Additional Maths Revision Notes

First Edition © C Morris. Only authorised for use by students at Reading School.

O CR Additio n al M a t h e m a t ics FSM Q (6993) Re visi o n N o t es

V e rsio n 1 . 1 M a rch 2 0 0 8

1

Cliv e M o r r is (E & O E )


Index Index...................................................................................................................................... 2 Syllabus for OCR FSMQ in Additional Mathematics (6993) ...................................... 4 Formulae............................................................................................................................... 7 Algebra.................................................................................................................................. 8 Rationalising Surds ............................................................................................................8 Manipulation of Algebraic Expressions ...........................................................................8 Addition and subtraction of polynomials .....................................................................8 Multiplication of polynomials.......................................................................................9 Division of polynomials ................................................................................................9 Remainder Theorem.........................................................................................................12 Factor Theorem ................................................................................................................13 Solution of Equations.......................................................................................................15 Solving Equations Reducing to Quadratics................................................................15 Solving Simultaneous Linear and Quadratic Equations (A Reminder)....................16 Completing the Square ................................................................................................17 Another (shorter) way of Completing the Square......................................................20 Sketching Quadratics using Completing the Square .................................................21 Finding the maximum and minimum point for a Quadratic Curve ..........................21 Solving Quadratic Equations by Completing the Square..........................................23 Solving Cubic Equations Using The Factor Theorem...............................................24 Solving cubic and cubic inequalities ..........................................................................26 Discriminant.................................................................................................................27 The Binomial Expansion .................................................................................................28 Pascal’s Triangle ..........................................................................................................28 Application to Probability – Binomial Distribution.......................................................30 Co-ordinate Geometry...................................................................................................... 32 The Straight Line..............................................................................................................32 Gradient of a Straight Line..........................................................................................32 Mid-point of a Line Segment ......................................................................................32 Length of a Line Segment ...........................................................................................33 Finding the Equation of a Straight Line .....................................................................33 Parallel and Perpendicular Gradients..........................................................................35 The Coordinate Geometry of Circles..............................................................................38 Equation of a Circle .....................................................................................................38 Finding the Centre and Radius of a Circle .................................................................39 Useful Properties in Circle Problems .........................................................................41 Finding the Equation of a Tangent to a Circle ...........................................................42 Finding the Equation of a Normal to a Circle ............................................................43 Finding the Closest Distance of a Given Point from a Circle ...................................43 When do circles meet?.................................................................................................44 Regions .............................................................................................................................45 Applications to Linear Programming .............................................................................47

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Trigonometry ..................................................................................................................... 50 IGCSE Revisited ..............................................................................................................50 Applications......................................................................................................................50 Graphs of Sine, Cosine and Tangent for Any Angle .....................................................51 Trigonometric Identities ..................................................................................................54 Solving simple trigonometric equations.........................................................................55 Trigonometry and Pythagoras in 3 Dimensions.............................................................59 Angle between a line and a plane ...............................................................................59 Line of greatest slope...................................................................................................59 Angle between two planes...........................................................................................60 Calculus............................................................................................................................... 61 Differentiation ..................................................................................................................61 Notation ........................................................................................................................61 Gradient Function ........................................................................................................61 Differentiation of powers of x and constant multiples, sums and differences. ........62 Equations of tangents and normals .............................................................................63 Location and Nature of Stationary Points ..................................................................64 Sketching Curves .........................................................................................................67 Practical Maximum and Minimum Problems ............................................................68 Integration.........................................................................................................................69 Integration as the Reverse of Differentiation .............................................................69 Indefinite Integration of powers of n, constant multiples, sums and differences ....69 Finding the constant of integration using given conditions ......................................70 Definite Integrals..............................................................................................................71 Area between a curve and the x axis ..........................................................................71 Area between two curves.............................................................................................74 Application to Kinematics ...............................................................................................75 Motion in a Straight Line ............................................................................................75 SUVAT Equations (Constant Acceleration Formulae) .............................................79 Displacement-time and Velocity-time Graphs...........................................................83

These Revision Notes contain the material that is additional to the IGCSE syllabus. Material already covered in the IGCSE Revision Notes will not be repeated here.

3


Syllabus for OCR FSMQ in Additional Mathematics (6993) Those statements in bold are in the Additional Mathematics syllabus but are not on the IGCSE syllabus. Algebra Manipulation of algebraic expressions

Be able to simplify expressions including algebraic fractions, square roots and polynomials.

The remainder theorem

Be able to find the remainder of a polynomial up to order 3 when divided by a linear factor.

The factor theorem

Be able to find linear factors of a polynomial up to order 3.

Solution of equations

Be confident in the use of brackets. Be able to solve a linear equation in one unknown. Be able to solve quadratic equations by factorisation, the use of the formula and by completing the square. Be able to solve a cubic equation by factorisation. Be able to solve two linear simultaneous equations in 2 unknowns. Be able to solve two simultaneous equations in 2 unknowns where one equation is linear and the other is quadratic. Be able to set up and solve problems leading to linear, quadratic and cubic equations in one unknown, and to simultaneous linear equations in two unknowns.

Inequalities

Be able to manipulate inequalities. Be able to solve linear and quadratic inequalities algebraically and graphically.

The binomial expansion

Understand and be able to apply the binomial expansion of (a + b)n where n is a positive integer.

Application to probability

Recognise probability situations which give rise to the binomial distribution. Be able to identify the binomial parameter, p, the probability of success. Be able to calculate probabilities using the binomial distribution.

4


Co-ordinate Geometry The straight line

Know the definition of the gradient of a line. Know the relationship between the gradients of parallel and perpendicular lines. Be able to calculate the distance between two points. Be able to find the mid-point of a line segment. Be able to form the equation of a straight line. Be able to draw a straight line given its equation. Be able to solve simultaneous equations graphically.

The co-ordinate geometry of circles

Know that the equation of a circle, centre (0,0), radius r is x2 + y2 = r2. Know that (x – a) 2 + (y – b)2 = r2 is the equation of a circle with centre (a, b) and radius r.

Inequalities

Be able to illustrate linear inequalities in two variables. Be able to express real situations in terms of linear inequalities. Be able to use graphs of linear inequalities to solve 2-dimensional maximisation and minimisation problems, know the definition of objective function and be able to find it in 2-dimensional cases.

Trigonometry Ratios of any angles and their graphs

Be able to use the definitions of sin , cos and tan for any angle (measured in degrees only). Be able to apply trigonometry to right angled triangles. Know the sine and cosine rules and be able to apply them. Be able to apply trigonometry to triangles with any angles. Know and be able to use the identity that Know and be able to use the identity sin 2

sin cos

tan cos 2

1.

Be able to solve simple trigonometrical equations in given intervals. Be able to apply trigonometry to 2 and 3 dimensional problems.

5


Calculus Differentiation

Be able to differentiate kxn where n is a positive integer or 0, and the sum of such functions. dy gives the gradient dx of the curve and measures the rate of change of y with respect to x.

Know that the gradient function

Know that the gradient of the function is the gradient of the tangent at that point. Be able to find the equation of a tangent and normal at any point on a curve. Be able to use differentiation to find stationary points on a curve. Be able to determine the nature of a stationary point. Be able to sketch a curve with known stationary points. Integration

Be aware that integration is the reverse of differentiation. Be able to integrate kxn where n is a positive integer or 0, and the sum of such functions. Be able to find a constant of integration. Be able to find the equation of a curve, given its gradient function and one point.

Definite integrals

Know what is meant by an indefinite and a definite integral. Be able to evaluate definite integrals. Be able to find the area between a curve, two ordinates and the x-axis. Be able to find the area between two curves.

Application to kinematics

Be able to use differentiation and integration with respect to time to solve simple problems involving variable acceleration. Be able to recognise the special case where the use of constant acceleration formulae is appropriate. Be able to solve problems using these formulae.

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Formulae You have been spoilt by not having to learn many mathematical formulae. In the Additional Mathematics examination you are not given a formula sheet and so the only formulae you will have with you are the ones that you have taken in there in your head! You are advised to make your own list of things to learn from the work covered during the Additional Mathematics course. You are also reminded to learn the formulae from the IGCSE formula sheet including b 2 c 2 a2 any rearrangements of formulae on it e.g. cos A 2bc

7


Algebra Rationalising Surds As well as the surds at IGCSE Level you may be asked to simplify more complicated surds using the difference of two squares as an aid. 2

2

2

3

2

3

2

3

5

3

5

3

5

3

5

3

5

3

5

3

2

3

4 2 3 22

3

9 6 5 2

3

4 2 3 4 3

2

5

5

2

4 2 3 1

14 6 5 9 5

4 2 3

14 6 5 4

7 3 5 2

Manipulation of Algebraic Expressions A polynomial is an expression that only contains positive integer powers of x and constants. A polynomial is therefore of the form a0 a1 x a2 x 2 a3 x3

an x n .

The numbers a1, a2 ,... are called the coefficients of x, x 2 etc. The degree of a polynomial is the highest power of x that occurs. A polynomial of degree 0 is a constant. A polynomial of degree 1 is linear A polynomial of degree 2 is quadratic. A polynomial of degree 3 is cubic. Addition and subtraction of polynomials When adding or subtracting polynomials combine the terms with the same powers. Examples

(3 2 x 2 x 2 3x 3 ) (7 5 x 3x 2

x 3 ) 3 7 2 x 5 x 2 x 2 3x 2 3 x 3 x 3 10 3x 5 x 2 4 x 3

(6 2 x 4 x 3 ) (4 4 x 7 x 2 ) 6 4 2 x ( 4 x ) 7 x 2 4 x3 2 6 x 7 x 2 4 x3

8

7 2

3 5 2


Multiplication of polynomials This is exactly like expanding linear brackets at GCSE. Multiply each term in the second bracket by each term in the first bracket and then simplify. Laying your work out systematically can avoid silly slips being made as shown in the example below. (3 4 x 2 x 3 )(1 x x 2

x3 )

3 3x 3x2 3x3 4 x 4 x2 4 x3 4 x4 3

x 7 x2

2 x 3 2 x4 2 x5 2 x6 5 x 3 6 x4 2 x5 2 x6

Division of polynomials There are 2 principal methods of dividing one polynomial by another. Example Find (2 x 3 2 x2

x into 2x 3 goes 2x 2 times

x 3) ( x 2)

x into 2x 3 goes 2x 2 times

Method 1 (Long Division)

x 2

2x 2 2 x 2 x2 x

3

2x

2 x3 4 x 2 2 x2

5 3

2x 2 times ( x 2) is (2 x3 4 x2 )

Take (2 x3 4 x2 ) from (2 x3 2 x2 ) and bring down the x

x

2 x2 4 x 5x

2x times ( x 2) is (2 x 2 4 x )

3

5 x 10 7

Take (2 x 2 4 x ) from (2 x 2

That is to say

(2 x 3 2 x2

( x 2) does not go into 7 so this is your remainder

x 3) ( x 2) (2x2

2 x 5) remainder 7

Other ways of writing this are

2 x3 2 x 2 x 3 x 2

2x 2 2 x 5

7 x 2

or

2 x3 2 x 2

x) and bring down the 3

x 3 ( x 2)(2x2

2 x 5) 7

9


Method 2 (Comparison of Coefficients) From Method 1 it is clear that when you divide a cubic by a linear term you will obtain a quadratic plus a constant remainder. Using the last form from above:

(2 x 3 2 x2

x 3) ( x 2)(ax 2 bx c) d where d is a constant.

Expanding on the right hand side gives (2 x 3 2 x 2

x 3)

ax3 bx 2

cx 2ax 2

(2 x 3 2 x 2

x 3)

ax3 bx 2

2ax2

So equating coefficients a

cx 2bx 2c d

(comparing coefficients of x 3 )

2

b 2a

2

b 4

2

b

2bx 2c d

(comparing coefficients of x 2 )

2

c 2b 1 (comparing coefficients of x ) c 4 1 c 5 2c d

3 (comparing constants)

10 d

3

d (2 x 3 2 x 2

7

x 3) ( x 2) (2x2

2 x 5) remainder 7

10


Another method (related to Method 2) is to do the following. Method 3

(2 x 3 2 x2

x 3) ( x 2)(ax 2 bx c) d where d is a constant.

We can work through this is stages In order to get the 2x 3 term a is clearly 2.

(2 x 3 2 x2

x 3) ( x 2)(2 x2 bx c) d

The next stage is to look at the coefficient of x 2 . The terms that will have an x 2 in come from x bx and 2 2x 2 . These together must give 2x 2 . b 4 b

2 2

(2 x 3 2 x2

x 3) ( x 2)(2 x2 2 x c) d

The terms that will have an x in come from 2 2x and cx . These together must give x. c 4 1 c 5

(2 x 3 2 x2

x 3) ( x 2)(2 x2 2 x 5) d

Equating the constant terms on both sides gives 2 5 d

3

10 d

3

d

7

So

(2 x 3 2 x2

x 3) ( x 2) (2x2

2 x 5) remainder 7

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Remainder Theorem There is a much easier way of finding the remainder when you divide by a linear term. This is called the remainder theorem. The remainder when f ( x) is divided by ( x a ) is f (a) The remainder when f ( x) is divided by (bx a ) is f

a b

Example 1 The remainder when f ( x) x 3 2 x 2 7 x 8 is divided by ( x 3) is given by f (3) 33 2 32 7 3 8 27 18 21 8 32 . Example 2 The remainder when f ( x) 3 x3 2 x 2 6 x 8 is divided by ( x 2) is given by f ( 2) 3 ( 2)3 2 ( 2)2 6 ( 2) 8 24 8 12 8 52 . Example 3 The remainder when f ( x) 8 x3 4 x 2

f ( 12 ) 8 8

1 3 2 1 8

1 2 2

4 4

1 4

2

1 2

2 x 1 is divided by (2 x 1) is given by

1.

1 1

1 1 1 1 2

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Factor Theorem This is a special case of the remainder theorem. If f (a) 0 then ( x a ) is a factor of f ( x) If f

a b

0 then (bx a ) is a factor of f ( x)

That is to say dividing f ( x) by ( x a ) or ( bx a) respectively leaves no remainder! Example 1 Show that x 3 and x 2 are factors of x 3 3x 2 10 x 24 . Solution Let f ( x )

x3 3 x 2 10 x 24

f (3) 33 3 32 10 3 24 27 27 30 24 0 x 3 is a factor of f ( x ) x 2

x ( 2)

f ( 2) ( 2)3 3 ( 2)2 10 ( 2) 24 8 12 20 24 0 x 2 is a factor of f ( x ) When looking for factors in a polynomial Check to seek whether x or powers of x are factors Start by looking for smaller values of a – a good strategy is check 1, then 1 , then 2, then 2 etc. If the coefficient of the highest power of x in the polynomial is a 2 or a 3 etc. then one of the factors will start (2 x...) or (3x...) but it is better to leave this to naturally appear as example 2 shows rather than looking for it directly.

13


Example 2 Use the factor theorem to factorise the cubic polynomial 2 x 3 9 x 2 7 x 6 Solution

Let f ( x ) 2 x 3 9 x 2 7 x 6 f (1) 2 9 7 6 6 0 so ( x 1) is not a factor of f ( x) f ( 1)

2 9 7 6

12 0 so ( x 1) is not a factor of f ( x )

f (2) 16 36 14 6 0 so ( x 2) is a factor of f ( x) 2 x 3 9 x 2 7 x 6 ( x 2)(2 x 2 bx 3) In the quadratic bracket on the right the x 2 coefficient must be 2 to give 2x 3 when you multiply out. In the quadratic bracket on the right the constant must be 3 to give 6 when you multiply out. To find the value of b look at the x 2 term on the right hand side when you multiply out and compare this with the x 2 term on the left hand side. 2 2 x 2 bx 2

9 x2

b 4

9

b

5

2 x 3 9 x 2 7 x 6 ( x 2)(2 x2 5x 3) As a check look at the x term which should be the same on both sides. On the right hand side this is

2

5 x 3x

7 x which agrees with the left hand side.

If the quadratic factorises we can now complete the factorisation. In this case it does and leads to

2 x 3 9 x 2 7 x 6 ( x 2)( x 3)(2 x 1)

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Solution of Equations Solving Equations Reducing to Quadratics Example 1 By first making the substitution y

x 2 solve the equation x 4 5 x 2 36 0 .

Solution x 4 5 x 2 36 0 ( x 2 )2 5( x 2 ) 36 0 y 2 5 y 36 0 ( y 9)( y 4) 0 y 9 0 or y 4 0 y

9 or

4

x2

9 or

4

3 since it is not possible for x 2 to be equal to

x Example 2

Solve the equation 2 x 7 Solution 2x 7 2 x2

4 x

7x 4

(2 x 1)( x

4)

2x 1 x

4 x

0

0 0 (multiplying through by x) 0 0 or x 4 1 2

0

or 4

15

4


Solving Simultaneous Linear and Quadratic Equations (A Reminder) Example 1 Find where the circle x 2

y2

25 and the line x y

7 meet.

Solution Where the graphs meet x 2 (7 x )2 x2

49 14 x x 2

25 (subtituting for y in terms of x from x y

7)

25 (expanding)

2 x 2 14 x 24 0 x 2 7 x 12 0 (divide through by 2 to make life easier) ( x 3)( x 4) 0 x 3 or 4 y

4 or 3

So the line meets the circle at (3, 4) and (4, 3) . Note that the x and y values must be paired in the final answer otherwise you may lose marks. Example 2 By solving the equations simultaneously find where the line y y x 2 x 2 meet and comment on your answer.

5 x 6 and the curve

Solution Where the graphs meet x2

x 2 5 x 6 (equating y values)

x2 4 x 4 0 ( x 2)2

0

x

2

y 5 2 6 4 So since there is one repeated solution y x 2 and y 4 i.e. at the point (2, 4) .

5 x 6 is a tangent to y

16

x2

x 2 when


Completing the Square There are quicker methods for those good with mental gymnastics but this basic routine is always effective. Another way of approaching things is to be found at the end of this section. Example 1 4 x 3 in the form ( x a)2 b .

Write x 2 Solution

x2

4 x 3 ( x a) 2 b

x2

4 x 3 ( x a)( x a ) b

x2

4x 3

x 2 2ax a 2 b

Then compare the bits on the 2 sides. 2a a

4 (comparing the x terms) 2

a2 b

3 (comparing the numbers)

22 b

3

4 b

3

b

3 4

7

So

x2

4 x 3 ( x 2) 2 7

17


Example 2 Write 7 6x x 2 in the form a ( x b)2 . Solution

7 6 x x2

a ( x b) 2

7 6 x x2

a ( x b)( x b)

7 6 x x2

a ( x 2 2bx b2 )

7 6 x x2

a b2 2bx x 2

Then compare the bits on the 2 sides. 2b

6 (comparing the x terms)

b a b2 a ( 3)2

3 7 (comparing the numbers) 7

a 9 7 a 16 So

7 6 x x2

16 ( x 3)2

18


Example 3 Write 2 x 2 8 x 11 in the form a( x b)2 c . Solution

2 x 2 8 x 11 a( x b)2 c 2 x 2 8 x 11 a( x b)( x b) c 2 x 2 8 x 11 a( x 2 2bx b2 ) c 2 x 2 8 x 11 ax 2 2abx ab2 c Then compare the bits on the 2 sides.

a

2 (comparing the x 2 terms)

2ab 6 (comparing the x terms) 2 2 b 8 b

2

ab 2 c 11 (comparing the numbers) 2 22 c 11 8 c 11 c 3 So

2 x 2 8 x 11 2( x 2)2 3

19


Another (shorter) way of Completing the Square Since

( x a) 2

x2 2ax a 2

( x a)( x a)

we can use the fact that

x 2 2ax ( x a) 2 a 2 In the brackets with the x is half the number of x’s in the original expression. Example 1

x2

4x 3

2)2

22

(x

2)2

2( x 2

8x )

7

2 (x

4)2

( 4)2

7

2 (x

4)2

16

(x

3

7

Example 2

2 x2

16 x

7

2( x

4)2

7

25

Example 3 3

6x

2 x2

3

2( x 2

3x )

3

2 (x

3 2 2

15 2

2( x

3 2 2

)

3 2 2

)

20


Sketching Quadratics using Completing the Square For example the curve y x 2 4 x 3 i.e. y ( x 2)2 7 is the curve y x 2 translated 2 units in the negative x direction and translated 7 units in the negative ydirection. Therefore the vertex of the curve (in this case the lowest point) is at ( 2, 7) .

y 10 8 6 4 2 -6 -5 -4 -3 -2 -1-2 -4 -6 -8 ( – 2, – 7) -10

2

y = x + 4x – 3

1 2 3 4 5

x

Finding the maximum and minimum point for a Quadratic Curve If you have completed the square on a quadratic it is easy to decide where the maximum (or minimum) point on the curve is. Example 1 Complete the square on y 2 x 2 8 x 2 and hence find the coordinates of the maximum point on the curve. Solution

y

2 x2 4 x

y

2( x 2)2 2 2 ( 2)2

y

2( x 2)2 6

2

Since the smallest value of 2( x 2) 2 6 will be when x 2 the minimum value of y 2 x 2 8 x 2 will be y 6 when x 2 . The minimum point will therefore have coordinates (2, 6) .

21


Example 2 Complete the square on y 3 8x 4 x 2 and hence find the coordinates of the maximum point on the curve. Solution

y

4 x2

y

4( x 1) 2 3 4 12

y

2x

3

7 4( x 1)2

Since the largest value of 7 4( x 1)2 will be when x y 3 8x 4 x 2 will be y 7 when x 1.

1 the maximum value of

The maximum point will therefore have coordinates ( 1, 7) .

22


Solving Quadratic Equations by Completing the Square Example 1 Solve the equation x 2

4 x 3 0 by first completing the square.

Solution x2

4x 3 0

( x 2)2 22 3 0 ( x 2)2 7

0

( x 2)2

7

x 2

7

x

2

7

Sometimes you are asked to give answers in surd form (which will be exact as no decimal approximation will have taken place) but if you have to give decimal answers you can obtain them easily from here. If you needed answers to 3 decimal places they would be 4.646 and 0.646. Example 2 Solve the equation 16 12 x 2 x 2

0 by completing the square.

Solution x 2 6 x 8 0 (dividing by

2 to obtain an equation starting x2 ..)

( x 3)2 ( 3)2 8 0 ( x 3) 2 17 ( x 3)2

0 17

x 3 x 3 x

17 17 1.12 or 7.12 (3 sf)

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Solving Cubic Equations Using The Factor Theorem Example 1 Show that x 1 is a factor of x 3 1 and hence factorise x 3 1 . Solution Let f ( x )

x3 1

f (1) 13 1 0 x 1 is a factor of x 3 1

x 3 1 ( x 1)(ax 2 bx c) ax 3 (b a) x 2 (c b) x c Comparing coefficients we have a 1 b a b 1 0

b 1

c b c 1 0

c 1

x 3 1 ( x 1)( x2

x 1)

This cannot be factorised further as x 2

x 1 does not factorise.

24


Example 2 Factorise fully 2 x 3 5x 2

x 6.

Solution

Let f ( x ) 2 x 3 5 x 2 f (1) 2 5 1 6 f ( 1)

x 6 2 0 so ( x 1) is not a factor

2 5 1 6 0 so ( x 1) is a factor

You can then go through two different routes. Either take out x 1 as a factor (as in the example on the next page) or find another factor. Pursuing this route gives

Let f ( x ) 2 x3 5 x 2

x 6

f (2) 16 20 2 6 0 so ( x 2) is a factor

2 x3 5 x2

x 6 ( x 1)( x 2)(2 x 3)

This final factor comes from observing that it must start with a 2x to give 2x 3 and must end up with 3 to get 6 .

25


Solving cubic and cubic inequalities Example Solve the equation 2 x 3 7 x 2 3x 18 0 . Hence solve the inequality 2 x 3 7 x 2 3x 18 0 . Solution Let f ( x )

2 x3 7 x 2 3 x 18

f (1) 10 so ( x 1) is not a factor of f ( x ) f ( 1) 12 so ( x 1) is not a factor of f ( x) f (2) 0 so ( x 2) is a factor of f ( x) Either by long division or by comparing coefficients

f ( x ) ( x 2)(2 x 2 3x 9) f ( x ) ( x 2)(2 x 3)( x 3) f ( x ) 0 when x 2 or 3 or

Consider the graph of y

3 2

2 x 3 7 x 2 3x 18.

f ( x) y

y = (x – 2)(2x + 3)(x – 3) 15 10 5 -5

-4

-3

-2

-1

1

2

3

4

5

6

x

-5 -10

f ( x)

0 when the curve is on or above the x axis so f ( x )

0 when

NB See also the inequalities section of the IGCSE Revision Notes.

26

3 2

x 2 or x 3.


Discriminant When solving the quadratic equation

ax 2 bx c

0, a

0

You know that the solutions (if there are any) are given by the quadratic equation formula x

b

b2 4ac 2a

The part underneath the square root sign is called the discriminant, often given the symbol . So b2 4ac . The discriminant gives quite a lot of information about the solutions of a quadratic equation and whether the quadratic factorises. Value of

Information given

0

Two distinct solutions

0

One (repeated) solution

0

Solutions

0

No solutions

a perfect square

The quadratic factorises

Examples 3x 2

2 x 4 0 has two solutions because

22

4 x2

4 x 1 0 has one (repeated) solution because

3x 2

2 x 3 0 has no solutions because

22

6 x2

x 2 factorises because

( 1)2

4 6

27

4 3

4 42

4 3 3

2

0.

52

4 4 1 32

0.

0.

49 which is a perfect square.


The Binomial Expansion Pascal’s Triangle The coefficients for expanding (1 x)n come from the rows of Pascal’s Triangle. 1 1

1

1

2

1

3

1 1 1 1

6 10

6 7

3

4 5

1 4

1

10

15 21

1

20 35

5 15

35

1 6

21

The numbers in Pascal’s Triangle also come from nCr

1 7

1

n! using appropriate r !(n r )!

values of r and n. For example the entries in the row beginning 1, 6, 15, 20, 15, 6, 1 come from 6

C0 , 6C1 , 6C2 , 6C3 , 6C4 , 6C5 , 6C6 respectively.

So we have that for positive integer values of n only. Binomial Expansions Type 1 (1 x) n

Where nCr

1

n

C1 x

n

C2 x 2

n

C3 x 3

n

Cr x r

n! . r !(n r )!

Note that nCr is sometimes written as

n (NB no fraction line!) r

28

xn


Binomial Expansions Type 2 x )n

(a

an

n

C1 a n 1 x

n

C2 a n 2 x 2

n

C3 a n 3 x3

n

Cr a n r xr

xn

Example 1 Find the first five terms in the expansion of (1 x)10 in ascending powers of x. Solution

(1 x)10

1

10

10

C2 x 2

C1 x

10

C3 x 3

10

C4 x 4 ...

1 10 x 45x 2 120 x 3 210 x 4 ... Example 2 Find the terms in the expansion of (1 2 x)7 in ascending powers of x up to and including the term in x 3 . Solution

(1 2 x )7 1

7

C1 ( 2 x )

7

C2 ( 2 x ) 2

7

C3 ( 2 x )3

1 14 x 84 x 2 280 x 3 ... Example 3 Find the binomial expansion of (3 2 x)5 and use your expansion to estimate 3.0025 correct to 1 decimal place. Solution (3 2 x)5

35

5

C1 34 (2 x)

5

C2 33 (2 x )2

5

C3 32 (2 x )3

243 810 x 1080 x 2 720 x 3 240 x4 3.0025

5

C4 31 (2 x )4

(2 x )5

32 x5

(3 2 0.001)5 243 810 0.001 (since higher power terms will not affect first dp) 243.8 (1 decimal place)

29


Example 4 Use the answer to example 2 to find the expansion in ascending powers of x up to and including the term in x 3 of (2 3x)(1 2 x)7 . Solution

(2 3 x)(1 2 x )7

(2 3x )(1 14 x 84 x 2 280 x3 ...) 2 28 x 168 x 2 560 x3 3 x 42 x 2

252 x3 ...

2 25 x 126 x 2 308 x3 ... Application to Probability – Binomial Distribution If X is the number of successes in n independent trials each of which has probability p of success and probability q ( 1 p ) of failure then X is said to have a Binomial Distribution with parameters n and p and we write X B(n, p) . P( X

r)

n

Cr p r q n

P( X

r)

n

Cr p r (1 p)n

Mean of X

r

r

np

Conditions for use of a Binomial Distribution A binomial distribution can be used to model a situation if Each trial has two possible outcomes (usually referred to as success or failure) There is a fixed number of trials The probability of success in each trial is constant The outcome of each trial is independent of the outcomes of all the other trials In the formula P( X

r)

n

Cr p r (1

p)n

r

there must be

r successes giving rise to the p r part of the formula n r failures giving rise to the (1 p)n r part of the formula The r success can occur in any of n Cr ways. Note that the powers of p and 1 p always add up to n.

30


Example 1 It is known that in a certain population 15% are left handed, Find the probability that in a sample of 9 people (a) exactly 5 are left handed (b) at most 3 are left handed (c) at least 1 is left handed Find also (d) the mean number of people who are left handed Solution X

B (9, 0.15)

(a) P( X

5)

9

C5 0.155 0.854

(b) P( X

3)

P( X

0 or 1 or 2 or 3)

P( X

3)

0.859

9

C1 0.151 0.858

(c) P( X 1) 1 P( X P( X

(d) Mean

0.00499 (3 sf)

9

C2 0.152 0.857

9

C3 0.153 0.856

0.966 (3 sf)

0)

1) 1 0.859

0.768 (3 sf)

9 0.15 1.35

Example 2 How many fair cubical dice must be rolled for there to be a 99% chance of obtaining at least one six? Solution P(at least one six) 1 P(no sixes) 1

5 n 6

5 n 6

1 0.99

5 n 6

0.01

0.99

5 25 6

0.01048... 0.01

5 26 6

0.008735... 0.01 (using trial and improvement)

At least 26 dice must be rolled for there to be a probability of at least one six 31


Co-ordinate Geometry The Straight Line Gradient of a Straight Line y

( x2 , y 2 )

x

Gradient

y2 x2

y1 x1

( x1 , y1 )

Example Find the gradient of the line joining the points ( 3, 7) and (2, 13) .

Gradient

13 7 2 ( 3)

20 5

4

Mid-point of a Line Segment y

Mid point M is ( x2 , y 2 )

M x

( x1 , y1 )

Example Find the midpoint of the line joining the points ( 3, 7) and (2, 13) . The midpoint is

3 2 7 ( 13) , 2 2

1 2

, 3

32

x1 2

x2 y1 ,

y2 2


Length of a Line Segment y

( x2 , y 2 )

Distance between ( x1 , y1 ) and ( x2 , y2 )

( x2

x1 )2

x

( x1 , y1 )

Example Find the length of the line segment joining the points ( 3, 7) and (2, 8) . Distance

(2 ( 3))2 ( 8 7) 2 52 ( 15)2 250 25 10 5 10

Finding the Equation of a Straight Line The equation of a straight line is of the form y is the y-intercept.

mx

c where m is the gradient and c

Remember that the equation must be in this form before you can read off the gradient. If it is not you must rearrange the equation first. Be careful that you give the answer in the required form. Sometimes a question will ask you to give your answer a specific way e.g. in the form ax by c 0 where a, b and c are integers. There are several approaches each of which needs you to have a gradient and a point that the line goes through. If you are given 2 points you can obviously find the gradient from this.

33

( y2

y1 ) 2


Example 1 The gradient of the line 3x 4 y 7

3 since rearranging the equation gives 4

0 is

3x 4 y 7 4y

0 3x 7 3 7 x 4 4

y Example 2

Find an equation of the straight line with gradient coordinates ( 3, 7) .

3 going through the point with

Approach 1 y

3x

c

Line goes through ( 3, 7) so y

7 when x

7

3 ( 3)

c

2

y

3x

2

Approach 2 (preferred) Gradient is

y 7 x ( 3)

3

y

7

3( x

y

7

3x 9

y

3x 2

34

3)

3 c


Example 3 Find an equation of the straight line through the points with coordinates (2, 5) and ( 3,15) . y ( 5) x 2

15 ( 5) 3 2

y x

5 2

20 5

y

5

4( x 2)

y

5

4x

8

y

4x

3

Gradient

4

Parallel and Perpendicular Gradients Two lines are parallel if and only if they have the same gradient. Two lines with gradients m1 and m2 are perpendicular if and only if

m2

1 i.e. m1m2 m1

NB If in an exam you are asked to show that two lines are perpendicular, show that when you multiply their gradients you get 1 .

1.

Example The equations of 5 lines are given below. Which lines are parallel to L and which lines are perpendicular to L. L : y 2x 5 M : 4x 2 y 8 N: P:

2y x 4 2x 4 y 5

R:

4x 2 y 3 0

Rearranging the lines give

L: y M: y

2x 5 2x 4

N:

y

1 2

x 2

P:

y

1 2

x

5 2

R:

y

2x

3 2

This shows that M and R are parallel to L and N are perpendicular to L. 35


Example Find the equation of the line that is parallel to y ( 3, 3) .

2 x 7 and goes through the point

Solution The gradient of y

2 x 7 is 2 so the equation of the required line is given by

y 3 2 x ( 3) y 3 2( x 3) y 3 2x 6 y

2x 9

Example Find the equation of the line that is perpendicular 3x 2 y 2 0 and goes through the point (2, 5) giving your answer in the form ax by c 0 where a, b and c are integers. Solution 3x 2 y 2

0 can be rearranged to give y

3 2

x 1 and therefore has gradient

The gradient of the line perpendicular to 3x 2 y 2

1

2 3

3 2

The equation of the perpendicular line is therefore

y ( 5) x 2

2 3

3( y 5)

2( x 2)

3 y 15

2x 4

2 x 3 y 11 0

36

0 is therefore given by

3 2

.


Reminder Remember that lines parallel to the x-axis are of the form y k where k is a constant and lines parallel to the y-axis are of the form x k where k is a constant. All lines perpendicular to a line of the form y where k1 and k2 are constants and vice versa.

k1 will therefore be of the form x

k2

Example Find the equation of the line perpendicular to x 4 and going through the point (2, 3) . Solution The line must be of the form y it must be y 3 .

k and since it goes through a point with y-coordinate 3

37


The Coordinate Geometry of Circles Equation of a Circle The equation of a circle centre the origin and with radius r is given by

x2

y2

r2

The equation of a circle centre ( a, b) and with radius r is given by

( x a) 2 ( y b) 2

r2

Notes For a circle the x and y coefficients must be the same. There can never be an xy term in the equation of a circle. The value of r 2 must be positive Examples

x2

y2

121 is a circle centre the origin and radius 11.

3x 2 3 y 2 147 is a circle centre the origin and radius 7 since it can be rewritten as x 2 y 2 49 by dividing throughout by 3. ( x 2)2 ( y 3)2

36 is a circle with centre (2, 3) and radius 6.

( x 3)2 ( y 1)2 16 is a circle with centre ( 3,1) and radius 4.

38


Finding the Centre and Radius of a Circle The recommended approach to determine whether a given equation is a circle and to determine its centre and radius is simply to complete the square from first principles on the x and y terms. Example 1

x2

y2 8x 6 y 5 0

x2

y2 8x 6 y 5 0

( x 4)2 42 ( y 3)2 ( 3)2 5 0 ( x 4)2 ( y 3)2 20 ( x 4) 2 ( y 3)2

0 20

This is a circle centre ( 4, 3) and radius

20

39

4 5

2 5.


Example 2 Which of the following are equations of circles and why (i)

x 2 2 y2 6 x 8 y 36

(ii)

x2

(iii)

( x y )2 ( x y)2

(iv)

y2

x2 4 x 6 y 12

(v)

x2

y 2 10 x 2 y 50 0

y 2 2 xy 6 x 12 y 11 0 50

Solutions (i)

This is not a circle because the coefficients of x 2 and y 2 are not the same.

(ii)

This is not a circle because there is a term in xy .

(iii)

When this is expanded you get

x 2 2 xy

y2

x 2 2 xy

y2

50

x2

y2

25

This is a circle centre the origin and radius 5. (iv)

When rearranged this gives

x2 4 x

y 2 6 y 12

( x 2)2 ( y 3)2 12 22 32

25

This is a circle centre (2, 3) and radius 5. (v)

When rearranged this gives

x2

y 2 10 x 2 y 50 0 ( x 5)2 ( y 1) 2

50 52 12

24

This is not a circle because the right hand can’t be a radius squared since it is negative.

40


Useful Properties in Circle Problems The diameter is twice the length of the radius. The angle in a semicircle is a right angle. The tangent to a circle at a point is perpendicular to the radius at that point. The perpendicular from the centre to a chord bisects the chord. Do not forget Pythagoras’ Theorem! Example Find the equation of the circle that has a diameter with endpoints (2, 4) and ( 4,8) Solution The centre of the circle is at

2 ( 4) 4 8 , 2 2

1, 2

The diameter of the circle is given by

( 4 2)2 (8 ( 4) 2

( 6)2 12 2

The radius of the circle is therefore

6 5 2

180 3 5

The equation of the circle is therefore x ( 1) x 1

2

2

y 2 y 2

2

2

(3 5) 2

45

41

36 5

6 5


Finding the Equation of a Tangent to a Circle This can be found by using the fact that the tangent to a circle is perpendicular to the radius of the circle. Example Find the equation of the tangent to the circle x 2 coordinates (6, 8) . x2

y2 6 x 8 y

( x 3)2 ( y 4)2

y2 6 x 8 y 0 at the point with

0 ( 3) 2

42

25

Centre is (3, 4), Radius is 5 Gradient of radius to (6, 8) Gradient of tangent at (6, 8)

4 ( 8) 3 6 1 4 3

4 3

3 4

The equation of the tangent is therefore as before y ( 8) x 6

3 4

4( y 8) 3( x 6) 3x 4 y 50

42


Finding the Equation of a Normal to a Circle The equation of a normal to a circle can be tackled in a similar fashion to the equation of the tangent. Remember that the equation of a normal to a circle at a particular point, P say, has the same gradient as the gradient of the line joining P to the centre of the circle. Example In the scenario outlined above, the equation of the normal at the point with coordinates (6, 8) on the circle with equation x 2 y2 6 x 8 y 0 is given by y ( 8) x 6

4 3

3( y 8)

4( x 6)

4x 3 y 0

Finding the Closest Distance of a Given Point from a Circle Essentially this reduces to finding the coordinates of the point, P, where a line through the given point, A and the centre of the circle, C, meet the circle and use this point to calculate the required distance. y 4

Find the centre of the circle 2

C

Find the distance AC. P

Find AP

A

2

4

x

43

AC radius .


Example Find the point on the circle with equation ( x 2)2 ( y 7)2 point with coordinates (8,19) .

5 that is closest to the

Solution The circle has centre (2, 7) and radius

5.

The distance from to the centre (2, 7) to (8,19) is given by

(8 2)2 (19 7) 2

62 122

180

36 5

6 5

The distance from (8,19) to the circle is therefore given by

6 5

5

5 5

When do circles meet? If r1 and r2 are the radii of two circles, r1 centres then (i) Circles touch externally

d

(ii) Circles touch internally

r1 r2

(iii) Circles do not intersect

d

r2 and d is the distance between their

d

r2 r1

(iii) Circles intersect at two distinct points

r1 r2

r2 r1

44

d

r1 r2


Regions Solid lines are used for inequalities that include = and dashed lines otherwise. You are usually expected to shade the regions that are excluded e.g. when representing the inequality x 3 you would shade the region that is NOT x 3 y 5 4 3 2 1

y

x>3

1 2 3 4 5 x

-5 -4 -3 -2 -1-1 -2 -3 -4 -5

x

2

5 4 3 2 1

y 5 4 3 2 1

y 5 4 3 2 1

y> –2

1 2 3 4 5 x

-5 -4 -3 -2 -1-1 -2 -3 -4 -5

1 2 3 4 5 x

-5 -4 -3 -2 -1-1 -2 -3 -4 -5

y

y

1

y

5 4 3 2 1 -5 -4 -3 -2 -1-1 -2 -3 -4 -5

1 2 3 4 5 x

-5 -4 -3 -2 -1-1 -2 -3 -4 -5

5 4 3 2 1 1 2 3 4 5 x y
-5 -4 -3 -2 -1-1 -2 -3 -4 -5

45

1 2 3 4 5 x x – 2y

2


If you are not sure which way to shade, just pick a test point off the line and see whether it satisfies the inequality. You will know which way to go then. In the last diagram, for example, if you use the point (3, 2) x 2 y 3 2 2 7 2 so (3, 2) does satisfy the inequality.

you get

When drawing a graph such as 3x 2 y 24 a very quick way is to find out where it meets the axes by putting x 0 to get y 12 and by putting y 0 to get x 8 . Questions often require you to shade regions that satisfy a number of inequalities. Example Use shading to show the region (often called the feasible region) that satisfies the following inequalities. You should shade the region that is not required. x

2

y

2

x y

3

Solution y

4 x=2 y=2 2

-6

-4

-2

0

2

-2

-4 x+y=

46

3

4

6

x


Applications to Linear Programming Work on regions can be extended to encompass problems involving linear programming. This is where an objective function is maximised or minimised subject to certain constraints that are usually able to be expressed as inequalities. Consider the feasible region defined by the inequalities x 2y

6

2x y 8 x

2

y

3

Suppose we wish to maximise the objective function x y . Such maximum points usually occur at a vertex (corner) of the region. The slope of lines of the form x y k indicates that in this case this would be where x 2 y 6 and 2 x y 8 meet. By solving simultaneously this pair of equations it can be shown that these meet at 3 13 ,1 13 . The maximum value of x y is therefore 3 13 1 13

4 23 .

y

4 x + 2y = 6 2 x= -4

-2

2 0

2

4

x

2x + y = 8 -2 y= -4

47

3


Example Grizelda is going to make some small cakes to sell at school and raise money for charity. She has decided to make some chocolate muffins and some yummy munchies. She would like to make as many cakes as possible but discovers that she only has 2 kg of flour and 750 g of butter. She has more than enough of the other ingredients. The cakes must be made in batches: For 12 muffins she needs 300 g of flour and 50 g of butter For 16 yummy munchies she needs 200 g of flour and 125 g of butter. (i)

Using x to represent the number of batches of muffins and y to represent the number of batches of yummy munchies, write down and simplify two inequalities relating to the available ingredients.

(ii)

Illustrate the region satisfied by these inequalities, using the horizontal axis for x and the vertical axis for y, and shading the unwanted region.

(iii)

Write down the objective function for the total number of cakes and find the greatest number of cakes that Grizelda can make.

48


Solution (i)

300 x 200 y 3x 2 y 50 x 125 y

2000 (restriction on flour) Using m to represent the number 20 750 (restriction on butter)

2 x 5 y 30

(ii)

y 20

15

10

3x + 2y = 20

5 The coordinates (4, 4) give the largest value of 12 x 16 y in the feasible region i.e. 112 cakes.

(iii)

2x + 5y = 30 0

The objective function is N

5

10

15 N = 12x + 16y

20

12 x 16 y .

Remember that lines of the form 12 x 16 y

k are all parallel to each other.

Look for the largest value of N that satisfies the conditions from (i) and (ii) with both x and y whole numbers (this is required from the context since the number of batches of each must be a whole number). Remember that this will usually be at or near a corner of the region formed by the constraints. The largest value of the objective function is 112 when x cakes is the largest number that can be made.

49

4, y

4 so 112

x


Trigonometry IGCSE Revisited You obviously need to be familiar with all the work from IGCSE to do with Trigonometry and Pythagoras in right angled triangles. This includes finding sides and angles. Sine and cosine rule. This includes finding sides and angles and may include the ambiguous case for use of the sine rule to find an angle. Finding areas of triangles including use of the formula

1 ab sin C 2

Remember that you will not have any of the formulae and will have to have learnt them! Applications You are much more likely to be asked application of trigonometry. Likely contexts include the use of terms such as Angle of elevation

Angle of elevation x°

Horizontal

Angle of depression

x°

Horizontal

Angle of depression

Problems involving bearings and other real life situations are also very likely so make sure that you revise this.

50


Graphs of Sine, Cosine and Tangent for Any Angle y y = sin x

1 0.5 -360

-270

-180

0

-90

90

-0.5 -1

Observe that sin 0 sin 90

sin180

sin 360

1

sin 270

1

sin 30

sin 150

sin(180 x ) sin( x )

0

0.5

sin x

sin x

The sine graph repeats itself every 360 . Be prepared to sketch this graph to help you solve trigonometric equations.

51

180

270

360 x


y y = cos x

1 0.5 -360

-270

-180

-90

0

90

180

-0.5 -1

Observe that cos 0

cos360

1

cos180

1

cos90

cos 270

0

cos 60

cos300

0.5

cos120

cos 60

cos(180 x) cos( x )

0.5

cos x

cos x

The cosine graph repeats itself every 360 . The cosine graph is the sine graph moved 90 to the right. Be prepared to sketch this graph to help you solve trigonometric equations.

52

270

360 x


y 10 y = tan x

8 6 4 2 -360

-270

-180

-90

0 -2

90

180

-4 -6 -8 -10

Observe that tan 0 tan 45

tan180

tan 360

0

1

tan135

tan 45

1

tan 90 tan 270 tan(180 x ) tan( x )

tan x tan x

The tangent graph repeats itself every 180 .

53

270

360 x


Trigonometric Identities The following results are true for all values of

sin 2

cos 2

sin cos

.

1

tan

It is also worth remembering that in a right angled triangle

z

y

90 – x

sin

x z

cos(90

)

cos

y z

sin(90

)

tan

x y

1 y x

54

1 tan(90

)


Solving simple trigonometric equations Example 1 Solve the equation sin

0.4 for 0

360 .

Solution One value can be found from

sin 1 (0.4)

23.6 (1 dp) .

From the symmetry of the sine graph there will also be a solution at 180

23.6

156.4 (1 dp)

Example 2 Solve the equation tan

1.2 for 180

180 .


tan 1 ( 1.2)

50.2 (1 dp) .

Since the tangent graph repeats itself every 180 there will also be a solution at 180 ( 50.2) 129.8 (1 dp)

Example 3 Solve the equation cos

0.2 for 180

360 .


cos 1 ( 0.2) 101.5 (1 dp) .

Another can be found from the symmetry of the cosine graph about

0 i.e.

101.5 (1 dp)

There is a third that can be found using the symmetry of the cosine graph about 180 . 360

101.5

258.5 (1 dp)

55


Example 4 Solve sin

2 cos

360 .

where 0

Solution sin cos

2

tan

2 63.4 or (63.4 180)

(using periodicity of tan graph)

63.4 or 243.4

Example 5 Solve sin

2 cos

360 .

0 where 0

Solution After a simple rearrangement this is the same as Example 4! Example 5 Solve 2 sin 2

1 where 0

360 .

Solution sin 2

1 2 1 2

sin When sin

1 2 45 or 135

When sin

1 2 225 or 315

56


Example 6 Solve 3 cos 2

5 0 for 0

5sin

360 .

Solution 3 cos 2

5sin

5 0

5sin

5 0

3 3sin 2

5sin

5 0

3sin 2

5sin

2 0

3 1 sin 2

Let y sin 3y2 5 y 2

0

3 y 2 ( y 1)

0

3y 2 y

0 or y 1 0 2 3

or y 1

sin

2 3

or sin

sin

2 3

leads to

1

41.8 or 180 41.8 138.2 sin

1 leads to 90

Example 7 Solve the equation tan cos

1 for 0

360 .

Solution

tan cos sin cos

1

cos

1

sin

1 270

57


Example 8 Solve the equation tan

2 sin

for 0

360 .

Solution

sin

tan

2sin

sin cos

2sin

sin

2 sin cos

2sin cos

0

sin (1 2cos ) 0 sin

0 0 or 180 or 360

or 1 2 cos

0

cos

1 2 60 or 300

Example 9 Solve the equation cos 2

0.3 for 0

360 .

Solution If 0

360

0 2 cos 2

720

0.3

2

72.5423.. or 360 72.5423.. or 360 72.5423.. or 360 (360 72.5423..)

2

72.5423.. or 287.4576.. or 432.5423.. or 647.4576..) 36.3 or 143.7 or 216.3 or 323.7

58


Trigonometry and Pythagoras in 3 Dimensions There are likely to be problems involving trigonometry and Pythagoras’ Theorem in 3 Dimensions and these may well be more involved and be more demanding that at IGCSE. You are more likely to find situations where the use of the sine or cosine rules will be more efficient. Remember the key skills of identifying which angles you are working with and extracting suitable triangles to work with. You also need to be familiar with some additional terms that you might not have met before just in case they come up. Angle between a line and a plane You should drop a perpendicular line down from the line to the plane to form a rightangled triangle. Plane Line

Angle between line and plane

Line of greatest slope This is the steepest line down a slope. It is essentially the path a ball would take if released on the slope!

Line of Greatest Slope

59


Angle between two planes This is the angle between lines in the two planes that meet at right angles to where the two planes meet.

Angle Between the Planes

60


Calculus Differentiation Notation

dy is the (first) derivative of y with respect to x. dx It is the rate of change of y with respect to x.

d2 y is the second derivative of y with respect to x. dx 2 dy This is the derivative of with respect to x. dx The first derivative of f ( x) is written as f '( x ) and the second derivative is written as f ''( x ) When differentiating something you will often see it written as

d something . dx

For example if you are differentiating x 3 3x 2 3x 2 you will often see this written as d 3 ( x 3 x 2 3 x 2) . dx Gradient Function

dy is also called the gradient function as it gives the gradient of a curve at the point dx ( x , y ) when the x-coordinate of the point is substituted into it.

61


Differentiation of powers of x and constant multiples, sums and differences.

d ax n dx

d ax n dx

nax n

bx m

1

Multiply by the power and reduce the power by 1.

where a is a constant.

nax n

1

mbx m

1

where a and b are constants,

Remember to write functions as powers of x before you differentiate and make sure that you simplify expressions first. Examples

d 4 x 3 3x 2 4 x 7 dx

The gradient function of something of the form mx c is simply m from the work covered on the gradient of a straight line.

12 x 2 6 x 4

The gradient function of a number (e.g. 5) is 0 because we know that y 5 is a straight line horizontal to the x-axis and hence has gradient 0.

d (2 x 1)(3x 2 2) dx

d x 7 x11 dx x3

d 4 x dx

d 6 x 3 3x2 dx 18 x 2 6 x 4

4x 2 You must write anything to be differentiated as powers of x before you begin, multiplying or dividing out before you begin.

x8

4 x3 8 x7

62


Equations of tangents and normals The normal to a curve at a point is the line perpendicular to the tangent at that point as shown by the diagram below. Normal

y

Tangent

x

Example Find the equation of the tangent and the normal to the curve y where x 2 . dy dx

This simply means put the value x 2 into the gradient function.

When x dy dx

Equation of the tangent is

6x

2 x at the point

2

2, y

3 22

6 2

2

2 2

16

14

x 2

y 16 x 2

14

y 16

14 x 28

y

14 x 12

y 16 Equation of the normal is x 2 14( y 16) 14 y

3x 2

x

1 14 ( x 2) 226

63

The gradient of the normal is 1 divided by the gradient of the tangent.


Location and Nature of Stationary Points A stationary point is a point on a curve for which

dy dx

0.

Stationary points can be maximum points, minimum points (these are both called turning points) or points of inflexion (sometimes spelt inflection). y

Maximum Point

Point of Inflexion

Minimum Point x

In order to determine the nature of a stationary point (i.e. to find out what sort of stationary point it is) you can either dy (a) find the gradient either side of the point, or dx d2 y (b) use the second derivative dx 2

If you use the first approach don’t go too far away from the stationary point or you might move past another stationary point and draw an incorrect conclusion.

Conclusion

dy to the left dx

dy at the point dx

0

Maximum point

+ ve /

0 –––

\

0

Minimum point

ve \

0 –––

/ + ve

Point of inflection

+ ve /

0 –––

/ + ve

Point of inflection

ve \

0 –––

\

Value of

0

d2 y dx 2

Check

dy either side dx

64

dy to the right dx ve

ve


Note: The examples here are harder than you will meet in the examination but they illustrate the principles well. Example 1 Find the coordinates and nature of the stationary points on the curve y Solution dy dx

3x 2 4 x 4 (3 x 2)( x 2)

d2 y dx 2

6x 4

dy dx

0 when x

When x d2 y dx 2 d2 y dx 2

2, y

2 or x

2 3

8; when x

2 3

,y

40 27

8 0, so there is a minimum at 2, 8 x 2

8 0, so there is a maximum at x

2 3

65

2 3

40 , 27

x3 2 x2 4 x .


Example 2 Find the coordinates of any stationary points on the curve y their nature.

x3 ( x 1)2 and determine

Solution y

x 3 ( x 1)2

x3 ( x 1)( x 1)

x 3 ( x 2 2 x 1) x5 2 x4 dy dx d2 y dx 2 dy dx

When

5 x 4 8x 3 3 x 2 20 x 3 24 x 2 6 x 5 x 4 8x 3 3 x 2

x 2 (5 x 2 8x 3)

0

x 2 (5 x 3)( x 1)

0

x d2 y dx 2 dy dx

0 or

3 5

2(10 x3 12 x2 3 x) 0

or 1

0 (no information gained) x 0

0.0385 0 x

dy dx d2 y dx 2

x3

0.1

0.0225 0 so a point of inflexion at (0,0) (same sign on both sides of st pt) x 0.1

0.72 0 so a maximum at (0.6, 0.03456) x 0.6

d2 y dx 2

2 0 so a minimum at (1, 0) x 1

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Sketching Curves You may be expected to sketch simple curves using any of the mathematics included in the syllabus. Useful things to bear in mind (although not all would be required on any particular sketch, nor should you work them all out unless you are asked to) are Where the curve meets the x-axis (i.e. when y

0)

Where the curve meets the y-axis (i.e. when x 0 ) The coordinates of any stationary points that you might know The general shapes of particular types of curves that you might know such as

y

x2

y

x3

y

1 x

y

1 x2

67


Practical Maximum and Minimum Problems Example A cylindrical tin has a volume of 128 cm3. Find the dimensions necessary for the tin to have the minimum possible surface area, and find the minimum possible surface area.

r

h

Let the base radius be r and the height be h.

2 r2

2 rh

Circular ends

Curved surface

Surface area, S

r 2 h 128

Volume, V h

128 r2

S

2 r2

2 r

2 r2

256 r

dS dr d2S dr 2 dS dr d2 S dr 2

256 r2

4 r 4

128 r2

512 r3

0 when 4 r 4

8

12

256 r2

0 i.e. when r 3

0 i.e. S is a minimum value when r

r 4

So the radius, r Smin

4 cm and the height, h 2

42

256 4

64 i.e. when r

128 42

96 cm 2

68

8 cm

4

4 cm


Integration Integration as the Reverse of Differentiation

dy dx dx

y

c where c is an arbitrary constant of integration.

Indefinite Integration of powers of n, constant multiples, sums and differences

ax n 1 n 1

ax n dx a f ( x) dx [f ( x)

a

c

n

1.

Remember c once you have integrated. Make sure that you have written things as powers of x before you integrate.

f ( x) dx

g( x )]dx

f ( x) dx

g( x) dx

Examples

3x 7 dx

3 8 x c 8

( x 2 2)( x 3 1)dx

( x5 x6 6

x10 5 x8 dx x4

x3 x4 4

x 2 2)dx x3 3

2x c

( x 6 5 x 4 )dx x7 7 x7 7

5 x5 5

c

x5 c

69


Finding the constant of integration using given conditions Example 1 Suppose that a curve y

f ( x) is such that

dy dx

2 x 1 and the curve passes through the

point (1, 2) . Solution y

x2

(2 x 1) dx

2 12 1 c c

4

so y

x2

x c

since the curve passes through (1, 2)

x 4

Example 2 Suppose that

dy dx

Given that y

3x 2 4 x b where b is a constant.

7 when x 1 and that y 12 when x

Solution dy dx

3x 2 4 x b (3x 2

y

x3 y

4 x b)dx

4 x2 2

bx c

x 3 2 x 2 bx c

7 13 2 12 b 1 c b c 4

(1)

12 23 2 22 b 2 c 2b c

4

(2)

b

8

(2) (1)

c 12 y

substituting for b in (1)

x 3 2 x 2 8 x 12

70

2 , find y in terms of x.


Definite Integrals b

If f ( x ) dx F( x ) c then f ( x ) dx F(b) F(a) . a

Area between a curve and the x axis y

y = f(x)

a

To find the area between the curve y calculate

x

b

f ( x) the x-axis and the lines x

b

b

y dx a

f ( x ) dx F(b) F(a ) a

For example the area between the x-axis, the curve y and x 3 is given by

x2

x 2 and the lines x 1

3

Area

( x2

x 2) dx

x3 3

x2 2

3

2

9

3 2 9 2

6

2x 1

13 3

2 3 1 2

12 2

y=x –x+2

10 9 8 7 6 5 4 3 2 1

3

1 3

2

y

1

3 3

a and x b we

2 1

2

8 23

-3 -2 -1

71

1

2

3

4

5 x


Note that sometimes areas can be calculated using more straightforward methods. For example, when working out the area under a straight line you can use the area of a triangle or trapezium. Note also that using symmetry can occasionally make calculations quicker. Example Find the shaded area between the curve y

x 3 and the line y

4x .

Solution The curve and line meet when

x3

3

y

y=x

y = 4x

10 8 6 4 2

4x

x3 4 x 0 x ( x 2 4) 0 x 0, 2

-3

-2

The area between 2 and 0 is clearly the same as the area between 0 and 2 so we require twice the area under the line take away the area under the curve.

-1 -2 -4 -6 -8 -10

1

There are two ways of proceeding. The first is completely by integration. 2

Area

2

2

2 4 x dx 2 x3 dx 2 (4 x x 3 ) dx 0

0

x4 4

2 2 x2

0 2

2 8 0

16 4

0 0

8 square units

The second uses the fact that the area under the line is the area of a triangle. When x 2 , y

8. 2

Area

2

1 2

2 8 2 x 3 dx 0 4

16 2

x 4

2

16 2 0

16 0 4

8 square units

72

2

3 x


A warning! You are strongly advised to draw a sketch of any curve before you find the area. This is because areas below curves are negative. If a curve has part above the x-axis and part below the x-axis you need to consider the two parts separately and combine the sizes of the areas. Example

x 2 9 , the x-axis and the lines

For example consider the area between the curve y x 2 and x 5 .

A sketch shows that part of the area is below the x-axis and part is above. y

2

y=x –9

12 8 4 -2 -1 -4

1 2 3 4 5 6 7 8 9 10 x

-8

3

(x

2

9) dx

2

The

3

x3 3

9x

9 27

8 3

18

2 32

2

sign is because the area is below the axis.

The actual area is 2 23 . 5

(x

2

9) dx

3

Area required

5

x3 3

9x

125 3

45

9 27

14 32

3

2 23 14 23 17 13 5

NB Had we done ( x 2 9) dx then we would have obtained that the (incorrect) answer 12! 2

This is because the integral would have given

73

2 23 14 23 12.


Area between two curves To find the area between two curves y f ( x) and y g( x ) which meet at x a and x b which are such that f ( x) g( x ) between these values find b

f ( x ) g( x) dx a

Example To find the area between the curves y

x 2 2 and y

x2 6 .

y 10 8 6 4 2 -6

-4

-2

2

y=x –2

2

-2 -4 -6 -8 -10

4

x

6

2

y= –x +6

First find where the curves meet.

x2 2 2x2

x2 6 8

x

2

We therefore need to evaluate the integral of the top curve take away the bottom one. 2

x2 6

Area

x2 2

dx Note that if you had subtracted the integrals the wrong way round you would have obtained the answer 21 13 .

2 2

2 x 2 8 dx 2

2 3 x 8x 3

2

16

16 3

16 3

Note also that because of the symmetry of both curves we could have found instead 2

2

x2 6

2

16

0

21 13

74

x2 2

dx


Application to Kinematics Motion in a Straight Line You need to be able to apply your knowledge of differentiation to motion in a straight line. The standard notation is t s

v a

time distance

ds dt dv dt

velocity acceleration

Maximum/minimum velocity occurs when maximum of graph i.e.

dv dt

0 i.e. when

a 0.

Remember also that

s

v dt

v

a dt

Don’t forget the constant of integration when integrating an indefinite integral – you will often need to use information you are given in the question to find its value.

75


Example 1 The distance of a particle, s metres, from a fixed point O after time t seconds is given by the formula s t 3 t 2 5t 2 . Find formulae, in terms of t, for (i) the velocity (ii) the acceleration Use your answers to find (iii) the velocity after 2 seconds (iv) the acceleration after 2 seconds (v) the time at which the velocity is 0 Solution (i) v

ds dt

3t 2 2t 5

dv dt

(ii) a

(iii) When t

2, v

3 22

(iv) When t

2, a

6 2 2 10 m/s 2

(v)

v 3t 2 2t 5 0

6t 2

2 2 5 12 4 5 3 m/s

(3t 5)(t 1) 0 3t 5 0 since t 1 0 gives a negative time which is impossible t 1 23 seconds

76


Example 2 A particle moves such that its displacement s metres after time t seconds is given by s

t3 t2 t 1.

(a)

Determine the value of t for which the velocity is 0.

(b)

Show that its acceleration is never 0.

Solution

v

(a)

ds dt

When v (b)

a

dv dt

3t 2 2t 1 (3t 1)(t 1) 0, t

1 3

since t must be positive.

6t 2

The only value of t that gives a 0 is t the acceleration can never be 0.

1 3

and since time is always positive

Example 3 A particle has acceleration given by a 9t 2 1 where t is the time since the particle started moving. Find the velocity in terms of t given that its initial velocity is 7 ms-1. Solution a

9t 2 1 (9t 2 1) dt 3t 3 t c

v When t

0, v 7 so 7 3 02 0 c

c 7 v 3t 3 t 7

77


Example 4 The velocity of a particle is given by v t 2 t at time t seconds. Find the distance moved in the 3rd second (i.e. between 2 and 3 seconds). Solution t 2 t dt

s s

t3 3

t2 2

c

When t

2, s

8 3

When t

3, s

27 3

Distance travelled

4 14 c c 2 3

27 c 2

9 c 2

27 c 2

14 c 3

8 56 metres

Another alternative approach is to use a definite integral. 3

t 2 t dt

s 2

t3 3

s

27 3

s

t2 2

3

2

9 2

Distance travelled 8 56 metres

78

8 3

4 2


SUVAT Equations (Constant Acceleration Formulae) These are for motion with constant acceleration in a straight line (horizontal or vertical) only.

v u at

s

v2

u2

2as

s

ut

1 2

s

u v t 2

s

vt

1 2

displacement

u initial velocity

at 2

v final velocity

at 2

a

acceleration

t

time

Remember: As these are vector quantities they have both magnitude and direction, so the signs of the quantities in these equations can be positive or negative. Choose a direction to be positive and then quantities in the opposite direction are negative. Slowing down, retardation and deceleration are all terms for negative acceleration. Usually g, the acceleration due to gravity is taken as 9.8 ms-2 (positive In order to use these equations successfully keep in mind: You should choose the equation that fits the information you know and the quantity that you are trying to calculate. If there are two unknowns to find it may be that you need to form and solve a pair of simultaneous equations. If a particle is projected up and then falls down again you can look at the motion in two stages, but this is not required, as long as you are careful with directions. Other useful results are

Total Distance Total Time

Average Speed

Average Velocity

Total Displacement Total Time

When projecting a particle vertically, at maximum height the velocity will be zero.

79


Example 1 A car begins to accelerate at 0.5 ms-2 for a distance of 500 m. At the end of this the car is travelling at 30 ms-1, calculate the initial speed of the car. Solution We know:

a = 0.5 s = 500 v = 30 u=? v2 = u2 + 2as

so use: 302

u2

2 0.5 500

u2

30 2 2 0.5 500

u2

400 20 ms-1

u

Example 2 A particle is projected upwards with a velocity of 30 ms-1. What is the maximum height it will reach? Solution Because the particle is only acting under its own weight, its acceleration will be due to gravity. Because it reaches maximum height the final velocity at that height will be zero. We therefore know:

s=? u = 30 v=0 a = - 9.8

v2 = u2 + 2as

so use:

0 302 2 ( 9.8) s 19.6 s 900 s

45.9 m (3 sf)

80


Example 3 Daisy is cycling at a steady speed of 6 ms to slow down at a rate of 0.5 ms 2 .

1

when she comes to a hill which causes her

(a)

How far up the hill does she travel before coming to a rest?

(b)

How long does it take her speed to be reduced to 2 ms 1 .

Solution (a) u

6

v

0

a Using v 2 02

0.5 u 2 2as 62 2

0.5 s

0 3 s s

36 She travels 36 metres up the hill

(b) u

6

v

2

a

0.5

Using v u at 2 6 0.5t t

0.5 t

4 4 0.5

8

She takes 8 seconds to slow down to 2 ms

81

2


Example 4 A particle is projected with a velocity of 20 ms-1. For how long will it be above a height of 16m? Solution We know:

u = 20 a = 9.8 s = 16 t=?

so use:

s = ut + ½ at2 16 = 20t – 4.9t 2 0 = 4.9t 2 – 20t + 16

use the quadratic formula

t = 1.09 s t = 2.99 s

(3 s.f.)

Two answers were expected because the particle will reach 16m both on the way up and on the way down during its motion. The time above 16m will be 2.99 1.09 1.9 seconds .

82


Displacement-time and Velocity-time Graphs These are often useful as tools in your armoury to tackle problems involving constant (straight lined velocity-time graphs) or variable acceleration (curved velocity-time graphs). (t, s) [Displacement-time] The gradient of this graph represents velocity. (t, v) [Velocity-time] The gradient of this graph represents acceleration. The area under this graph represents displacement when taking direction into account and distance if only the total of the magnitudes of the areas are considered. Example 1 Displacement (metres)

4

A

-6

B

C

D 2

4

5

F 6.5

8.5

time (seconds)

E

Between points A and B on the graph the particle is travelling with a constant velocity of 2 ms-1 away from its starting point. Between B and C the particle is stationary. Between C and D the particle travels back towards its starting point with constant velocity –4 ms-1 i.e. in the opposite direction to its initial motion. Between D and E the particle travels 6 m away from its starting point in the opposite direction, still with constant velocity –4 ms-1. For the final part of the journey between E and F the particle travels with constant velocity 3 ms-1 in its original direction until it returns to its starting point. 83


Example 2 Velocity -1 (ms )

4

A

The total of areas ABCD and DEF represents the distance travelled.

C

B

D 2

4

-6

5

6.5

F 8.5

time (seconds)

E

At the point on the graph marked A, the particle starts at rest. Between points A and B on the graph the particle is travelling with a constant acceleration of 2 ms-2 away from its starting point. Between B and C the particle is travelling with constant velocity 4 ms 1 away from its starting point. Between C and D the particle decelerates with deceleration 4 ms-2 until at D it is instantaneously stationary. It continues to decelerate with deceleration 4 ms-2 (and travel away from its starting point in the opposite direction) until at E it is instantaneously travelling at –6 ms-1 away from its starting point. It then accelerates with acceleration 3 ms-2 (N.B. an object travelling with negative velocity and positive acceleration is slowing down) until at F it is stationary again. The total distance travelled can be calculated by evaluating areas ABCD and DEF.

Area ABCD

Area DEF

(5 2) 4 14 2

(This is a trapezium)

3.5 6 10.5 2

Total distance travelled 14 10.5 24.5 metres

The displacement from its original position needs to take into account the direction of the motion i.e. you need the area of ABCD the area of DEF 14 10.5 2.5 metres

Notice that Examples 1 and 2 are quite different even though the graphs look the same!!!

84

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