of order 1. Example 9.28 : Find the order of each element of the group (G, .) where G = {1, − 1, i, − i}. Solution: In the given group, the identity element is 1. ∴ 0(1) = 1. 0(− 1) = 2 [Q we have to multiply − 1 two times (minimum) to get 1 i.e., (− 1) (− 1) = 1] 0(i) = 4 [Q we have to multiply i four times to get 1, i.e., (i) (i) (i) (i) = 1] 0(− i) = 4 [Q we have to multiply − i four times to get 1]. Example 9.29 : Find the order of each element in the group G = {1, ω, ω2}, consisting of cube roots of unity with usual multiplication. Solution: We know that the identity element is 1. ∴ 0(1) = 1. 0(ω) = 3. Since ω . ω . ω = ω3 = 1 0(ω2) = 3 since (ω2) (ω2) (ω2) = ω6 = 1 Example 9.30 : Find the order of each element of the group (Z4, +4) Solution: Z4 = {[0], [1], [2], [3]} is an abelian group under the addition modulo 4. The identity element is [0] and note that [4] = [8] = [12] = [0] ∴ 0([0]) = 1 0([1]) = 4 [Q we have to add [1] four times to get [4] or [0]] 0 ([2]) = 2 [Q we have to add [2] two times to get [4] or [0]] 0 ([3]) = 4 Q we have to add [3] four times to get [12] or [0]
9.2.5 Properties of Groups :
Theorem : The identity element of a group is unique. Proof : Let G be a group. If possible let e1 and e2 be identity elements in G. Treating e1 as an identity element we have e1 * e2 = e2 Treating e2 as an identity element, we have e1 * e2 = e1 From (1) and (2), e1 = e2 ∴ Identity element of a group is unique. Theorem : The inverse of each element of a group is unique. Proof : Let G be a group and let a ∈ G. If possible, let a1 and a2 be two inverses of a.
188
… (1) … (2)
Treating a1 as an inverse of ‘a’ we have a * a1 = a1 * a = e. Treating a2 as an inverse of ‘a’, we have a * a2 = a2 * a = e a1 = a1 * e = a1 * (a * a2) = (a1 * a) * a2 = e * a2 = a2 ⇒ Inverse of an element is unique. Theorem : (Cancellation laws) Let G be a group. Then for all a, b, c ∈ G, (i) a * b = a * c ⇒ b = c (Left Cancellation Law) (ii) b * a = c * a ⇒ b = c (Right Cancellation Law) Now
Proof : (i)
a * b = a * c ⇒ a−1 * (a * b) = a−1 * (a * c) ⇒ (a−1 * a) * b = (a−1 * a) * c ⇒ e*b=e*c ⇒ b=c
(ii)
b * a = c * a ⇒ (b * a) * a−1 = (c * a) * a−1 ⇒ b * (a * a −1) = c * (a * a−1) ⇒ b*e=c*e ⇒ b=c
Theorem : In a group G, (a−1) Proof :
−1
= a for every a ∈ G. −1
We know that a−1 ∈ G and hence (a−1) a−1 * (a−1)
−1
∈ G. Clearly a * a−1 = a−1 * a = e
−1
= (a−1) * a−1 = e
⇒ a * a−1 = (a−1)
−1
⇒ a = (a−1) Theorem : (Reversal law)
−1
* a−1 (by Right Cancellation Law)
Let G be a group a, b ∈ G. Then (a * b)− 1 = b−1 * a−1. Proof : It is enough to prove b−1 * a−1 is the inverse of (a * b) ∴ To prove (i) (a * b) * (b−1 * a−1) = e (ii) (b−1 * a−1) * (a * b) = e (i) (a * b) * (b−1 * a−1) = a * (b * b−1) * a−1 = a * (e) * a−1 = a * a−1 = e
189
(ii)
(b−1 * a−1) * (a * b) = = =
b−1 * (a−1 * a) * b b−1 * (e) * b b−1 * b = e
∴ b−1 * a−1 is the inverse of a * b i.e., (a * b)−1 = b−1 * a−1 EXERCISE 9.4 (1) Let S be a non-empty set and o be a binary operation on S defined by xoy = x ; x, y ∈ S. Determine whether o is commutative and associative. (2) Show that the set N of natural members is a semi-group under the operation x * y = max {x, y}. Is it a monoid? (3) Show that the set of all positive even integers forms a semi-group under the usual addition and multiplication. Is it a monoid under each of the above operations? 1 0 0 1 (4) Prove that the matrices , form a group under matrix 0 1 1 0 multiplication. (5) Show that the set G of all positive rationals forms a group under the ab composition * defined by a * b = 3 for all a, b ∈ G. 1 0 ω 0 ω2 0 0 1 0 ω2 0 ω (6) Show that , , , , , 0 1 0 ω2 0 ω 1 0 ω 0 ω2 0 where ω3 = 1, ω ≠ 1 form a group with respect to matrix multiplication. (7) Show that the set M of complex numbers z with the condition | z | = 1 forms a group with respect to the operation of multiplication of complex numbers. (8) Show that the set G of all rational numbers except − 1 forms an abelian group with respect to the operation * given by a * b = a + b + ab for all a, b ∈ G. (9) Show that the set {[1], [3], [4], [5], [9]} forms an abelian group under multiplication modulo 11. (10) Find the order of each element in the group (Z5 − {[0]}, .5) a o (11) Show that the set of all matrices of the form , a ∈ R − {0} forms o o an abelian group under matrix multiplication. (12) Show that the set G = {2n / n ∈ Z} is an abelian group under multiplication.
190
10. PROBABILITY DISTRIBUTIONS 10.1 Introduction : In XI Standard we dealt with random experiments which can be described by finite sample space. We studied the assignment and computation of probabilities of events. In the Sciences one often deals with variables as a ‘quantity that may assume any one of a set of values’. In Statistics we deal with random variables - variables whose observed value is determined by chance.
10.2. Random Variable : The outcomes of an experiment are represented by a random variable if these outcomes are numerical or if real numbers can be assigned to them. For example, in a die rolling experiment, the corresponding random variable is represented by the set of outcomes {1, 2, 3, 4, 5, 6} ; while in the coin tossing experiment the outcomes head (H) or tail (T) can be represented as a random variable by assuming 0 to T and 1 to H. In this sense a random variable is a real valued function that maps the sample space into the real line. Let us consider the tossing of two fair coins at a time. The possible results are {HH, TH, HT, TT}. Let us consider the variable X which is “the number of heads obtained” while tossing two fair coins. We could assign the value X = 0 to the outcome of getting no heads, X = 1 to the outcome of getting only 1 head and X = 2 to the out come of getting 2 heads. Therefore X (TT) = 0, X(TH) = 1, X (HT) = 1 and X (HH) = 2. Therefore X takes the values 0,1,2. Thus we can assign a real number X(s) to every element s of the sample space S. Definition : If S is a sample space with a probability measure and X is a real valued function defined over the elements of S, then X is called a random variable. A random variable is also called a chance variable or a stochastic variable. Types of Random variables : (1) Discrete Random variable (2) Continuous Random variable
10.2.1 Discrete Random Variable : Definition : Discrete Random Variable If a random variable takes only a finite or a countable number of values, it is called a discrete random variable. Note : Biased coins may have both sides marked as tails or both sides marked as heads or may fall on one side only for every toss, whereas a fair or unbiased coin means, it has equal chances of falling on heads and tails. Similarly biased dice may have repeated numbers on several sides ; some numbers may be missing. For a fair die the probability of getting any number from one to six will be 1/6.
191
Example : 1. The number of heads obtained when two coins are tossed is a discrete random variable as X assumes the values 0, 1 or 2 which form a countable set. 2. Number of Aces when ten cards are drawn from a well shuffled pack of 52 cards. The random variable X assumes 0, 1, 2, 3 or 4 which is again a countable set. i.e., X (No aces) = 0, X (one ace) = 1, X (two aces) = 2, X (three aces) = 3, X (four aces) = 4 Probability Mass Function : The Mathematical definition of discrete probability function p(x) is a function that satisfies the following properties : (1) The probability that X can take a specific value x is p(x) ie., P(X = x) = p(x) = px. (2) p(x) is non – negative for all real x. (3) The sum of p(x) over all possible values of X is one. That is ∑pi = 1 where j represents all possible values that X can have and pi is the probability at X = xi If a1, a2, . . . am, a, b1, b2, . . bn, b be the values of the discrete random variable X in ascending order then (i) P(X ≥ a) = 1 − P(X < a) (ii) P(X ≤ a) = 1 − P(X > a) (iii) P(a ≤ X ≤ b) = P(X = a) + P(X = b1) + P(X = b2) + . . . . . . + P(X = bn) + P(X = b). Distribution function : (Cumulative Distribution function) The distribution function of a random variable X is defined as F(x) = P(X ≤ x) = ∑ p(xi) : (− ∞ < x < ∞). xi ≤ x
Properities of Distribution function : 1)
F(x) is a non-decreasing function of x
2)
0 ≤ F(x) ≤ 1, − ∞ < x < ∞
192
3) 4) 5)
Lt F(x) = 0 x→−∞ Lt F(∞) = F(x) =1 x →+∞ P(X = xn) = F(xn) − F(xn −1) F(− ∞) =
Illustration : Find the probability mass function and cumulative distribution function for getting number of heads when three coins are tossed once. Solution : Let X be the random variable “getting number of Heads”. Sample space when three coins are tossed is S = HHH HHT HTH THH HTT THT TTH TTT ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ R : 3 2 2 2 1 1 1 0 (No.of Heads)
Since X is the random variable getting the number of heads, X takes the values 0, 1,2 and 3. (X : S → R). 1 P (getting no head) = P (X = 0) = 8 3 P (getting one head) = P (X = 1) = 8 3 P (getting two heads) = P (X = 2) = 8 1 P (getting three heads) = P (X = 3) = 8 P(x) ∴ probability mass function is given by 1 1/8 if x = 0 3/8 if x = 1 OR P (X= x) = 3/8 if x = 2 1/8 if x = 3 X 0 1 2 3 1/ P(X = x) 1/8 3/8 3/8 1/8 8 0
To find cumulative distribution function.
1
2
3
Fig. 10.1 x
We have F(x) =
∑ P(X =
xi = − ∞
193
xi)
x
1 When X = 0, F(0) = P(X = 0) = 8 1
∑ P(X =
When X = 1, F(1) =
xi)
i=−∞
3 4 1 1 = P(X = 0) + P(X = 1) = 8 + 8 = 8 = 2 2
∑ P(X =
When X = 2, F(2) =
xi)
i=−∞
= P (X = 0) + P(X = 1) + P(X = 2) 1 3 3 7 = 8 + 8 + 8 =8 3
When X = 3, F(3) =
∑ P(X =
xi)
i=−∞
= P (X = 0) + P(X = 1) + P(X = 2) + P(X = 3) 3 3 1 1 =8 + 8 + 8 + 8 = 1 Cumulative distribution function is 0 if − ∞ < x < 0 1/8 if 0 ≤ x < 1 1/2 if 1 ≤ x < 2 F(x) = 7/8 if 2 ≤ x < 3 1 if 3 ≤ x < ∞
F(x)
1O 7/8 O 1/2 O 1/8 O 0
O
O
O
1
2
3
Fig. 10.2
x
Example 10.1 : Find the probability mass function, and the cumulative distribution function for getting ‘3’s when two dice are thrown. Solution : Two dice are thrown. Let X be the random variable of getting number of ‘3’s. Therefore X can take the values 0, 1, 2.
194
25 P(no ‘3’) = P(X = 0) = 36
Sample Space (1,3) (1,4) (2,4) (2,3)
(1,5) (2,5)
(1,6) (2,6)
(3,4)
(3,5)
(3,6)
(4,1) (4,2) (4,4) (4,3) (5,1) (5,2) (5,4) (5,3) (6,1) (6,2) (6,3) (6,4) probability mass function is given by x 0 1 2 P(X = x) 25/36 10/36 1/36 Cumulative distribution function :
(4,5) (5,5) (6,5)
(4,6) (5,6) (6,6)
10 P(one ‘3’) = P(X = 1) = 36 1 P(two ‘3’s) = P(X = 2) = 36
(1,1) (2,1)
(1,2) (2,2)
(3,1)
(3,2)
(3,3)
x
We have F(x) =
∑ P(X =
xi)
xi = − ∞
25 F(0) = P(X = 0) = 36 10 35 25 F(1) = P(X = 0) + P(X = 1) = 36 + 36 = 36 25 10 1 36 F(2) = P(X = 0 ) + P(X = 1) + P(X =2) = 36 + 36 + 36 = 36 = 1 x 0 1 2 F(x) 25/36 35/36 1 Example 10.2 A random variable X has the following probability mass function x 0 1 2 3 4 5 6 P(X = x) k 3k 5k 7k 9k 11k 13k (1) Find k. (2) Evaluate P(X < 4), P(X ≥ 5) and P(3< X ≤ 6) 1 (3) What is the smallest value of x for which P (X ≤ x) > 2 . Solution : 6 (1) Since P(X = x) is a probability mass function ∑ P(X = x) = 1 x=0 ie.,P(X=0) + P(X = 1) +P(X = 2) +P(X = 3) +P(X = 4) +P(X = 5)+P(X = 6) = 1. 1 ⇒ k + 3k + 5k + 7k + 9k + 11k + 13k = 1 ⇒ 49 k = 1 ⇒ k = 49
195
(2)
P(X < 4) = P(X = 0) + P(X = 1 ) + P(X = 2) + P(X = 3) 3 5 7 16 1 = 49 + 49 + 49 + 49 = 49 11 13 24 P(X ≥ 5) = P(X = 5) + P(X = 6) = 49 + 49 = 49 9 11 13 33 P(3 < X ≤ 6) = P(X = 4) + P(X = 5) + P(X = 6) = 49 + 49 + 49 = 49 (3) The minimum value of x may be determined by trial and error method. 1 4 1 1 P(X ≤ 0) = 49 < 2 ; P(X ≤ 1) = 49 < 2 1 16 1 9 P(X ≤ 2) = 49 < 2 ; P(X ≤ 3) = 49 < 2 1 25 P(X ≤ 4) = 49 > 2 1
∴ The smallest value of x for which P(X ≤ x) > 2 is 4. Example 10.3 :An urn contains 4 white and 3 red balls. Find the probability distribution of number of red balls in three draws one by one from the urn. (i) with replacement (ii) without replacement Solution : (i) with replacement Let X be the random variable of drawing number of red balls in three draws. ∴ X can take the values 0,1,2,3. 3 P(Red ball) = 7 = P(R) 4 P(Not Red ball) = 7 = P(W) 4 4 4 64 Therefore P(X = 0) = P(www) = 7 × 7 × 7 = 343 P(X = 1) = P(Rww) + P(wRw) + P(wwR) 4 4 3 4 4 3 3 4 4 = 7 × 7 × 7 + 7 × 7 × 7 + 7 × 7 × 7 144 48 = 3 × 343 = 343 P(X = 2) = P(RRw) + P(RwR) + P(wRR) 3 4 4 3 3 3 3 3 4 = 7 × 7 × 7 + 7 × 7 × 7 + 7 × 7 × 7 3 4 36 108 3 = 3 × 7 × 7 × 7 = 3 × 343 = 343
196
3 3 3 27 P(X = 3) = P(RRR) = 7 × 7 × 7 = 343 The required probability distribution is X 0 1 2 P(X = x) 64/343 144/343 108/343 Clearly all pi’s are ≥ 0 and ∑pi = 1.
3 27/343
2) Without replacement : It is also treated a simultaneous case. Method 1 : Method 2 : Using combination Using Conditional Probability 3 2 4 (i) P(no red ball) (i) P(www) = 7 × 6 × 5 4c × 3c 3 0 4 P(X = 0) = 7c = 35 3 4 4×1 = 35 = 35 (ii) P(1 red ball) 4c × 3c P(X = 1) =
2
7c
1
3
6×3 18 = 35 = 35 (iii) P(2 red ball) 4c × 3c P(X = 2) =
1
7c
2
3
4×3 12 = 35 = 35 (iv)P(3 red ball) 4c × 3c P(X = 3) =
0
7c
(ii) P(Rww) + P(wRw) + P(wwR) 4 3 3 3 3 4 = 7 × 6 × 5 + 7 × 6 × 5 3 3 4 + 7 × 6 × 5 36 36 18 = 3 × 210 = 70 = 35 (iii) P(RRw) + P(RwR) + P(wRR) 2 4 4 2 3 3 = 7 × 6 × 5 + 7 × 6 × 5 3 2 4 + 7 × 6 × 5 24 12 = 3 × 210 = 35 (iv)
3
3
1 1×1 = 35 = 35
197
3 2 1 P(RRR) = 7 × 6 × 5 1 = 35
X
0
1
2
3
P(X = x)
4 35
18 35
12 35
1 35
Clearly all pi’s are ≥ 0 and ∑pi = 1
10.2.2 Continuous Random Variable : Definition : A Random Variable X is said to be continuous if it can take all possible values between certain given limits. i.e., X is said to be continuous if its values cannot be put in 1 − 1 correspondence with N, the set of Natural numbers. Examples for Continuous Random Variable The life length in hours of a certain light bulb. Let X denote the ph value of a chemical compound which is randomly selected. Then X is a continuous random variable because any ph value, between 0 and 14 is possible. If in the study of ecology of a lake, we make depth measurements at randomly chosen locations then X = the depth at such location is a continuous random variable. The limit will be between the maximum and minimum depth in the region sampled. Probability Density Function (p.d.f.) : The mathematical definition of a continuous probability function f(x) is a function that satisfies the following properties. (i) The probability that X is between two points a and b is b
P(a ≤ x ≤ b) = ⌡ ⌠f(x) dx a
(ii) It is non-negative for all real X. ∞ (iii) The integral of the probability function is 1 i.e., ⌠ ⌡f(x) dx = 1 −∞ Continuous probability functions are referred to as p.d.f. Since continuous probability function are defined for uncountable number of points over an interval, the probability at a single point is always zero. a i.e., P(X = a) = ⌡ ⌠f(x) dx = 0. a
198
The probabilities are measured over intervals and not at single points. That is, the area under the curve between two distinct points defines the probability for that interval. ∴ P(a ≤ x ≤ b) = P(a ≤ X < b) = P(a < x ≤ b) = P(a < x < b) Discrete Probability function are referred to as probability mass function and continuous probability function are referred to as probability density function. The term probability function covers both discrete and continuous distribution. Cumulative Distribution Function : If X is a continuous random variable, the function given by x F(x) = P(X ≤ x) = ⌠ f(t)dt for − ∞ < x < ∞ where f(t) is the value of the
⌡
−∞ probability density function of X at t is called the distribution function or cumulative distribution of X. Properties of Distribution function : (i) F(x) is a non-decreasing function of x (ii) 0 ≤ F(x) ≤ 1, − ∞ < x < ∞. x −∞ lt (iii) F(− ∞) = x → − ∞ ⌠ f(x) dx = ⌠ ⌡ ⌡f(x) dx = 0 −∞
lt (iv) F(∞) = x → ∞
−∞
∞
x
⌠ ⌡f(x) dx = ⌠ ⌡f(x) dx = 1
−∞ −∞ (v) For any real constant a and b and a ≤ b, d (vi) f(x) = dx F(x)
P(a ≤ x ≤ b) = F(b) − F(a)
i.e., F′(x) = f(x) Example 10.4 : A continuous random variable X follows the probability law, k x (1 − x )10, 0 < x < 1 f(x) = elsewhere 0 Find k ∞ Solution : Since f(x) is a p.d.f ⌠ ⌡f(x) dx = 1 −∞
199
1
By properties of definite integral a a f(x) dx = ⌠ ⌠ f(a − x)dx
10 ⌠ ⌡kx(1 −x) dx = 1
i.e.,
0
1
k(1 − x) [1 − (1 − x)]10dx = 1 ⌠ ⌡
i.e.,
⌡
⌡
0
0
0 1
i.e.,
k ⌠ (1 − x)x10dx = 1
⌡ 0
1
x11 x12 1 1 i.e., k ⌠ (x10 − x11)dx = 1 ⇒ k 11 − 12 = 1 ⇒ k 11 − 12 = 1 ⇒ k = 132 0 ⌡ 0 1
Example 10.5 : A continuous random variable X has p.d.f. f(x) = 3x2, 0 ≤ x ≤ 1, Find a and b such that. (i) P(X ≤ a) = P(X > a) and (ii) P(X > b) = 0.05 Solution : (i) Since the total probability is 1, [Given that P(X ≤ a) = P (X > a] P(X ≤ a) + P(X > a) = 1 i.e., P(X ≤ a) + P(X ≤ a) = 1 1 ⇒ P(X ≤ a) = 2 a
a
1 ⇒ ⌠ ⌡f(x) dx = 2
⇒
1
2 ⌠ ⌡3x dx = 2 0
0
1
a
3x3 1 1 1 3 i.e., 3 = 2 ⇒ a3 = 2 i.e., a = 2 0 (ii) P(X > b) = 0.05 1 1 2 ∴ ⌠ ⌡3x dx = 0.05 ⌡f(x) dx = 0.05 ∴ ⌠ b
b 1
3x3 3 = 0.05 ⇒ 1 − b3 = 0.05 b 1
95 19 3 b = 1 − 0.05 = 0.95 = 100 ⇒ b = 20 3
200
Example 10.6 : If the probability density function of a random variable is given k (1 − x2), 0 < x < 1 by f(x) = elsewhere 0 find (i) k (ii) the distribution function of the random variable. ∞ Solution: (i) Since f(x) is a p.d.f. ⌠ ⌡f(x) dx = 1 −∞
1
1
x3 1 ⇒ k x − 3 = 1 ⇒ k 1 − 3 = 1 0 0 3 2 ⇒ 3 k = 1 or k = 2 x (ii) The distribution function F(x) = ⌠ ⌡ f(t) dt 2 ⌠ ⌡k(1 − x ) dx = 1
−∞
(a) When x ∈ (− ∞, 0] x F(x) = ⌠ f(t) dt = 0
⌡
−∞ (b) When x ∈ (0, 1) x F(x) = ⌠ f(t) dt
⌡
−∞ x x 3 0 3 = ⌠ f(t) dt + ⌠ f(t) dt = 0 + ⌠ 2 (1 − t2) dt = 2 ⌡ ⌡ ⌡ 0 0 −∞
x3 x − 3
(c) When x ∈ [1, ∞) 1 x 13 x 0 F(x) = ⌠ f(t) dt = ⌠ f(t) dt + ⌠ f(t) dt + ⌠ f(t) dt = 0 + ⌠ 2 (1 − t2) dt + 0
⌡
⌡
−∞
−∞ 1
3 t3 = 2 t − 3 = 1 0
⌡
⌡
⌡
0
1
0
−∞
201
1 π + tan−1 x − ∞ < x < ∞ is a distribution π 2 function of a continuous variable X, find P(0 ≤ x ≤ 1) 1 π Solution: F(x) = 2 + tan−1 x π P(0 ≤ x ≤ 1) = F(1) − F(0) 1 π 1 π = + tan−1 1 − 2 + tan−1 0 π π 2 1 π π 1 π 1 π π π 1 = 2 + 4 − 2 + 0 = π π 2 + 4 − 2 = 4 π A, 1 < x < e3 is a probability density function of Example 10.8 : If f(x) = x 0, elsewhere a continuous random variable X, find p(x > e) ∞ Solution: Since f(x) is a p.d.f. ⌡ ⌠f(x) dx = 1 Example 10.7 : If F(x) =
−∞
e
3 3
⌠A dx = 1 ⇒ A[log x] e = 1 1 ⌡x 1
⇒ A[log e3 − log 1] = 1 ⇒ A[3] = 1 ⇒ A = 1/3 1 , 1 < x < e3 Therefore f(x)= 3x 0 elsewhere e3
3
e 1 1 1 P(x > e) = 3 ⌠ x dx =3 [log x] ⌡ e e
1 2 1 = 3 [log e3 − log e] = 3 [3 − 1] = 3 Example 10.9 :For the probability density function 2e−2x, x > 0 f(x)= , find F(2) ,x≤0 0 2 Solution : F(2) = P(X ≤ 2) = ⌠ ⌡f(x) dx −∞
202
2
e−2x e4 − 1 −2x = − [e−4 − 1] = 1 − e−4 = 4 =⌠ dx = 2 . ⌡2e −2 0 e 0 Example 10.10 : The total life time (in year) of 5 year old dog of a certain breed is a Random Variable whose distribution function is given by , for x ≤ 5 0 F(x) = 1 − 25 , for x > 5 Find the probability that such a five year old dog x2 will live (i) beyond 10 years (ii) less than 8 years (iii) anywhere between 12 to 15 years. Solution : (i) P(dog living beyond 10 years) P(X > 10) = 1 − P(X ≤ 10) 25 = 1 − 1 − 2 when x = 10 x 25 3 1 = 1 − 1 − 100 = 1 − 4 = 4 (ii) P(dog living less than 8 years ) P(X < 8) = F(8) [since P(X < 8) = P(X ≤ 8) for a continuous distribution] 25 25 39 = 1 − 2 = 1 − 64 = 64 8 2
(iii) P(dog living any where between 12 and 15 years ) = P(12 < x < 15) 25 25 1 = F(15) − F(12) = 1 − 2 − 1 − 2 = 16 15 12 EXERCISE 10.1 (1) Find the probability distribution of the number of sixes in throwing three dice once. (2) Two cards are drawn successively without replacement from a well shuffled pack of 52 cards. Find the probability distribution of the number of queens. (3) Two bad oranges are accidentally mixed with ten good ones. Three oranges are drawn at random without replacement from this lot. Obtain the probability distribution for the number of bad oranges. (4) A discrete random variable X has the following probability distributions. X 0 1 2 3 4 5 6 7 8 P(x) a 3a 5 a 7 a 9 a 11 a 13 a 15 a 17 a (i) Find the value of a (ii) Find P(x < 3) (iii) Find P(3 < x < 7)
203
(5) Verify that the following are probability density functions. 2x , 0 ≤ x ≤ 3 1 1 , −∞ < x < ∞ (b) f(x) = (a) f(x) = 9 π (1 + x2) 0 elsewhere 3, 0
α − 1 −β xα e , x, α, β > 0 . Find (i) k (ii) P(X > 10) f(x) = kx 0 , elsewhere x<0 0 2 (8) For the distribution function given by F(x) = x 0 ≤ x ≤ 1 1 x>1 find the density function. Also evaluate (i) P(0.5 < X < 0.75) (ii) P(X ≤ 0.5) (iii) P(X > 0.75) (9) A continuous random variable x has the p.d.f defined by ce−ax, 0 < x < ∞ f(x) = 0 elsewhere . Find the value of c if a > 0. (10) A random variable X has a probability density function k , 0 < x < 2π f(x) = 0 elsewhere π 3π π Find (i) k (ii) P 0 < X < 2 (iii) P 2 < X < 2
10.3 Mathematical Expectation : Expectation of a discrete random variable : Definition : If X denotes a discrete random variable which can assume the values x1, x2, . . . . . . xn with respective probabilities p1, p2, . . . pn then the mathematical expectation of X, denoted by E(X) is defined by E(X) = p1 x1 + p2 x2 + . . . . . + pnxn =
204
n
n
i=1
i=1
∑ pi xi where ∑ pi = 1
Thus E(X) is the weighted arithmetic mean of the values xi with the
weights p(xi) ∴ X = E(X) Hence the mathematical Expectation E(X) of a random variable is simply the arithmetic mean. Result : If ϕ(X) is a function of the random variable X, then E[ϕ (X)] = ∑ P(X = x) ϕ (x). Properties : Result (1) :
E(c) = c where c is a constant
Proof :
E(X) = ∑ pi xi ∴ E(c) = ∑ pi c = c ∑ pi = c as ∑ pi = 1
Result (2) : Proof :
Result (3) : Proof :
∴ E(c) = c E(cX) = c E(X) E(cX) = ∑ (cxi)pi = (c x1) p1 + (c x2) p2 + . . . (c xn) pn = c( p1 x1 + p2x2 +. . . . pn xn) = E(aX + b) = E(aX + b) = =
c E(X) a E(X) + b. ∑ (a xi+ b) pi (a x1+ b) p1 + (a x2 + b)p2 + (a xn + b) pn
= a( p1 x1 + p2x2 +. . . . pn xn) + b∑ pi = a E(X) + b. Similarly E(aX − b) = aE(X)− b Moments : Expected values of a function of a random variable X is used for calculating the moments. We will discuss about two types of moments. (i) Moments about the origin (ii) Moments about the mean which are called central moments. Moments about the origin : If X is a discrete random variable for each positive integer r (r = 1, ...) the th r moment µr′ = E(Xr) = ∑ pi xir First moment : µ1′ = E(X) = ∑ pi xi This is called the mean of the random variable X. Second moment : µ2′ = E(X2) = ∑ pi xi2
205
Moments about the Mean : (Central Moments) For each positive integer n, (n = 1, 2, ...) the nth central moment of the discrete random variable X is µn = E(X − X )n = ∑(xi −x− )n pi First moment about the Mean µ1 = E(X − X )1 = ∑(xi −x− )1 pi µ = ∑ x p − −x ∑ p = ∑ x p − − x (1) as ∑ p = 1 1
i i
i
i i
i
= E(X) − E(X) = 0 The algebraic sum of the deviations about the arithmetic mean is always zero 2nd moment about the Mean µ2 = E(X − X )2 2
2
= E(X2 + X − 2 X X ) = E(X2) + X −2 X E(X) (‡ X is a constant) = E(X2) + [E(X)]2 − 2E(X) E(X) µ2 = E(X2) −[E(X)]2 = µ2′ − (µ1′)2 Second moment about the Mean is called the variance of the random variable X µ2 = Var (X) = E(X − X )2 = E(X2) − [E(X)]2
Result (4) : Proof :
Var (X ± c) = Var X where c is a constant. w.k.t. Var (X) = E(X − X )2
Var (X + c) = E[(X + c) − E (X + c)]2 = E[X + c − E(X) − c]2
= E[X − X ]2 = Var X Similarly Var (X − c) = Var (X) ∴ Variance is independent of change of origin. Result (5) :
Var (aX) = a2 Var (X)
Proof :
Var (aX) = E[aX − E(aX)]2 = E[aX − aE(X)]2 = E[a {X − E(X)}]2
= a2 E[X − E(X)]2 = Change of scale affects the variance
206
a2 Var X
Result (6) :
Var (c) = 0 where c is a constant. Proof : Var (c) = E[c − E(c)]2 = E[c − c]2 = E(0) = 0 Example 10.11 : Two unbiased dice are thrown together at random. Find the expected value of the total number of points shown up. Solution : Let X be the random variable which represents the sum of the numbers shown in the two dice. If both show one then the sum total is 2. If both show six then the sum is 12. The random variable X can take values from 2 to 12. (1, 1) (1, 2) (2, 1) (1, 3) (2, 2) (3, 1) (1, 4) (2, 3) (3, 2) (4, 1) (1, 5) (2, 4) (3, 3) (4, 2) (5, 1) (1, 6) (2, 5) (3, 4) (4, 3) (5, 2) (6, 1) (2, 6) (3, 5) (4, 4) (5, 3) (6, 2) (3, 6) (4, 5) (5, 4) (6, 3) (4, 6) (5, 5) (6, 4) (5, 6) (6, 5) (6, 6) ∴ The probability distribution is given by. X 2 3 4 5 6 7 8 9 10 11 12 2 3 4 5 6 5 4 3 2 1 1 P(X = x) 36 36 36 36 36 36 36 36 36 36 36 E(X) = ∑ pi xi = ∑ xi pi 1 2 3 1 252 = 2 × 36 + 3 × 26 + 4 × 36 + . . . .+ 12 × 36 = 36 = 7 Example 10.12 : The probability of success of an event is p and that of failure is q. Find the expected number of trials to get a first success. Solution: Let X be the random variable denoting ‘Number of trials to get a first success’. The success can occur in the 1st trial. ∴ The probability of success in the 1st trial is p. The success in the 2nd trial means failure in the 1st trial. ∴ Probability is qp. Success in the 3rd trial means failure in the first two trials. ∴ Probability of success in the 3rd trial is q2p. As it goes on, the success may occur in the nth trial which mean the first (n −1) trials are failures. ∴ probability = qn−1p.
207
∴ The probability distribution is as follows X 1 2 3 ... P(x)
p
qp
2
q p
...
n ... n−1
q p... ∴ E(X) = ∑ pi xi = 1 . p + 2qp + 3q2p + . . . + nqn−1 p . . = p[1 + 2q + 3q2 + . . .+ nqn−1 + ... ] p 1 = p[1 − q]−2 = p(p)−2 = 2 = p p Example 10.13 : An urn contains 4 white and 3 Red balls. Find the probability distribution of the number of red balls in three draws when a ball is drawn at random with replacement. Also find its mean and variance. Solution : The required probability distribution is [Refer Example 10.3] X 0 1 2 3 144 108 27 64 P(X = x) 343 343 343 343 Mean E(X) = ∑ pi xi 9 64 144 108 27 = 0 343 + 1 343 + 2 343 + 3 343 = 7 Variance = E(X2) − [E(X)]2 E(X2) = ∑ pi xi2 117 64 144 108 27 = 0 343 + 12343 + 22343 + 32 343 = 49 36 9 2 117 Variance= 49 − 7 = 49 Example 10.14 :A game is played with a single fair die, A player wins Rs. 20 if a 2 turns up, Rs. 40 if a 4 turns up, loses Rs. 30 if a 6 turns up. While he neither wins nor loses if any other face turns up. Find the expected sum of money he can win. Solution : Let X be the random variable denoting the amount he can win. The possible values of X are 20,40, − 30 and 0. 1 P[X = 20] = P(getting 2) = 6 1 P[X = 40] = P(getting 4) = 6 1 P[X = − 30] = P(getting 6) = 6
208
1 The remaining probability is 2 X 20 40 0 −30 P(x) 1/6 1/6 1/6 1/2 Mean E (X) = ∑ pi xi 1 1 1 1 = 20 6 + 40 6 + (−30) 6 + 0 2 = 5 Expected sum of money he can win = Rs. 5 Expectation of a continuous Random Variable : Definition : Let X be a continuous random variable with probability density function f(x). Then the mathematical expectation of X is defined as ∞ E(X) = ⌡ ⌠ xf(x) dx −∞ Note : If ϕ is function such that ϕ(X) is a random variable and E [ϕ (X)] exists then ∞ E[ϕ (X)] = ⌠ ⌡ϕ (x) f(x) dx −∞
∞
2 E(X ) = ⌠ ⌡ x f(x) dx −∞ Variance of X = E(X2) − [E(X)]2 Results : (1) E(c) = c where c is a constant ∞ ∞ E(c) = ⌡ c f(x) dx = c ⌠ ⌠f(x) dx = c ⌡ 2
−∞
(2)
−∞
∞ as ⌠ ⌡f(x) dx = 1 −∞
E(aX ± b) = a E(X) ± b ∞ ∞ ⌠(ax ± b) f(x) dx = ⌡ E(aX ± b) = ⌡ ⌠ax f(x) dx −∞
∞ ± ⌠ ⌡b f(x) dx −∞
−∞ ∞ =a ⌠ ⌡x f(x) dx ± b ⌠ ⌡f(x) dx = a E(X) ± b
∞
−∞
−∞
209
Example 10.15 : In a continuous distribution the p.d.f of X is 3 x (2 − x) 0< x < 2 f(x)=4 , otherwise. 0 Find the mean and the variance of the distribution. 2 ∞ ⌠ 3 E(X) = ⌠ Solution : ⌡ x f(x) dx = x. 4 x(2 − x) dx
⌡
−∞ 2
0 2
3 3 2 2 3 = 4⌠ ⌡ x (2 − x) dx = 4 ⌠ ⌡(2x − x ) dx 0
3 =4
0
x x 3 2 16 2 3 − 4 = 4 3(8) − 4 = 1 0
∴ Mean = 1 ∞ 2 ⌠ x2f(x) dx = E(X ) = ⌡ −∞
3
4
2
2
⌠ x2 3 x(2 − x) dx ⌡ 4 0
2
2 3 3 x4 x5 3 16 32 6 ⌠ (2 x3 − x4) dx = 4 2 4 − 5 = 4 2 − 5 = 5 = 4⌡ 0 0
6 1 Variance = E(X2) − [E(X)]2 = 5 − 1 = 5 Example 10.16 : Find the mean and variance of the distribution 3e−3x,0 < x < ∞ f(x) = ,elsewhere 0 Solution : ∞ E(X) = ⌡ ⌠ x f(x) dx
∞
−∞
∞
n
xn e− αx dx = n + 1 ⌠ ⌡ α ∞
1 1 −3x −3x dx = 3. 2 = 3 =⌠ ⌡ x (3e ) dx = 3 ⌠ ⌡xe 3 0 0
0 When n is a positive integer
∞ ∞ 2 2 2 −3x 2 −3x ⌠ x E(X2) = ⌠ (3e ) dx = 3 dx = 3 . 3 = 9 ⌡ ⌡x e 3 0 0
210
2 1 2 1 Var(X) = E[X2] − E[X]2 = 9 − 3 = 9
1 ∴Mean = 3
1 ; Variance = 9 EXERCISE 10.2
(1) A die is tossed twice. A success is getting an odd number on a toss. Find the mean and the variance of the probability distribution of the number of successes. (2) Find the expected value of the number on a die when thrown. (3) In an entrance examination a student has to answer all the 120 questions. Each question has four options and only one option is correct. A student gets 1 mark for a correct answer and loses half mark for a wrong answer. What is the expectation of the mark scored by a student if he chooses the answer to each question at random? (4) Two cards are drawn with replacement from a well shuffled deck of 52 cards. Find the mean and variance for the number of aces. (5) In a gambling game a man wins Rs.10 if he gets all heads or all tails and loses Rs.5 if he gets 1 or 2 heads when 3 coins are tossed once. Find his expectation of gain. (6) The probability distribution of a random variable X is given below : X P(X = x)
0 0.1
1 0.3
2 0.5
3 0.1
If Y = X2 + 2X find the mean and variance of Y. (7) Find the Mean and Variance for the following probability density functions
1 ,−12 ≤ x ≤ 12 (i) f(x) = 24 0 ,otherwise xe−x
(iii) f(x) =
0
αe−α x
(ii) f(x) =
0
, if x > 0 ,otherwise
211
, if x > 0 ,otherwise
10.4 Theoretical Distributions : The values of random variables may be distributed according to some definite probability law which can be expressed mathematically and the corresponding probability distribution is called theoretical distribution. Theoretical distributions are based on expectations on the basis of previous experience. In this section we shall study (1) Binomial distribution (2) Poisson distribution (3) Normal distribution which figure most prominently in statistical theory and in application. The first two distributions are discrete probability distributions and the third is a continuous probability distribution. Discrete Distributions : Binomial Distribution : This was discovered by a Swiss Mathematician James Bernoulli (1654−1705) Bernoulli’s Trials : Consider a random experiment that has only two possible outcomes. For example when a coin is tossed we can take the falling of head as success and falling of tail as failure. Assume that these outcomes have probabilities p and q respectively such that p + q =1. If the experiment is repeated ‘n’ times independently with two possible outcomes they are called Bernoulli’s trials. A Binomial distribution can be used under the following condition. (i) any trial, result in a success or a failure (ii) There are a finite number of trials which are independent. (iii) The probability of success is the same in each trial. Probability function of Binomial Distribution : Let n be a given positive integer and p be a given real number such that 0 ≤ p ≤ 1. Also let q = 1 − p. Consider the finite probability distribution described by the following table. xi 0 1 2 ... n P(xi)
qn
nc1pqn−1
nc2 p2qn−2
...
pn
The table shown above is called the Binomial distribution. The 2nd row of the table are the successive terms in the binomial expansion of (q + p)n. Binomial probability function B(n,p,x) gives the probability of exactly x successes in ‘n’ Bernoullian trials, p being the probability of success in a trial. The constants n and p are called the parameters of the distribution.
212
Definition of Binomial Distribution : A random variable X is said to follow Binomial distribution if its probability mass function is given by
n px qn −x, x = 0, 1, . . .n P(X = x) = p(x) = Cx 0 otherwise Constants of Binomial Distribution : Mean = np Variance = npq Standard deviation =
variance = npq
X ∼ B(n, p) denotes that the random variable X follows Binomial distribution with parameters n and p. Note : In a Binomial distribution mean is always greater than the variance. Example 10.17 : Let X be a binomially distributed variable with mean 2 and 2 standard deviation . Find the corresponding probability function. 3
Solution :
np = 2 ∴ npq = 4/3 npq ∴ q = np
;
npq
=
2 3
4 2 6 =3 1 2 ∴ p = 1−q = 1−3 = 3 1 np = 2 ∴ n 3 = 2 ⇒ n = 6 ∴ The probability function for the distribution is 1 x 2 6−x , x = 0, 1, 2, P[X = x] = 6C 3 3 x 4/3
= 2
=
…6
Example 10.18 : A pair of dice is thrown 10 times. If getting a doublet is considered a success find the probability of (i) 4 success (ii) No success. Solution : n = 10 . A doublet can be obtained when a pair of dice thrown is {(1,1), (2,2) (3,3), (4,4), (5,5) (6,6)} ie., 6 ways. Probability of success is getting a doublet 6 1 1 5 ∴ p = 36 = 6 ; q = 1 − p = 1 − 6 = 6
213
Let X be the number of success. We have P[X = x] = nC px qn−x x
1 4 5 6 (a) P(4 successes) = P[X = 4] = 10C 6 6 4 = (b)
35 5 6 210 × 56 = 216 6 10 6
P (no success) = P(X = 0) = 10C 0
5 10 = 5 10 6 6
Example 10.19 : In a Binomial distribution if n = 5and P(X = 3) = 2P(X = 2) find p Solution :
P(X = x) = nC px qn−x x
P(X = 3) = 5C p3q2 and P(X = 2) = 5C p2q3 3 2 ∴ 5C p3q2 = 2 5C p2q3 3
(
)
2
∴ p = 2q 2 ⇒ 3p = 2 ; p = 3 Example 10.20 : If the sum of mean and variance of a Binomial Distribution is 4.8 for 5 trials find the distribution. np + npq = 4.8 ⇒ np(1 + q) = 4.8 Solution : 5 p [1 + (1 − p) = 4.8 p = 2 (1 − p)
p2 − 2p + 0.96 = 0 ⇒ p = 1.2 , 0.8 ∴ p = 0.8 ; q = 0.2 [‡p cannot be greater than 1] ∴ The Binomial distribution is P[X = x] = 5C (0.8)x (0.2)5−x, x = 0 to 5 x
Example 10.21 : The difference between the mean and the variance of a Binomial distribution is 1 and the difference between their squares is 11.Find n. Solution : Let the mean be (m + 1) and the variance be m from the given data.[Since mean > variance in a binomial distribution] (m +1)2 − m2 = 11 ⇒ m = 5 ∴ mean = m + 1 = 6 5 1 ⇒ np = 6 ; npq = 5 ∴ q = 6 , p = 6
214
⇒ n = 36.
EXERCISE 10.3 (1) The mean of a binomial distribution is 6 and its standard deviation is 3. Is this statement true or false? Comment. (2) A die is thrown 120 times and getting 1 or 5 is considered a success. Find the mean and variance of the number of successes. (3) If on an average 1 ship out of 10 do not arrive safely to ports. Find the mean and the standard deviation of ships returning safely out of a total of 500 ships (4) Four coins are tossed simultaneously. What is the probability of getting (a) exactly 2 heads (b) at least two heads (c) at most two heads. (5) The overall percentage of passes in a certain examination is 80. If 6 candidates appear in the examination what is the probability that atleast 5 pass the examination. (6) In a hurdle race a player has to cross 10 hurdles. The probability that he 5 will clear each hurdle is 6. What is the probability that he will knock down less than 2 hurdles.
10.4.2 Poisson Distribution : It is named after the French Mathematician Simeon Denis Poisson (1781 − 1840) who discovered it. Poisson distribution is also a discrete distribution. Poisson distribution is a limiting case of Binomial distribution under the following conditions. (i) n the number of trials is indefinitely large ie., n → ∞. (ii) p the constant probability of success in each trial is very small ie., p → 0. (iii) np = λ is finite where λ is a positive real number. When an event occurs rarely, the distribution of such an event may be assumed to follow a Poisson distribution. Definition : A random variable X is said to have a Poisson distribution if the e−λ λx probability mass function of X is P(X = x) = , x = 0,1,2, …for some λ > 0 x The mean of the Poisson Distribution is λ, and the variance is also λ. The parameter of the Poisson distribution is λ.
215
Examples of Poisson Distribution : (1) The number of alpha particles emitted by a radio active source in a given time interval. (2) The number of telephone calls received at a telephone exchange in a given time interval. (3) The number of defective articles in a packet of 100, produced by a good industry. (4) The number of printing errors at each page of a book by a good publication. (5) The number of road accidents reported in a city at a particular junction at a particular time. Example 10.22 : Prove that the total probability is one. ∞
Solution :
∞ e−λ λx e−λ λ0 e−λ λ1 e−λ λ2 = + + + ... x 0 1 2 x=0
∑ p(x) = ∑ x=0
λ2 = e−λ [1 + λ + + . . . ] = e−λ . eλ 2
= e0 = 1
Example 10.23 : If a publisher of non-technical books takes a great pain to ensure that his books are free of typological errors, so that the probability of any given page containing atleast one such error is 0.005 and errors are independent from page to page (i) what is the probability that one of its 400 page novels will contain exactly one page with error. (ii) atmost three pages with errors. [e−2 = 0.1353 ; e−0.2. = 0.819]. n = 400 , p = 0.005 Solution : ∴ np = 2 = λ (i) P(one page with error) = P(X = 1) =
e−λ λ1 e−221 = = 0.1363 × 2 = 0.2726 1 1
(ii)P(atmost 3 pages with error) = P(X ≤ 3)
e−λ λx 3 e−2(2)x 2 22 23 =∑ = e2 1 + + + x 1 2 3 x=0 x 0 19 = e−2 3 = 0.8569 3
= ∑
216
Example 10.24 : Suppose that the probability of suffering a side effect from a certain vaccine is 0.005. If 1000 persons are inoculated, find approximately the probability that (i) atmost 1 person suffer. (ii) 4, 5 or 6 persons suffer. [e−5 = 0.0067] Solution : Let the probability of suffering from side effect n = (i)
be p
λ = np = 5.
1000 , p = 0.005 ,
P(atmost 1 person suffer) = p(X ≤ 1) = p(X = 0) + p(X = 1) =
e−λ λ0 e−λλ1 + = e−λ [1 + λ] 0 1
= e−5 (1 + 5) = 6 × e−5 = 6 × 0.0067 = 0.0402 (ii) P(4, 5 or 6 persons suffer) = p(X = 4) + p(X = 5) + p(X = 6) =
e−λ λ4 e−λ λ5 e−λ λ6 e−λ λ4 λ λ2 1 + 5 + 30 + + = 4 5 6 4
=
5 25 e−5 54 17 10625 e−5 54 1 + + 5 30 = 24 6 = 144 × 0.0067 24
= 0.4944 Example 10.25 : In a Poisson distribution if P(X = 2) = P(X = 3) find P(X =5) [given e−3 = 0.050]. Solution : Given P(X = 2) = P(X = 3) ∴
e−λ λ2 e−λ λ3 = 2 3 ⇒ 3λ2 = λ3
⇒ λ2 (3 − λ) = 0 P(X = 5) =
As
λ ≠ 0. λ = 3
e−λ λ5 e−3 (3)5 0.050 × 243 = = = 0.101 120 5 5
217
Example 10.26 : If the number of incoming buses per minute at a bus terminus is a random variable having a Poisson distribution with λ=0.9, find the probability that there will be (i) Exactly 9 incoming buses during a period of 5 minutes (ii) Fewer than 10 incoming buses during a period of 8 minutes. (iii) Atleast 14 incoming buses during a period of 11 minutes. Solution : λ for number of incoming = 0.9 (i) buses per minute ∴ λ for number of incoming buses per 5 minutes = 0.9 × 5 = 4.5 P exactly 9 incoming buses e−λ λ9 during 5 minutes = 9 i.e., P(X = 9) = (ii)
e−4.5 × (4.5)9 9
fewer than 10 incoming buses during a period of 8 minutes = P(X <10) Here λ = 0.9 × 8 = 7.2 e−7.2 × (7.2)x x x=0 9
∴ Required probability = ∑ (iii)
P atleast 14 incoming buses during a period of 11 minutes = P(X ≥ 14) = 1 − P(X < 14) Here λ = 11 × 0.9 = 9.9 e−9.9 × (9.9)x x x=0 13
∴ Required probability = 1 − ∑
(The answer can be left at this stage). EXERCISE 10.4 (1) Let X have a Poisson distribution with mean 4. Find (i) P(X ≤ 3) (ii) P(2 ≤ X < 5) [e−4 = 0.0183]. (2) If the probability of a defective fuse from a manufacturing unit is 2% in a box of 200 fuses find the probability that (i) exactly 4 fuses are defective (ii) more than 3 fuses are defective [e−4 = 0.0183].
218
(3) 20% of the bolts produced in a factory are found to be defective. Find the probability that in a sample of 10 bolts chosen at random exactly 2 will be defective using (i) Binomial distribution (ii) Poisson distribution. [e−2 = 0.1353]. (4) Alpha particles are emitted by a radio active source at an average rate of 5 in a 20 minutes interval. Using Poisson distribution find the probability that there will be (i) 2 emission (ii) at least 2 emission in a particular 20 minutes interval. [e−5 = 0.0067]. (5) The number of accidents in a year involving taxi drivers in a city follows a Poisson distribution with mean equal to 3. Out of 1000 taxi drivers find approximately the number of drivers with (i) no accident in a year (ii) more than 3 accidents in a year [e−3 = 0.0498].
10.4.3 Normal Distribution : The Binomial and the Poisson distribution described above are the most useful theoretical distribution for discrete variables i.e., they relate to the occurrence of distinct events. In order to have mathematical distribution suitable for dealing with quantities whose magnitude is continuously varying, a continuous distribution is needed. The normal distribution is also called the normal probability distribution, happens to be the most useful theoretical distribution for continuous variables. Many statistical data concerning business and economic problems are displayed in the form of normal distribution. In fact normal distribution is the ‘corner stone’ of Modern statistics. Like the Poisson distribution, the normal distribution may also be regarded as a limiting case of binomial distribution. Indeed when n is large and neither p nor q is close to zero the Binomial distribution is approximated by the normal distribution inspite of the fact that the former is a discrete distribution, where as the later is a continuous distribution. Examples include measurement errors in scientific experiments, anthropometric measurements of fossils, reaction times in psychological experiment, measurements of intelligence and aptitude, scores on various tests and numerous economic measures and indication. Definition : A continuous random variable X is said to follow a normal distribution with parameter µ and σ (or µ and σ2) if the probability function is 1 x − µ2
−2 1 σ f(x) = e σ 2π
; −∞ < x < ∞, − ∞ < µ < ∞, and σ > 0.
219
X ∼ N(µ, σ) denotes that the random variable X follows normal distribution with mean µ and standard deviation σ. Note : Even we can write the normal distribution as X∼ N(µ, σ2) symbolically. In this case the parameters are mean and variance. The normal distribution is also called Gaussian Distribution. The normal distribution was first discovered by De-Moivre (1667 − 1754) in 1733 as a limiting case of Binomial distribution. It was also known to Laplace not later than 1744 but through a historical error it has been credited to Gauss who first made reference to it in 1809. Constants of Normal distribution : Mean = µ Variance = σ2 Standard deviation = σ ∞ -∞ x =µ The graph of the normal curve is z =0 shown above. Fig. 10.3 Properties of Normal Distribution : (1) The normal curve is bell shaped (2) It is symmetrical about the line X = µ ie., about the mean line. (3) Mean = Median = Mode = µ 1 is the (4) The height of the normal curve is maximum at X = µ and σ 2π maximum height (probability). (5) It has only one mode at X = µ. ∴ The normal curve is unimodal (6) The normal curve is asymptotic to the base line. (7) The points of inflection are at X = µ ± σ (8) Since the curve is symmetrical about X = µ, the skewness is zero. (9) Area property : P(µ −σ < X < µ + σ) = 0.6826 P(µ −2σ < X < µ + 2σ) = 0.9544 P(µ −3σ < X < µ + 3σ) = 0.9973 (10) A normal distribution is a close approximation to the binomial distribution when n, the number of trials is very large and p the probability of success is close to 1/2 i.e., neither p nor q is so small. (11) It is also a limiting form of Poisson distribution i.e., as λ → ∞ Poisson distribution tends to normal distribution.
220
Standard Normal Distribution : A random variable X is called a standard normal variate if its mean is zero and its standard deviation is unity. A normal distribution with mean µ and standard deviation σ can be converted into a standard normal distribution by performing change of scale and origin. The formula that enables us to change from the x scale to the z – scale and X−µ vice versa is Z = σ The probability density function of the standard normal variate Z is given by 1
− 2 z2 1 ϕ(z) = e ; −∞ < z < ∞ 2π The distribution does not contain any parameter. The standard normal distribution is denoted by N(0,1). The total area under the normal probability curve is unity. ∞ 0 ∞ i.e., ⌠ ⌠ ⌠ ϕ (z)dz = 1 ⇒ f(x) dx = ⌡ ⌡ ϕ (z)dz = ⌡ −∞ −∞ −∞ Area Property of Normal Distribution :
∞
⌠ ⌡ϕ (z)dz = 0.5 0
The Probability that a random variable X lies in the interval (µ − σ, µ + σ) is given by µ +σ
P(µ −σ < X < µ + σ) = ⌠ ⌡ f(x) dx µ −σ
substituting X = µ − σ and X = µ + σ in Z =
X−µ σ
1
P(−1< Z< 1)= ⌠ ⌡ ϕ (z)dz −1
-∞
1
= 2⌠ ⌡ ϕ (z)dz (by symmetry)
-1
0
1
Fig. 10.4
0
= 2 × 0.3413, (from the area table) = 0.6826
221
∞
Also P(µ −2σ < X < µ + 2σ) µ +2σ
= ⌠ ⌡ f(x) dx µ −2σ 2
-∞
-2
P(−2 < Z < 2) = ⌠ ⌡ϕ (z)dz
0
∞
2
Fig. 10.5
−2
2
= 2⌠ ⌡ϕ (z)dz , (by symmetry) 0
= 2 × 0.4772 = 0.9544 Similarly P(µ −3σ < X < µ + 3σ) =
µ +3σ
3
µ −3σ
−3
⌠ ⌡ϕ (z)dz ⌡ f(x) dx = ⌠
-∞
-3
= 2 × 0.49865 = 0.9973
0
3
∞
Fig. 10.6
Therefore the probability that a normal variate X lies outside the range µ ± 3σ is given by P( | X − µ | > 3σ) = P( | Z | >3) = 1 − p(−3 < Z < 3) = 1 − 0.9973 = 0.0027 Note : Since the areas under the normal probability curve have been tabulated interms of the standard normal variate Z, for any problem first convert X to Z. The entries in the table gives the areas under the normal curve between the mean (z = 0) and the given value of z as shown below : Therefore entries corresponding to negative values are unnecessary because the normal curve is symmetrical. For -∞ ∞ z 0 example P(0 ≤ Z ≤ 1.2) = P(−1.2 ≤ Z ≤ 0) Fig. 10.7 Example 10.27 : Let Z be a standard normal variate. Calculate the following probabilities. (i)
P(0 ≤ Z ≤ 1.2)
(ii) P(−1.2 ≤ Z ≤ 0)
(iii) Area to the right of Z = 1.3
(iv) Area to the left of Z = 1.5
(v) P(−1.2 ≤ Z ≤ 2.5) (vi) P(−1.2 ≤ Z ≤ − 0.5)
222
(vii) P(1.5 ≤ Z ≤ 2.5)
Solution : (i) P(0 ≤ Z ≤ 1.2) P(0 ≤ Z ≤ 1.2) = =
area between Z = 0 and Z = 1.2 0.3849
-∞
z =0 z =1.2
∞
Fig. 10.8 (ii) P(−1.2 ≤ Z ≤ 0) P(−1.2 ≤ Z ≤ 0)
= P(0 ≤ Z ≤ 1.2) by symmetry = 0.3849 (iii) Area to the right of Z = 1.3 P(Z > 1.3) = area between Z = 0 to Z = ∞ − area between Z = 0 to Z = 1.3 = P(0 < Z < ∞) − P(0 ≤ Z <1.3) = 0.5 − 0.4032 = 0.0968 (iv) Area of the left of Z = 1.5 = P(Z < 1.5) = P(−∞ < Z< 0) + P(0 ≤ Z < 1.5) = 0.5 + 0.4332 = 0.9332 (v) P(−1.2 ≤ Z < 2.5) = P(− 1.2 < Z < 0) + P(0 < Z < 2.5) = P(0 ≤ Z < 1.2) + P(0≤ Z ≤ 2.5) [by symmetry] = 0.3849 + 0.4938 = 0.8787 (vi) P(−1.2 ≤ Z ≤ −0.5) = P(−1.2 < Z < 0) − P(−0.5 < Z < 0) = P(0 < Z < 1.2) − P(0 < Z < 0.5) [due to symmetry] = 0.3849 − 0.1915 = 0.1934
223
-∞
∞
z =-1.2 z =0
Fig. 10.9
-∞
z =0 z =1.3
∞
Fig. 10.10
-∞
z =0 z =1.5
∞
Fig. 10.11
-∞
-1.2 z =0
2.5
∞
Fig. 10.12
-∞
z =0 z =-1.2 z =-.5
Fig. 10.13
∞
(vii) P(1.5 ≤ Z ≤ 2.5) Required area = P(0 ≤ Z ≤ 2.5) − P(0 ≤ Z ≤ 1.5) = 0.4938 − 0.4332 = 0.0606 Example 10.28 : Let Z be a standard normal variate. Find the value of c in the following problems. (i) P(Z < c) = 0.05 (iii) P(Z > c) = 0.05
(ii) P(−c < Z < c) = 0.94 (iv) P(c < Z < 0) = 0.31
Solution : (i) P(Z < c) = 0.05 i.e., P(− ∞ < Z < c) = 0.05 As area is < 0.5, c lies to the left of Z = 0. From the area table Z value for the area 0.45 is 1.65. ∴c = − 1.65
. 45 -∞
c
Fig. 10.14
(ii) P(−c < Z < c) = 0.94 As Z = −c and Z = +c lie at equal distance from Z = 0, 0.94 ∴ We have P(0 < Z < c) = 2 = 0.47. Z value for the area 0.47 from the table is 1.88 ∴ c = 1.88 and − c = − 1.88
∞
0
. 94 . 47
. 47 -∞
z =-c
∞
z =c
z =0
Fig. 10.15 (iii) P(Z > c) = 0.05 ⇒ P(c < Z < ∞) = 0.05 From the data it is clear that c lies to the right of Z = 0 The area to the right of Z = 0 is 0.5 P(0 < Z < ∞) − P(0 < Z < c) = 0.05 0.5 − P(0 < Z < c) = 0.05
. 45
∴ 0.5 − 0.05 = P(0 < Z < c)
-∞
0.45 = P(0 < Z < c) From the area table Z value for the area 0.45 is 1.65
z =0
0.05 z =c
∞
Fig. 10.16 ∴ c = 1.65
(iv) P(c < Z < 0) = 0.31 As c is less than zero, it lies to the left of Z = 0. From the area table the Z value for the area 0.31 is 0.88. As it in to the left of Z = 0, c = − 0.88 Example 10.29 : If X is normally distributed with mean 6 and standard deviation 5 find. (i) P(0 ≤ X ≤ 8) (ii) P( | X − 6 | < 10)
224
Solution : Given µ = 6, σ = 5 (i) P(0 ≤ X ≤ 8) X−µ σ 0−6 −6 -∞ ∞ When X = 0, Z = 5 = 5 = − 1.2 z =-1.2 z =0 z =.4 8−6 2 Fig. 10.17 When X = 8, Z = 5 = 5 = 0.4 ∴ P(0 ≤ X ≤ 8) = P(−1.2 < Z < 0.4) = P(0< Z <1.2) + P(0 < Z < .4) (due to symmetry) = 0.3849 + 0.1554 = 0.5403 (ii) P( | X − 6| < 10) = P(−10 < (X − 6) < 10) ⇒ P(−4 < X < 16) −10 −4 − 6 = 5 = −2 When X = −4, Z = 5 -∞ ∞ 16 − 6 10 z =-2 z =2 z =0 When X = 16, Z = 5 = 5 = 2 Fig. 10.18 P(− 4 < X < 16) = P(−2 < Z < 2) = 2 P(0 < Z < 2) (due to symmetry) = 2 (0.4772) = 0.9544 Example 10.30 : The mean score of 1000 students for an examination is 34 and S.D is 16. (i) How many candidates can be expected to obtain marks between 30 and 60 assuming the normality of the distribution and (ii) determine the limit of the marks of the central 70% of the candidates. Solution : µ = 34, σ = 16, N = 1000 X−µ (i) P(30 < X < 60) ; Z = σ 30 − µ 30 − 34 = ∴ X = 30, Z1 = 16 σ -∞ ∞ −4 x =30 x =34 x =60 = 16 = −0.25 z1 z2 Z1 = −0.25 Fig. 10.19 60 − 34 26 Z2 = 16 = 16 = 1.625 Z2 ≈ 1.63 (app.) We know that Z =
P(−0.25 < Z < 1.63) =P(0 < Z< 0.25) + P(0 < Z < 1.63) (due to symmetry)
225
= 0.0987 + 0.4484 = 0.5471 No of students scoring between 30 and 60 = 0.5471 × 1000 = 547. (ii) limit of central 70% of Candidates : Value of Z1 from the area table = − 1.04 for the area 0.35 [ as Z1 lies to left of Z = 0] -∞
Similarly Z2 = 1.04
. 35 z1
. 35 z =0
z2
∞
Fig. 10.20 X − 34 Z2 = 16 = − 1.04 X2− 34 = − 1.04 × 16 + 34 X2 = − 16.64 + 34
X − 34 = 1.04 16 X1 = 16 × 1.04 + 34 Z1 =
= 16.64 + 34 X1 = 50.64
X2 = 17.36
∴ 70% of the candidate score between 17.36 and 50.64. Example 10.31 : Obtain k, µ and σ2 of the normal distribution whose probability distribution function is given by 2 f(x) = k e−2x + 4x
−∞ < X < ∞
Solution : Consider −2x2 + 4x = −2 (x2 − 2x) = −2 [(x −1)2 − 1] = −2 (x − 1)2 + 2 2 2 ∴ e−2x + 4x = e2. e−2(x −1) 1 (x −1) −2 1/4
= e2. e Comparing it with f(x) we get −2x2 + 4x
ke 2
⇒ ke
1 x −1 2 −2 1 e /2
2
1 x −1 2 −2 1/2
= e2. e
1 x −µ −2 1 = e σ σ 2π
2
1 x −µ 2
−2 1 = e σ σ 2π
1 2e−2 2 1 1 ,σ =4 we get σ = 2 = µ = 1 and k = 1 . e−2 = 2π 2 2π
226
Example 10.32 : The air pressure in a randomly selected tyre put on a certain model new car is normally distributed with mean value 31 psi and standard deviation 0.2 psi. (i) What is the probability that the pressure for a randomly selected tyre (a) between 30.5 and 31.5 psi (b) between 30 and 32 psi (ii) What is the probability that the pressure for a randomly selected tyre exceeds 30.5 psi ? Solution : Given µ = 31 and σ = 0.2 (i) (a) P(30.5 < X < 31.5) ; Z =
X−µ σ
When X = 30.5, Z =
30.5 − 31 −0.5 = 0.2 = −2.5 0.2
When X = 31.5, Z =
31.5 − 31 0.5 = 0.2 = 2.5 0.2
-∞
x =30.5
µ =31
x =31.5
∞
Fig. 10.21
∴ Required probability P(30.5 < X < 31.5) = P( −2.5 < Z < 2.5) = 2 P(0< Z < 2.5) [since due to symmetry] = 2(0.4938) = 0.9876 (b) P(30 < X < 32) When X = 30,
Z=
30 − 31 −1 0.2 = 0.2 = − 5
When X = 32,
Z=
32 − 31 1 0.2 = 0.2 = 5
-∞
µ =31
∞
Fig. 10.22
P(30 < X < 32) = P(−5 < Z < 5) = area under the whole curve = 1 (app.) (ii)
When X = 30.5 , Z =
30.5 − 31 −0.5 = 0.2 = −2.5 0.2
P(X > 30.5) = P(Z > − 2.5) = 0.5 + p(0 < Z < 2.5) = 0.5 + 0.4938 = 0.9938
-∞
∞
z =-2.5
Fig. 10.23
227
EXERCISE 10.5 (1) If X is a normal variate with mean 80 and standard deviation 10, compute the following probabilities by standardizing. (i) P(X ≤ 100)
(ii) P(X ≤ 80)
(iii) P(65 ≤ X ≤ 100)
(iv) P(70 < X)
(v) P(85 ≤ X ≤ 95) (2) If Z is a standard normal variate, find the value of c for the following (i) P(0 < Z < c) = 0.25
(ii) P(−c < Z < c) = 0.40
(iii) P(Z > c) = 0.85 (3) Suppose that the amount of cosmic radiation to which a person is exposed when flying by jet across the United States is a random variable having a normal distribution with a mean of 4.35 m rem and a standard deviation of 0.59 m rem. What is the probability that a person will be exposed to more than 5.20 m rem of cosmic radiation of such a flight. (4) The life of army shoes is normally distributed with mean 8 months and standard deviation 2 months. If 5000 pairs are issued, how many pairs would be expected to need replacement within 12 months. (5) The mean weight of 500 male students in a certain college in 151 pounds and the standard deviation is 15 pounds. Assuming the weights are normally distributed, find how many students weigh (i) between 120 and 155 pounds (ii) more than 185 pounds. (6) If the height of 300 students are normally distributed with mean 64.5 inches and standard deviation 3.3 inches, find the height below which 99% of the student lie. (7) Marks in an aptitude test given to 800 students of a school was found to be normally distributed. 10% of the students scored below 40 marks and 10% of the students scored above 90 marks. Find the number of students scored between 40 and 90. (8) Find c, µ and σ2 of the normal distribution whose probability function 2 is given by f(x) = c e−x + 3x , − ∞ < X < ∞.
228
