Sample Final Exam #2 Math 3C

Math 3C
Sample Final Exam #2
Laney College, Fall 2011
Fred Bourgoin
1. Find the local maxima, local minima, and saddle points of
f (x, y) = x3 + y 3 − 6y 2 − 3x + 9 .
2. Let C be the unit circle in the xy-plane, centered at the origin, and
oriented counterclockwise when viewed from the positive z-axis. Compute
Z h
i
(yz 2 − y)~i + (xz 2 + x) ~j + 2xyz ~k · d~r
C
in two ways.
(a) Directly.
(b) Using Stokes’ Theorem.
3. Find ∂z/∂u, where z = (x + y)ey , x = u2 + v 2 , and y = u2 − v 2 . Express
your answer in terms of u and v.
4. A box of side 3 in the first octant has faces in the planes x = 0, x = 3,
y = 0, y = 3, z = 0, z = 3, and is oriented outward. Remove the face
in the yz-plane to get an open surface S with boundary C, oriented
appropriately. Let F~ = (x + y)~i − z ~j + y ~k.
Z
F~ · d~r.
(a) Compute
C
Z
~
F~ · dA.
(b) Compute
S
5. Let F~ = 2x~i − 4y ~j + (2z − 3) ~k,
Z and let C be the line segment from
~ · d~r.
F
(1, 1, 1) to (2, 3, −1). Compute
C
6. Consider the given contour diagram for a function f (x, y).
(a) Is f (x, y) a linear function? Justify.
(b) Find an expression for f (x, y).
1
7. Find the angle between the planes 5(x − 1) + 3(y + 2) + 2z = 0 and
x + 3(y − 1) + 2(z + 4) = 0.
8. Find the derivative of f (x, y, z) = 3x2 y 2 + 2yz at (−1, 0, 4) in the
direction of −~i + 3~j + 3~k.
9. Let S be the 2 × 2 × 2 rectangular surface centered at the origin and
oriented outward. Compute the given integral two ways.
Z
~
(x3 ~i + 2y ~j + 3~k) · dA
S
(a) Directly.
(b) Using the Divergence Theorem.
10. Find a vector normal to the surface z 2 − 2xyz = x2 + y 2 at (1, 2, −1).
11. Optimize f (x, y) = 3x − 4y subject to the constraint x2 + y 2 = 5.
Z Z √
Z √
3
12. Evaluate
0
9−z 2
√
− 9−z 2
9−y 2 −z 2
−
√
x2 dx dy dz.
9−y 2 −z 2
13. Let F~ = x~i + y ~j + (z 2 + 3) ~k, and let S be the rectangle z = 4,
0Z ≤ x ≤ 2, 0 ≤ y ≤ 3, oriented in the positive z-direction. Compute
~
F~ · dA.
S
14. Let S be the upper half of the sphere of radius 5, centered at the
~ = (x + cos y)~i + (y + sin x) ~j + (z + 3) ~k.
origin, oriented upward. Let F
Compute
Z
~.
F~ · dA
S
2
Math 3C
Sample Final Exam #2 Solutions
Laney College, Fall 2011
Fred Bourgoin
1. Find the local maxima, local minima, and saddle points of
f (x, y) = x3 + y 3 − 6y 2 − 3x + 9 .
Solution. First, find the critical points:
fx (x, y) = 3x2 − 3 = 0
2
fy (x, y) = 3y − 12y = 0
=⇒
=⇒
x = ±1
y = 0, 4
So, the critical numbers are (−1, 0), (−1, 4), (1, 0), and (1, 4). Let’s
compute more derivatives:
fxx (x, y) = 6x ,
fyy (x, y) = 6y − 12 ,
fxy (x, y) = 0 ,
and D(x, y) = 36x(y − 2). Then:
D(−1, 0) = 72 > 0 and fxx (−1, 0) = −6 < 0
=⇒
local max
D(−1, 4) = −72 < 0
=⇒
saddle point
D(1, 0) = −72 < 0
=⇒
saddle point
D(1, 4) = 72 > 0 and fxx (−1, 0) = 6 > 0
=⇒
local min
2. Let C be the unit circle in the xy-plane, centered at the origin, and
oriented counterclockwise when viewed from the positive z-axis. Compute
Z h
i
2
2
~
~
~
(yz
−
y)
i
+
(xz
+
x)
j
+
2xyz
k
· d~r
I=
C
in two ways.
(a) Directly.
Solution. Parameterize the circle: ~r(t) = cos t~i + sin t ~j with
0 ≤ t ≤ 2π. Then ~r 0 (t) = − sin t~i + cos t ~j and
Z 2π
Z 2π
dt = 2π .
(− sin t~i + cos t ~j) · (− sin t~i + cos t ~j) dt =
I=
0
0
3
(b) Using Stokes’ Theorem.
Solution. Let D be the disk, oriented upward, whose boundary
~ = 2 ~k,
is C. Since curl F
Z
~ = 2 (area of D) = 2π .
2 dA
I=
D
3. Find ∂z/∂u, where z = (x + y)ey , x = u2 + v 2 , and y = u2 − v 2 . Express
your answer in terms of u and v.
Solution. Use the chain rule, of course.
∂z ∂x ∂z ∂y
∂z
=
·
+
·
∂u
∂x ∂u ∂y ∂u
= (ey )(2u) + (x + y + 1) ey (2u)
= 2u(x + y + 2) ey = 4u(u2 + 1) eu
2 −v 2
.
4. A box of side 3 in the first octant has faces in the planes x = 0, x = 3,
y = 0, y = 3, z = 0, z = 3, and is oriented outward. Remove the face
in the yz-plane to get an open surface S with boundary C, oriented
appropriately. Let F~ = (x + y)~i − z ~j + y ~k.
Z
F~ · d~r.
(a) Compute
C
Solution. According to the right-hand rule, the square C (which
lies in the yz-plane) must be oriented clockwise when viewed from
the positive x-axis. Notice that in the yz-plane, the ~i component
of the vector field will contribute nothing to the line integral, so
~ = −z ~j + y ~k. Instead of parameterwe can assume the field is F
izing each edge of the square, let’s use Green’s Theorem, calling
the region inside the square R
Z
Z
Z
~
~
dA = 18 .
curl F dA = 2
F · d~r =
C
(b) Compute
Z
S
R
R
~
F~ · dA.
Solution. Let’s use the divergence theorem. Since div F~ = 1,
Z
Z
Z
~
~
~
dV = 27 ,
div F dV =
F · dA =
Sc
W
4
W
where Sc is the closed cube and W is the region inside it. Now,
the flux through the removed face B is
Z 3Z 3
Z 3Z 3
Z
27
~
~
~
~
~
~
F ·dA =
(y i−z j+y k)·(−i) dA = −
y dydz = − .
2
0 0
0 0
B
Hence, the flux through S is 27 − (−27/2) = 81/2.
5. Let F~ = 2x~i − 4y ~j + (2z − 3) ~k,
Z and let C be the line segment from
~ · d~r.
F
(1, 1, 1) to (2, 3, −1). Compute
C
Solution. Parameterize C:
~r (t) = (1 + t)~i + (1 + 2t) ~j + (1 − 2t) ~k
with
0 ≤ t ≤ 1.
Then ~r 0 (t) = ~i + 2 ~j − 2 ~k and
Z 1
Z
~
[(2 + 2t)~i − (4 + 8t) ~j − (1 + 4t) ~k] · (~i + 2 ~j − 2 ~k) dt
F · d~r =
0
C
Z 1
(4 + 6t) dt = −7 .
=−
0
6. Consider the given contour diagram for a function f (x, y).
(a) Is f (x, y) a linear function? Justify.
Solution. Yes, it is, because the
contours are parallel and equally
spaced.
(b) Find an expression for f (x, y).
Solution.
The plane goes
through the origin and the slopes
are m = 1 and n = 3, so its equation is z = x + 3y.
7. Find the angle between the planes 5(x − 1) + 3(y + 2) + 2z = 0 and
x + 3(y − 1) + 2(z + 4) = 0.
Solution. The angle between the planes is the same as the angle
between their normal vectors ~u = 5~i + 3 ~j + 2 ~k and ~v = ~i + 3 ~j + 2 ~k,
so
18
9
~u · ~v
=√ √ =√
=⇒
θ ≈ 38.7◦ .
cos θ =
k~uk k~v k
38 14
133
5
8. Find the derivative of f (x, y, z) = 3x2 y 2 + 2yz at (−1, 0, 4) in the
direction of −~i + 3~j + 3~k.
~ = 6xy 2 ~i + (6x2 y + 2z) ~j + 2y ~k, so
Solution. The gradient of f is ∇f
~ (−1, 0, 4) = 8 ~j. Normalize the direction vector:
∇f
1
~u = √ (−~i + 3~j + 3~k) .
19
Then the directional derivative is
~ (−1, 0, 4) · ~u = √24 .
f~u (−1, 0, 4) = ∇f
19
9. Let S be the 2 × 2 × 2 rectangular surface centered at the origin and
oriented outward. Compute the given integral two ways.
Z
~
(x3 ~i + 2y ~j + 3~k) · dA
S
(a) Directly.
Solution. The top and bottom faces are the graphs of z = 1 and
z = −1 over the region −1 ≤ x ≤ 1, −1 ≤ y ≤ 1, so
Z 1Z 1
Z
~
~
(x3 ~i + 2y ~j + 3~k) · ~k dx dy = 12 ,
F · dA =
−1 −1
ZT
~ · dA
~ = −12 .
F
B
For the other faces,
Z
Z
~
~
F · dA =
F
Z
~ · dA
~=
F
Z
B
−1 −1
Z 1Z 1
(~i + 2y ~j + 3~k) · ~i dy dz = 4 ,
(−~i + 2y ~j + 3~k) · (−~i) dy dz = 4 ,
~ · dA
~=
F
−1 −1
Z 1Z 1
(x3 ~i − 2 ~j + 3~k) · (−~j) dx dz = 8 ,
R
~ · dA
~=
F
Z
(x3 ~i + 2 ~j + 3~k) · ~j dx dz = 8 .
Z
S
L
Z
Hence,
1Z 1
−1 −1
1Z 1
−1 −1
~ = 12 − 12 + 4 + 4 + 8 + 8 = 24.
F~ · dA
6
(b) Using the Divergence Theorem.
~ = 3x2 + 2,
Solution. Since div F
Z
S
Z
~=
F~ · dA
2
(3x + 2) dV =
W
=6
Z
1Z 1
Z
1Z 1Z 1
(3x2 + 2) dx dy dz
−1 −1 −1
dy dz = 24 .
−1 −1
10. Find a vector normal to the surface z 2 − 2xyz = x2 + y 2 at (1, 2, −1).
Solution. We can think of the surface as one level surface of
g(x, y, z) = x2 − 2xyz − x2 − y 2 = −2xyz − y 2 .
~
Its gradient is ∇g(x,
y, z) = −2yz ~i − (2xz + 2y) ~j − 2xz ~k, so a vector
perpendicular to the surface at the given point is
~
∇g(1,
2, −1) = 4~i − 2 ~j + 2 ~k .
11. Optimize f (x, y) = 3x − 4y subject to the constraint x2 + y 2 = 5.
Solution. Use Lagrange multipliers.

3 = 2xλ

−4 = 2yλ
 2
x + y2 = 5
Equating λ from the first two equations yields y = − 43 x. When we
plug that into the third equation, we get the points ( √35 , − √45 ) and
(− √35 , − √45 ). Now,
√
3
4
f √ , −√
=5 5
5
5
√
3
4
f −√ , √
= −5 5
5
5
12. Evaluate I =
Z 3Z
0
√
9−z 2
−
√
9−z 2
Z √9−y2 −z 2
−
√
9−y 2 −z 2
=⇒
maximum
=⇒
minimum
x2 dx dy dz.
Solution. The region of integration is the sphere of radius 3 centered
7
at the origin, so we should use spherical coordinates.
Z 2πZ πZ 3
I=
ρ2 sin2 φ sin2 θ ρ2 sin2 φ dρ dφ dθ
0
0 0
Z 2πZ πZ 3
=
ρ4 sin4 φ sin2 θ dρ dφ dθ
0
0 0
Z Z
243 2π π 4
=
sin φ sin2 θ dφ dθ
5 0 0
Z
243π 2π 2
243π 2
=
.
sin θ dθ =
20 0
20
13. Let F~ = x~i + y ~j + (z 2 + 3) ~k, and let S be the rectangle z = 4,
0Z ≤ x ≤ 2, 0 ≤ y ≤ 3, oriented in the positive z-direction. Compute
~
F~ · dA.
S
Solution. Since the surface is flat,
Z
Z
Z
~
~
~
~
F · k dA = 19 dA = (19)(6) = 114 .
F · dA =
S
S
S
14. Let S be the upper half of the sphere of radius 5, centered at the
~ = (x + cos y)~i + (y + sin x) ~j + (z + 3) ~k.
origin, oriented upward. Let F
Compute
Z
~.
F~ · dA
S
~ = ~n dA are not
Solution. Parameterizing the surface or using dA
good ideas. Let’s instead let’s consider the closed surface S + D obtained by adding the disk D of radius 5 in the xy-plane centered at
the origin. Then, by the divergence theorem,
Z
Z
~
~
dV = 500π .
F · dA = 3
W
S+D
Now, we need to find the surface integral through D and subtract it
from what we just got. The orientation vector of D is ~n = −~k, so
Z
Z
Z
~ · ~n dA = −3
~=
dA = −75π .
F
F~ · dA
D
Hence,
Z
S
D
D
~ = 500π − (−75π) = 575π.
F~ · dA
8