Name
Sec
MATH 251
Final Exam
Sections 508
Spring 2008
Solutions
P. Yasskin
Multiple Choice: (5 points each. No part credit.)
1. Which of the following lines lies in the plane:
1-13
/65
14
/25
15
/15
Total
/105
2x − y − z = 0 ?
a. x, y, z = 1, 2, 3 + t1, 1, 1
b. x, y, z = 3, 2, 1 + t1, 1, 1
c. x, y, z = 2, 1, 3 + t1, 1, 1
Correct Choice
d. x, y, z = 3, 1, 2 + t1, 1, 1
e. x, y, z = 1, 3, 2 + t1, 1, 1
Plug each line into the plane:
(a) 2x − y − z = 2 − 2 − 3 + 2t − t − t = −3 ≠ 0
(b) 2x − y − z = 6 − 2 − 1 + 2t − t − t = 3 ≠ 0
(c) 2x − y − z = 4 − 1 − 3 + 2t − t − t = 0
(d) 2x − y − z = 6 − 1 − 2 + 2t − t − t = 3 ≠ 0
(e) 2x − y − z = 2 − 3 − 2 + 2t − t − t = −3 ≠ 0
2. Find the equation of the plane tangent to the graph of the function
fx, y = x 2 y + xy 2 at the point 2, 1. Then the z-intercept is
a. −12
Correct Choice
b. −6
c. 0
d. 6
e. 12
f = x 2 y + xy 2
f2, 1 = 6
f x = 2xy + y
2
z = f2, 1 + f x 2, 1x − 2 + f y 2, 1y − 1
f x 2, 1 = 5
= 6 + 5x − 2 + 8y − 1
f y = x 2 + 2xy
f y 2, 1 = 8
= 5x + 8y − 12
1
3. Find the arc length of the curve ⃗
rt = e t , 2t, 2e −t  between 1, 0, 2 and e, 2, 2e −1 .
Hint:
Look for a perfect square.
a. e − 2e −1
b. 1 + e − 2e −1
Correct Choice
c. e − 2e −1 − 1
d. e + 2e −1
e. e + 2e −1 − 3
⃗v = e t , 2, −2e −t 
|v⃗| = e 2t + 4 + 4e −2t = e t + 2e −t
⃗rt = 1, 0, 2 at t = 0
⃗rt = e, 2, 2e −1  at t = 1
L=
∫ 0 |v⃗| dt = ∫ 0 e t + 2e −t  dt = e t − 2e −t  10 = e − 2e −1  − 1 − 2 = 1 + e − 2e −1
1
1
4. Find the tangential acceleration a T of the curve ⃗
rt = e t , 2t, 2e −t .
Hint:
Which formula is easier?
a. e t + 2e −t
b. e t + 4e −t
c. e t − 2e −t
Correct Choice
d. e t − 4e −t
e. e 2t − 4e −2t
⃗v = e t , 2, −2e −t 
|v⃗| =
e 2t + 4 + 4e −2t = e t + 2e −t
aT =
d|v⃗|
= e t − 2e −t
dt
OR
T̂ =
1
e t , 2, −2e −t 
e t + 2e −t
⃗
a = e t , 0, 2e −t 
a T = T̂ ⋅ ⃗
a=
1
e 2t − 4e −2t  = e t − 2e −t
e t + 2e −t
2
5. The volume of a cone is V = 1 πr 2 h.
3
If the radius r is currently 3 cm and increasing at 2 cm/sec
while the height h is currently 4 cm and decreasing at 1 cm/sec,
is the volume increasing or decreasing and at what rate?
a. decreasing at 19π cm 3 / sec
b. decreasing at 13π cm 3 / sec
c. neither increasing nor decreasing
d. increasing at 13π cm 3 / sec
Correct Choice
e. increasing at 19π cm 3 / sec
dV = ∂V dr + ∂V dh = 2 πrh dr + 1 πr 2 dh = 2 π342 + 1 π3 2 −1 = 13π
3
3
3
3
dt
dt
dt
∂r dt
∂h dt
This is positive and so increasing.
6. Which of the following is a local maximum of fx, y = sinx siny?
a. 0, 0
b.
π ,0
2
c. π, π
d.
0, π
2
e. None of the above
Correct Choice
f x x, y = cosx siny
f y x, y = sinx cosy
π ,0
Since f x 0, π = f y π , 0 = 1, the points 0, π
and
are not even critical points.
2
2
2
2
f xx x, y = − sinx siny
f yy x, y = − sinx siny
f xy x, y = cosx cosy
Since f xx 0, 0 = 0
we have
2
D0, 0 = f xx f yy − f xy = −1
Since f xx π, π = 0
we have
f yy 0, 0 = 0
f yy π, π = 0
2
Dπ, π = f xx f yy − f xy = −1
f xy 0, 0 = 1
and
0, 0 is a saddle.
f xy π, π = 1
and
0, π
2
is a saddle.
3
3
x 2 + y = 17 at the point 3, 2, 1.
z2
z3
7. Find the equation of the plane tangent to the surface
Then the z-intercept is
Correct Choice
a. 0
b. −42
c. 42
d. −18
e. 18
3
2
y
Let f = x2 + 3 and P = 3, 2, 1. Then ⃗
∇f =
z
z
⃗ =⃗
N
∇f = 6, 12, −42
2
3
2x , 3y , −2x 2 + −3y
2
3
3
4
z
z
z
z
and
P
⃗⋅X = N
⃗⋅P
N
8. Compute
6x + 12y − 42z = 6 ⋅ 3 + 12 ⋅ 2 − 42 ⋅ 1 = 0
∫ F⃗ ⋅ ds⃗
If x = y = 0, then z = 0.
counterclockwise around the circle x 2 + y 2 = 4
⃗ =
for the vector field F
−x 4 y + 1 x 2 y 3 , xy 4 + x 3 y 2 .
3
Hint: Use Green’s Theorem and factor the integrand.
a. 4π
b.
c.
d.
e.
3
8π
3
16π
3
32π
3
64π
3
Correct Choice
∫ F⃗ ⋅ ds⃗ = ∫ P dx + Q dy
∫ F⃗ ⋅ ds⃗ = ∫∫
=
where P = −x 4 y + 1 x 2 y 3 and Q = xy 4 + x 3 y 2
3
∂Q
− ∂P dx dy =
∂x
∂y
∫∫y 4 + 3x 2 y 2  − −x 4 + x 2 y 2  dx dy = ∫∫x 4 + y 4 + 2x 2 y 2  dx dy
∫∫x 2 + y 2  2 dx dy = ∫ 0 ∫ 0 r 4 r dr dθ = 2π
2π
2
r6
6
2
9
= 64π
3
4
9. Find the mass of the half cylinder x 2 + y 2 ≤ 4 for 0 ≤ z ≤ 10 and y ≥ 0
if the density is ρ = x 2 + y 2 .
a. 10π
b. 20π
c. 40π
Correct Choice
d. 80π
e. 160π
M=
π
∫∫∫ ρ dV = ∫ 0 ∫ 0 ∫ 0 r 2 r dr dθ dz = 10π
10
2
r4
4
2
0
= 40π
10. Find the center of mass of the half cylinder x 2 + y 2 ≤ 4 for 0 ≤ z ≤ 10 and y ≥ 0
if the density is ρ = x 2 + y 2 .
a. 0, 1, 5
b.
c.
d.
e.
0, 8 , 5
5π
0, 5π , 5
8
0, 16 , 5
5π
0, 5π , 5
16
M xz =
Correct Choice
π
∫∫∫ yρ dV = ∫ 0 ∫ 0 ∫ 0 r sin θ r 2 r dr dθ dz = 10
10
ȳ = M xz = 128 = 16
M
5π
40π
2
By symmetry,
x̄ = 0
− cos θ
π
0
r5
5
2
0
= 102 32
5
= 128
z̄ = 5
5
11.
Find the area inside the cardioid r = 1 + cos θ
but outside the circle r = 1.
a.
π
4
b.
π
2
c. 2 − π
4
d. 2 + π
4
Correct Choice
e. 2 − π
2
A=
π/2
1+cos θ
∫∫ 1 dA = ∫ −π/2 ∫ 1
= 1
2
r dr dθ =
π/2
∫ −π/2 2 cos θ + cos 2 θ dθ =
sin2θ
= 1 2 sin θ + 1 θ +
2
2
2
12.
π/2
∫ −π/2
1
2
r2
2
π/2
∫ −π/2
π/2
−π/2
1+cos θ
π/2
dθ = 1 ∫
1 + cos θ 2 − 1 dθ
2 −π/2
r=1
1 + cos2θ
2 cos θ +
dθ
2
= 1 21 + 1
2
2
π
2
−π
2
− 1 2−1 + 1
2
2
Compute ∮ ⃗
∇f ⋅ ds⃗ counterclockwise once
y
= 2+ π
4
4
2
around the polar curve r = 3 + cos4θ
for the function fx, y = x 2 y.
-4
a. 2π
-2
2
4
x
-2
b. 4π
-4
c. 6π
d. 8π
e. 0
Correct Choice
By the FTCC,
∫ A ⃗∇f ⋅ ds⃗ = fB − fA.
B
However, since it is a closed curve, B = A, and
OR by Green’s Theorem,
∫ A ⃗∇f ⋅ ds⃗ = 0.
B
∮ ⃗∇f ⋅ ds⃗ = ∫∫ ⃗∇ × ⃗∇f ⋅ k̂ dA = 0
because ⃗
∇×⃗
∇f = 0 for any f.
6
13.
∫∫∫ ⃗∇ ⋅ F⃗ dV = ∫∫ F⃗ ⋅ dS⃗
Gauss’ Theorem states
∂H
H
Compute either integral for the solid hemisphere, H,
x 2 + y 2 + z 2 ≤ 4 with z ≥ 0
given by
⃗ = xz 2 , yz 2 , 0.
F
and the vector field
Notice that the boundary of the solid hemisphere ∂H consists of the hemisphere surface S
given by x 2 + y 2 + z 2 = 4 with z ≥ 0 and the disk D given by x 2 + y 2 ≤ 4 with z = 0.
a. 64π
b.
c.
d.
e.
15
128π
15
8 π2
3
32 π 2
3
64 π 2
3
Correct Choice
⃗
∇ ⋅ F = z 2 + z 2 + 0 = 2z 2 = 2ρ 2 cos 2 ϕ
The volume integral:
π/2
∫∫∫ ⃗∇ ⋅ F dV =
∫0 ∫0
H
2π
= 128π
5
dV = ρ 2 sin ϕ dρ dθ dϕ
2
5
∫ 0 2ρ cos ϕ ρ sin ϕ dρ dθ dϕ = 2π 2ρ5
2
2
2
2
0
− cos 3 ϕ
3
π/2
0
−0 − −1
3
3
= 128π
15
⃗ r, θ = r cos θ, r sin θ, 0
The surface integral over the disk:
R
⃗ R
⃗ r, θ
F
= 0, 0, 0
∫∫ F⃗ ⋅ dS⃗ = ∫ 0 ∫ 0 0 dr dθ = 0
2π
2
D
⃗ θ, ϕ = 2 sin ϕ cos θ, 2 sin ϕ sin θ, 2 cos ϕ
R
The surface integral over the hemisphere:
k̂
̂
î
⃗e θ = −2 sin ϕ sin θ, 2 sin ϕ cos θ,
0 
⃗e ϕ = 2 cos ϕ cos θ, 2 cos ϕ sin θ, −2 sin ϕ
⃗ = ⃗e θ × ⃗e ϕ = î−4 sin 2 ϕ cos θ − ̂ 4 sin 2 ϕ sin θ
N
+ k̂ −4 sin ϕ cos ϕ sin 2 θ − 4 sin ϕ cos ϕ cos 2 θ
= −4 sin 2 ϕ cos θ, −4 sin 2 ϕ sin θ, −4 sin ϕ cos ϕ
⃗ = 4 sin 2 ϕ cos θ, 4 sin 2 ϕ sin θ, 4 sin ϕ cos ϕ
Reverse N
⃗ R
⃗ r, θ = 8 sin ϕ cos 2 ϕ cos θ, 8 sin ϕ cos 2 ϕ sin θ, 0
F
π/2
∫∫ F⃗ ⋅ dS⃗ = ∫∫ F⃗ ⋅ N⃗ dϕ dθ = ∫ 0 ∫ 0
2π
S
Let
32 sin 3 ϕ cos 2 ϕ dϕ dθ = 2π ∫
S
u = cos ϕ
S
0
32 sin ϕ1 − cos 2 ϕ cos 2 ϕ dϕ
du = − sin ϕ dϕ
∫∫ F⃗ ⋅ dS⃗ = −2π ∫ 1 321 − u 2 u 2 du = −64π
0
π/2
u3 − u5
5
3
0
1
= 64π 1 − 1
5
3
= 128π
15
7
Work Out: (Part credit possible. Show all work.)
14.
∫∫ ⃗∇ × F⃗ ⋅ dS⃗ = ∮ F⃗ ⋅ ds⃗
(25 points) Verify Stokes’ Theorem
∂C
C
for the cone C given by z = 2 x 2 + y 2
for z ≤ 8
⃗ = yz, −xz, z.
oriented up and in, and the vector field F
Be sure to check and explain the orientations. Use the following steps:
⃗ r, θ = r cos θ, r sin θ, 2r
a. Note: The cone may be parametrized as R
Compute the surface integral by successively finding:
⃗, ⃗
⃗, ⃗
⃗ R
⃗ r, θ ,
⃗e r , ⃗e θ , N
∇×F
∇×F
∫∫ ⃗∇ × F⃗ ⋅ dS⃗
C
k̂
̂
î
⃗e r =
cos θ
sin θ,
⃗e θ = −r sin θ, r cos θ,
2
0
⃗ = ⃗e r × ⃗e θ = î−2r cos θ − ̂ 2r sin θ + k̂ r cos 2 θ + r sin 2 θ = −2r cos θ, −2r sin θ, r
N
oriented correctly
⃗
⃗ =
∇×F
î
∂
∂x
̂
∂
∂y
k̂
∂
∂z
yz,
−xz,
z
⃗
⃗ R
⃗ r, θ
∇×F
= î0 − −x − ̂ 0 − y + k̂ −z − z = x, y, −2z
= r cos θ, r sin θ, −4r
∫∫ ⃗∇ × F⃗ ⋅ dS⃗ = ∫∫ ⃗∇ × F⃗ ⋅ N⃗ dr dθ = ∫ 0 ∫ 0 −2r 2 cos 2 θ − 2r 2 sin 2 θ − 4r 2  dr dθ
2π
C
=
C
2π
4
∫ 0 ∫ 0 −6r 2  dr dθ = 2π−2r 3  40 = −256π
4
8
Recall:
⃗ = yz, −xz, z
F
b. Compute the line integral by parametrizing the boundary curve and successively finding:
∮ F⃗ ⋅ ds⃗
⃗ r⃗θ,
⃗rθ, ⃗v, F
∂C
⃗rθ = 4 cos θ, 4 sin θ, 8
⃗v = −4 sin θ, 4 cos θ, 0
oriented correctly
⃗ r⃗θ = yz, −xz, z = 32 sin θ, −32 cos θ, 8
F
∮ F⃗ ⋅ ds⃗ = ∫ 0
2π
⃗ ⋅ ⃗v dθ =
F
∂C
15.
∫ 0 −128 sin 2 θ − 128 cos 2 θ dθ = ∫ 0 −128 dθ = −256π
2π
2π
(15 points) A rectangular solid sits on the xy-plane with
its top four vertices on the paraboloid z = 4 − x 2 − 4y 2 .
Find the dimensions and volume of the largest such box.
The dimensions are 2x, 2y and z.
V = 2x2yz = 4xyz = 4xy4 − x 2 − 4y 2  = 16xy − 4x 3 y − 16xy 3
V x = 16y − 12x 2 y − 16y 3 = 0
x > 0,
(1)
y>0
and
4 − 3x 2 − 4y 2 = 0
3∗(1) - (2):
8 − 8x 2 = 0
V y = 16x − 4x 3 − 48xy 2 = 0
z > 0 to give positive, non-zero dimensions.
(2)
x2 = 1
4 − x 2 − 12y 2 = 0
x=1
y2 = 1
y= 1
4
2
2
2
z = 4 − x − 4y = 4 − 1 − 1 = 2
4y 2 = 4 − 3x 2 = 1
The dimensions are 2x = 2, 2y = 1 and z = 2.
V = 2x2yz = 4
9