MATHEMATICS 200 December 2011 Final Exam Solutions

x2 +4y 2
1. Consider the function f (x, y) = e .

(a) Draw a “contour map” of f , showing all types of level curves that occur.
(b) Find the equation of the tangent plane to the graph z = f (x, y) at the point where
(x, y) = (2, 1).
(c) Find the tangent plane approximation to the value of f (1.99, 1.01) using the tangent
plane from part (b).

Solution. (a) Observe that, for any constant C, the curve x2 + 4y 2 = C is the level
curve f = eC .

• If C = 0, then x2 + 4y 2 = C = 0 is the pair of lines y = ± x2 .

p
• If C > 0, then x2 + 4y 2 = C > 0 is the hyperbola y = ± 12 C + x2 .
p
• If C < 0, then x2 + 4y 2 = C < 0 is the hyperbola x = ± |C| + 4y 2 .

y
f “e9

f “e
f “e´9 f “e´1 f “e´1 f “e´9
f “1 x
f “e

f “e9

(b), (c) Since

x2 +4y 2
fx (2, 1) = 2xe = 4
(x,y)=(2,1)
x2 +4y 2
fy (2, 1) = 8ye =8
(x,y)=(2,1)

The tangent plane to z = f (x, y) at (2, 1) is

z = f (2, 1) + fx (2, 1) (x 2) + fy (2, 1) (y 1) = 1 4(x 2) + 8(y 1)

= 1 4x + 8y

and the tangent plane approximation to the value of f (1.99, 1.01) is

f (1.99, 1.01) ⇡ 1 4(1.99 2) + 8(1.01 1) = 1.12

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2. Suppose z = f (x, y) has continuous second order partial derivatives, and x = r cos t,
y = r sin t. Express the following partial derivatives in terms r, t, and partial derivatives
of f .
@z
(a) @t
@2z
(b) @t2

Solution. By definition z(r, t) = f (r cos t , r sin t).

(a) By the chain rule

@z @f @f
(r, t) = r sin t (r cos t , r sin t) + r cos t (r cos t , r sin t)
@t @x @y

(b) By linearity, the product rule and the chain rule

 
@ 2z @ @f @ @f
(r, t) = r sin t (r cos t , r sin t) + r cos t (r cos t , r sin t)
@t2 @t @x @t @y
@f
= r cos t (r cos t , r sin t)
@x
@ 2f @2 f
+ r2 sin2 t (r cos t , r sin t) r 2
sin t cos t (r cos t , r sin t)
@x2 @y@x
@f
r sin t (r cos t , r sin t)
@y
2 @2 f 2 2 @ f
2
r sin t cos t (r cos t , r sin t) + r cos t (r cos t , r sin t)
@x@y @y 2
@f @f
= r cos t r sin t
@x @y
2
@ f @2 f 2 @ f
2
+ r2 sin2 t 2r 2
sin t cos t + r 2
cos t
@x2 @x@y @y 2

with all of the partial derivatives of f evaluated at (r cos t , r sin t).

3. A bee is flying along the curve of intersection of the surfaces 3z + x2 + y 2 = 2 and

z = x2 y 2 in the direction for which z is increasing. At time t = 2, the bee passes
through the point (1, 1, 0) at speed 6.

(a) Find the velocity (vector) of the bee at time t = 2.

(b) The temperature T at position (x, y, z) at time t is given by T = xy 3x + 2yt + z.
Find the rate of change of temperature experienced by the bee at time t = 2.

Solution. (a) Let’s use v to denote the bee’s velocity vector at time t = 2.

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 • The bee’s direction of motion is tangent to the curve. That tangent is perpendicular
to both the normal vector to 3z + x2 + y 2 = 2 at (1, 1, 0), which is

h2x , 2y , 3i = h2 , 2 , 3i
(x,y,z)=(1,1,0)

and the normal vector to z = x2 y 2 at (1, 1, 0), which is

h2x , 2y , 1i = h2 , 2, 1i
(x,y,z)=(1,1,0)

So v has to be some constant times

2 3
ı̂ı ˆ| k̂
h2 , 2 , 3i ⇥ h2 , 2, 1i = det 42 2 3 5 = h4 , 8 , 8i
2 2 1

or, equivalently, some constant times h1 , 2 , 2i.

• Since the z–component of v has to be positive, v has to be a positive constant times
h 1 , 2 , 2i.
• Since the speed has to be 6, v has to have length 6.

As | h 1 , 2 , 2i | = 3
v = 2h 1, 2 , 2i = h 2 , 4 , 4i

(b) Suppose that the bee is at x(t), y(t), z(t) at time t. Then the temperature that the
bee feels at time t is

T x(t), y(t), z(t), t = x(t)y(t) 3x(t) + 2y(t)t + z(t)

Then the rate of change of temperature (per unit time) felt by the bee at time t = 2 is

d
T x(t), y(t), z(t), t = x0 (2)y(2) + x(2)y 0 (2) 3x0 (2) + 2y 0 (2)2 + 2y(2) + z 0 (2)
dt t=2

Recalling that, at time t = 2, the bee is at (1, 1, 0) and has velocity h 2 , 4 , 4i

d
T x(t), y(t), z(t), t = ( 2)(1) + (1)( 4) 3( 2) + 2( 4)2 + 2(1) + 4
dt t=2
= 10

4. Find the radius of the largest sphere centred at the origin that can be inscribed inside
(that is, enclosed inside) the ellipsoid

2(x + 1)2 + y 2 + 2(z 1)2 = 8

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Solution. In order for a sphere of radius r centred on the origin to be enclosed in the
ellipsoid, every point of the ellipsoid must be at least a distance r from the origin. So the
largest allowed r is the distance from the origin to the nearest point on the ellipsoid.
We have to minimize f (x, y, z) = x2 + y 2 + z 2 subject to the constraint g(x, y, z) =
2(x + 1)2 + y 2 + 2(z 1)2 8. By Theorem 2.10.2 in the CLP–III text, any local minimum
or maximum (x, y, z) must obey the Lagrange multiplier equations

fx = 2x = 4 (x + 1) = gx (E1)
fy = 2y = 2 y = gy (E2)
fz = 2z = 4 (z 1) = gz (E3)
2(x + 1) + y + 2(z 1)2 = 8
2 2
(E4)

for some real number .

By equation (E2), 2y(1 ) = 0, which is obeyed if and only if at least one of y = 0,
= 1 is obeyed.

• If y = 0, the remaining equations reduce to

x = 2 (x + 1) (E1)
z = 2 (z 1) (E3)
(x + 1)2 + (z 1)2 = 4 (E4)

Note that 2 cannot be 1 — if it were, (E1) would reduce to 0 = 1. So equation

(E1) gives

2 1
x= or x+1=
1 2 1 2
Equation (E3) gives

2 1
z= or z 1=
1 2 1 2
1 1
Substituting x + 1 = 1 2
and z 1= 1 2
into (E4) gives

1 1 1
2
+ 2
= 4 () =2
(1 2 ) (1 2 ) (1 2 )2
1 p
() =± 2
1 2
So we now have p
two candidates
p for the location of the
p max and p min, namely
(x, y, z) = 1 + 2, 0, 1 2 and (x, y, z) = 1 2, 0, 1 + 2 .

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 • If = 1, the remaining equations reduce to

x = 2(x + 1) (E1)
z = 2(z 1) (E3)
2(x + 1)2 + y 2 + 2(z 1)2 = 8 (E4)

Equation (E1) gives x = 2 and equation (E3) gives z = 2. Substituting these into
(E4) gives

2 + y 2 + 2 = 8 () y 2 = 4 () y = ±2

So we have the following candidates for the locations of the min and max
p p p p
point 1+ 2, 0, 1 2 1 2, 0, 1 + 2 ( 2, 2, 2) ( 2, 2, 2)
p p
value of f 2 3 2 2 2 3+2 2 12 12
min max max

Recalling that f (x, y, z) is the square of the distance

p from (x, y, z) to the origin, the
p
maximum allowed radius for the enclosed sphere is 6 4 2 ⇡ 0.59.

5. (a) Consider the iterated integral

Z 0 Z 2

p
cos(x3 ) dx dy
4 y

i. Draw the region of integration.

ii. Evaluate the integral.
(b) Evaluate the double integral
ZZ p
y x2 + y 2 dA
D

over the region D = (x, y) x2 + y 2  2, 0  y  x .

Solution. (a) i. On the domain of integration

• y runs from 4 to 0 and

p
• for each y in that range, x runs from y (when y = x2 ) to 2.

The figure on the left below provides a sketch of the domain of integration. It also shows
the generic horizontal slice that was used to set up the given iterated integral.

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 y p2,0q
y p2,0q
x x

p2,´4q p2,´4q
?
x“ ´y y“´x2

R2
(a) ii. The inside integral, p y cos(x3 ) dx looks nasty. So let’s reverse the order of
integration and use vertical, rather than horizontal, slices. From the figure on the right
above, on the domain of integration,

• x runs from 0 to 2 and

• for each x in that range, y runs from x2 to 0.

So the integral
Z 0 Z 2 Z 2 Z 0
3
p
cos(x ) dx dy = dx dy cos(x3 )
4 y 0 x2
Z 2  2
sin(x3 )
2 3
= dx x cos(x ) =
0 3 0
sin(8)
=
3
p
(b) Let’s switch to polar coordinates. In polar coordinates, the circle x2 +y 2 = 2 is r = 2
and the line y = x is ✓ = ⇡4 .

y y“x

x
x2 ` y 2 “ 2

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 In polar coordinates dA = r dr d✓, so the integral
p
p
ZZ p Z ⇡/4 Z 2 z }| {
y x2 +y 2
2 2
z}|{
y x + y dA = d✓ dr r r sin ✓ r
D 0 0
p
Z ⇡/4  2
r4
= d✓ sin ✓
0 4 0
h i⇡/4
= cos ✓
0
1
=1 p
2

6. Let R be the triangle with vertices (0, 2), (1, 0), and (2, 0). Let R have density ⇢(x, y) = y 2 .
Find ȳ, the y–coordinate of the center of mass of R. You do not need to find x̄.

Solution. Here is a sketch of R.

y
p0,2q

x“2´y

2´y
x“ 2 R

p1,0q p2,0q x

Note that

• the equation of the straight line through (2, 0) and (0, 2) is y = 2 x, or x = 2 y.

(As a check note that both points (2, 0) and (0, 2) are on x = 2 y.
2 y
• The equation of the straight line through (1, 0) and (0, 2) is y = 2 2x, or x = 2
.
(As a check note that both points (0, 2) and (1, 0) are on x = 2 2 y .

By definition, the y–coordinate of the center of mass of R is the weighted average of y

over R, which is RR RR 3
y ⇢(x, y) dA y dA
ȳ = RRR = RRR 2
R
⇢(x, y) dA R
y dA
On R,

• y runs from 0 to 2. That is, 0  y  2.

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 2 y
• For each fixed y in that range, x runs from 2
to 2 y. In inequalities, that is
2 y
2
 x  2 y.
Thus ⇢
2 y
R= (x, y) 0  y  2, x2 y
2
For both n = 2 and n = 3, we have
ZZ Z 2 Z 2 y
n
y dA = dy dx y n
2 y
R 0 2
Z 2
2 y
= dy y n
0 2
 n+1 2
1 2y y n+2
=
2 n+1 n+2 0

1 2n+2 2n+2
=
2 n+1 n+2
2n+1
=
(n + 1)(n + 2)
So
RR 3 24
y dA (4)(5) 6
ȳ = RRR 2 = 23
=
R
y dA (3)(4)
5
RRR
7. Evaluate the triple integral E
x dV , where E is the region in the first octant bounded
2
by the parabolic cylinder y = x and the planes y + z = 1, x = 0, and z = 0.
Solution. First, we need to develop an understanding of what E looks like. Here are
sketches of the parabolic cylinder y = x2 , on the left, and the plane y + z = 1, on the
right.
z z

y`z “1

y y

y “ x2
x x

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 E is constructed by using the plane y + z = 1 to chop the top o↵ of the parabolic cylinder
y = x2 . Here is a sketch.

z “ 1´y

y
y “ 1, z “ 0
y “ x2
x p1, 1, 0q

So
E= (x, y, z) 0  x  1, x2  y  1, 0  z  1 y
and the integral
ZZZ Z 1 Z 1 Z 1 y
x dV = dx dy dz x
E 0 2
Z 1 Zx 1 0

= dx dy x(1 y)
0 x2
Z 1  1
y2
= dx x y
0 2 x2
Z 1 
x x5
= dx x3 +
0 2 2
1 1 1
= +
4 4 12
1
=
12

8. The body of a snowman is formed by the snowballs x2 + y 2 + z 2 = 12 (this is its body)

and x2 + y 2 + (z 4)2 = 4 (this is its head).

(a) Find the volume of the snowman by subtracting the intersection of the two snow balls
from the sum of the volumes of the snow balls. [Recall that the volume of a sphere
of radius r is 4⇡
3
r3 .]

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(b) We can also calculate the volume of the snowman as a sum of the following triple
integrals:
1. Z 2⇡ Z Z
3
2⇡ 2
⇢2 sin ' d⇢ d✓ d'
0 0 0
2. Z Z p Z r
2⇡ 3 4 p
3

p
r dz dr d✓
0 0 3r

3. Z Z Z p
⇡ 2⇡ 2 3
⇢2 sin(') d⇢ d✓ d'
⇡
6
0 0

Circle the right answer from the underlined choices and fill in the blanks in the
following descriptions of the region of integration for each integral. [Note: We have
translated the axes in order to write down some of the integrals above. The equations
you specify should be those before the translation is performed.]
i. The region of integration in (1) is a part of the snowman’s
body/head/body and head.
It is the solid enclosed by the
sphere/cone defined by the equation
and the
sphere/cone defined by the equation .
ii. The region of integration in (2) is a part of the snowman’s
body/head/body and head.
It is the solid enclosed by the
sphere/cone defined by the equation
and the
sphere/cone defined by the equation .
iii. The region of integration in (3) is a part of the snowman’s
body/head/body and head.
It is the solid enclosed by the
sphere/cone defined by the equation
and the
sphere/cone defined by the equation .
p p
Solution. (a) As a check, the body of the snow man has radius 12 = 2 3 ⇡ 3.46,
which is between 2 (the low point of the head) and 4 (the center of the head). Here is a
sketch of a side view of the snowman.

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 z
x2 ` y 2 ` pz ´ 4q2 “ 4

z“3

x2 ` y 2 ` z 2 “ 12

We want to determine the volume of the intersection of the body and the head, whose
side view is the darker shaded region in the sketch.

• The outer boundary of the body and the outer boundary of the head intersect when
both x2 + y 2 + z 2 = 12 and x2 + y 2 + (z 4)2 = 4. Subtracting the second equation
from the first gives

z2 (z 4)2 = 12 4 () 8z 16 = 8 () z = 3

Then substituting z = 3 into either equation gives x2 + y 2 = 3. So the intersection

of the outer boundaries of the head and body (i.e. the neck) is the circle x2 + y 2 = 3,
z = 3.
• The top boundary of the intersection
p is part of the top half of the snowman’s body
and so has equation z = + 12 x2 y 2 .
• The bottom boundary of the intersection
p is part of the bottom half of the snowman’s
head, and so has equation z = 4 4 x2 y 2

The intersection of the head and body is thus

p p
V = (x, y, z x2 + y 2  3, 4 4 x2 y2  z  12 x2 y2

We’ll compute the volume of V using cylindrical coordinates

Z p3 Z 2⇡ Z p12 r2
Volume(V) = dr d✓ p
dz r
0 0 4 4 r2
Z p
3 ⇥p p ⇤
= dr 2⇡ r 12 r2 4+ 4 r2
0
 p
3
1 3/2 1 3/2
= 2⇡ 12 r2 2r2 4 r2
3 3 0

1 3/2 1 3/2 1 3/2 1 3/2
= 2⇡ 9 2(3) 1 + 12 + 2(0)2 + 4
3 3 3 3

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 
1 1 3/2 8
= 2⇡ 9 6 + 12 +
3 3 3

1 3/2 38
= 2⇡ 12
3 3

So the volume of the snowman is


4⇡ 3/2 4⇡ 3 1 3/2 38
12 + 2 2⇡ 12
3 3 3 3
2⇡ h 3/2
i
= 12 + 54
3

(b) The figure on the left below is another side view of the snowman. This time it is
divided into a lighter gray top part, a darker gray middle part and a lighter gray bottom
part. The figure on the right below is an enlarged view of the central part of the figure
on the left.

2⇡{3 p0,0,4q
z
2
?
x ` y ` pz ´ 4q “ 4
2 2 2
?
p 3,0,3q
3

z“3

?
2 3
x

x2 ` y 2 ` z 2 “ 12 ⇡{3 p0,0,0q

i. The top part is the Pac–Man

2⇡{3 p0,0,4q

2
?
p 3,0,3q
?
3

12
part of the snowman’s head. It is the part of the sphere
x2 + y 2 + (z 4)2  4
that is above the cone r
x2 + y 2
z 4=
3
p
(which contains the points (0, 0, 4) and ( 3, 0, 3)).
ii. The middle part is the diamond shaped
p0,0,4q

2
?
p 3,0,3q

?
2 3

p0,0,0q

part of the snowman’s head and body. It is bounded on the top by the cone
r
x2 + y 2
z 4=
3
p
(which contains the points (0, 0, 4) and ( 3, 0, 3)) and is bounded on the bottom by the
cone p
z = 3(x2 + y 2 )
p
(which contains the points (0, 0, 0) and ( 3, 0, 3)).
iii. The bottom part is the Pac–Man
?
p 3,0,3q
?
3

?
2 3

⇡{3 p0,0,0q

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part of the snowman’s body. It is the part of the sphere

x2 + y 2 + z 2  12

that is below the cone p

z= 3(x2 + y 2 )
p
(which contains the points (0, 0, 0) and ( 3, 0, 3)).

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