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MTh301-Calculus II Midterm Special 2006
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¾ A plane can be perfectly determined by o one point. o one point and normal vector. o one point and parallel vector. o one normal vector.
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¾ At critical points function can assume o maximum value. o minimum value. o both maximum and minimum value. o zero value
Solution: Let x = length y = width A = area
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¾ The length and width of a rectangle are measured with the errors of at most 4 % and 3%, respectively. Use differentials to approximate the maximum percentage error in the calculated area.
Then A = xy So ∂A =y ∂x
∂A =x ∂y Therefore,
∂A ≈
∂A ∂A + ∂y ∂x ∂y
= y∂x + x∂y
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We desire percentage change in A, which is relative change multiplied by 100 so let’s work out relative change first. This is given by ∂A y∂x x∂y ≈ + A A A
=
∂x ∂y + x y
Since
A = xy ∂y ∂x ≤ 0.04 ≤ 0.04 and −0.03 ≤ y x
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−0.04 ≤
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∂A given the above constraints. This happens A ∂y ∂x ∂A when = 0.04 and = 0.03 , giving = 0.07 . This is relative error, so the y x A percentage error is 7%.
The maximum possible value for
Convert the Cartesian coordinates (1, 2 , −2) into cylindrical coordinates.
Solution:
We are to convert (x,y,z) int (r, θ ,z) Given: x =1
y= 2 z = −2
We know that:
r = x2 + y2 Put value of x and y in the above equation:
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r = (1) 2 + ( 2 ) 2
r = 1+ 2 r= 3 tan θ =
y x y x
θ = tan −1 ( )
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2 ) 1
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θ = tan −1 (
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θ = tan −1 2 θ = 54.73 z = −2
So the cylindrical coordinates are ( 3 ,54.73,−2)
====================================================== ¾ A unit vector in the direction of
o o o o
G ^ ^ ^ a = i+ j − k
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1 1 −1 , , 3 3 3
1 1 −1 , , 3 3 3
1 −1 −1 , , 2 2 2 1 1 −1 , , 2 2 2
3 3 Find equation of the tangent plane to the surface z = x + y at the point (2,1,3) .
Solution: f x = 3x − y
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f y = 3y − x
f z = −1 fx ( P) = 6 −1 = 5 fy (P) = 3 − 2 = 1 f z ( P ) = −1
Equation of tangent line to the surface through P is:
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5 ( x − 2 ) + 1( y − 1) − 1( z − 3 ) = 0
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5 x − 10 + y − 1 − z + 3 = 0 Hence:
5x + y − z − 8 = 0
====================================================== ^ ^ ^ ^ ^ G ^ G a = i + 2 j + k and b = 2 i + 3 j + 2 k
Dot product of o 8 o 5 o 10 o 13
∫∫ x dA ;
is
2
¾ Evaluate the double integral R 16 y= , y = x , and x = 8 x . by
R is the region bounded
Solution: Area of R =
=∫
8
2
∫
x
16 x
∫∫
R
x 2 dA
x 2 dydx
x = ∫ ∫16 x 16 dx 2 x x 8
x
2
8
= ∫ x 3 − 16 xdx 2
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x 4 16 x 2 − 2 4 2 4 8 x =∫ − 8x2 2 4 ⎛ 84 ⎞ ⎛ 24 ⎞ = ⎜ − 8× 8⎟ − ⎜ − 8× 2 ⎟ ⎠ ⎝4 ⎠ ⎝ 4 =∫
8
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¾ Domain of x + y is o x+ y> 0 o x≥0,y≥0 o x > 0, y > 0 o x+ y≥ 0
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= 972
Solution:
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¾ Use double integrals to compute the volume of the solid bounded by the cylinder x 2 + y 2 = 9 and the planes y + z = 9 and z = 0
∫ ∫ (9 − y)da R
R as a type 1 region we find that: V =∫
2
∫
9− x2
−2 − 9 − x 2
( 9 − y ) dydx
9 − x2 1 2⎤ ⎡ = ∫ ⎢9 y − y ⎥ −2 2 ⎦ y = − 9 − x 2 dx ⎣ 2
2
= ∫ 18 9 − x 2 dx −2
⎛1 ⎞ = 18 ⎜ π 92 ⎟ = 729π ⎝2 ⎠
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