Name
Sec
MATH 253
Final Exam
Sections 200,501,502
Solutions
1-10
/50
14
/20
12
/10
15
/5
13
/20 Total
/105
Spring 2009
P. Yasskin
Multiple Choice: (5 points each. No part credit.)
1.
Compute
2e −2 ,4,8e 2
∫e
⃗ ⋅ ds⃗ where F
⃗ = (yz, xz, xy) along the curve ⃗r(t) = (te −t , t 2 , t 3 e t ).
F
−1 ,1,e 1
⃗ . Use the Fundamental Theorem of Calculus for Curves.
HINT: Find a scalar potential for F
a.
2e −3 − 4 + 8e 3
b.
2e −3 + 4 + 8e 3
c.
2e −2 − 6e −1 + 12e − 8e 2
d.
63
e.
64
Correct Choice
⃗ = (yz, xz, xy) = ∇
⃗ f for f = xyz. So
F
2e −2 ,4,8e 2
∫e
2.
−1 ,1,e 1
Compute
⃗ ⋅ ds⃗ =
F
2e −2 ,4,8e 2
∫e
−1 ,1,e 1
⃗ f ⋅ ds⃗ = f(2e −2 , 4, 8e 2 ) − f(e −1 , 1, e 1 ) = 64 − 1 = 63
∇
∮(sin(x 3 ) − 2x 2 y) dx + (2xy 2 + cos(y 3 )) dy
counterclockwise around the circle
x + y = 9.
HINT: Use Green’s Theorem.
2
2
a.
81π
b.
36π
c.
18π
d.
9π
e.
3π
Correct Choice
By Green’s Theorem,
∮(sin(x 3 ) − 2x 2 y) dx + (2xy 2 + cos(y 3 )) dy = ∫ ∫
=
2π
3
∂ (2xy 2 + cos(y 3 )) − ∂ (sin(x 3 ) − 2x 2 y) dx dy
∂x
∂y
∫ ∫(2y 2 + 2x 2 ) dx dy = ∫ 0 ∫ 0 2r 2 r dr dθ = 2π
r4
2
3
r=0
= 81π
1
3.
Find the equation of the line perpendicular to the hyperboloid xyz = 6 at the point (3, 2, 1).
a.
2x + 3y + 6z = 18
b.
3x + 2y + z = 18
c.
(x, y, z) = (3 + 2t, 2 − 3t, 1 + 6t)
d.
(x, y, z) = (2 + 3t, 3 + 2t, 6 + t)
e.
(x, y, z) = (3 + 2t, 2 + 3t, 1 + 6t)
P = (3, 2, 1)
Correct Choice
⃗ F = (yz, xz, xy)
∇
F = xyz
⃗ =∇
⃗F
N
⃗ = (3, 2, 1) + t(2, 3, 6) = (3 + 2t, 2 + 3t, 1 + 6t)
X = P + tN
4.
(3,2,1)
= (2, 3, 6)
Find the mass of the "twisted cubic" curve
⃗r(t) = 2 t 3 , t 2 , t for 0 ≤ t ≤ 1 if the density is ρ = 3xz + 3y 2 .
3
a.
b.
c.
d.
e.
M= 5
3
M = 17
7
M = 86
21
M = 111
35
Correct Choice
M=1
ρ = 3 2 t 3 t + 3(t 2 ) 2 = 5t 4
3
1
7
5
1
1
1
M = ∫ ρ ds = ∫ ρ |v⃗| dt = ∫ 5t 4 (2t 2 + 1) dt = 5 ∫ (2t 6 + t 4 ) dt = 5 2 t + t
=5 2 + 1
0
0
0
7
5 0
7
5
⃗v = (2t 2 , 2t, 1)
5.
|v⃗| =
4t 4 + 4t 2 + 1 = 2t 2 + 1
= 17
7
Find the z-component of the center of mass of the "twisted cubic" curve
⃗r(t) = 2 t 3 , t 2 , t for 0 ≤ t ≤ 1 if the density is ρ = 3xz + 3y 2 .
3
a.
b.
c.
d.
e.
z̄ = 1
2
z̄ = 204
175
z̄ = 175
204
z̄ = 25
12
z̄ = 12
25
Correct Choice
z-mom = M xy = ∫ z ρ ds =
8
6
=5 t + t
4
6
1
0
1
1
1
∫ 0 zρ |v⃗| dt = ∫ 0 t 5t 4 (2t 2 + 1) dt = 5 ∫ 0 (2t 7 + t 5 ) dt
=5 1 + 1
4
6
= 25
12
z̄ =
M xy
z-mom
=
= 25 7 = 175
M
M
12 17
204
2
6.
The point (1, 2) is a critical point of the function f(x, y) = 16x 4 + y 4 − 32xy.
Classify the point (1, 2) using the Second Derivative Test.
a.
Local Mininum
b.
Local Maximum
c.
Saddle Point
d.
Inflection Point
e.
Test Fails
f x (1, 2) = 0
f x = 64x 3 − 32y
f y = 4y 3 − 32x
f xx (1, 2) = 192 > 0
f xx = 192x 2
D = f xx f yy − f xy2
7.
Correct Choice
f yy = 12y 2
f y (1, 2) = 0
f yy (1, 2) = 48
D(1, 2) = 192 ⋅ 48 − 32 2 = 8192 > 0
f xy = −32
Local Minimum
Find the area inside the small loop of the limacon r = −1 + 2 cos θ.
a.
b.
c.
d.
e.
f xy (1, 2) = −32
2 3 − 2π
3
3 − 1π
3
1π+ 1 3 −2
2
2
π− 3 3
Correct Choice
2
π − 3 3
4
2
y
1
0
1
2
3
x
-1
 cos θ = 1
 θ = ±60° = ± π
2
3
−1+2 cos θ
2
π/3
−1+2 cos θ
π/3
π/3
2
r
1
(−1 + 2 cos θ) dθ
A = ∫∫ 1 dA = ∫
r dr dθ = ∫
dθ =
−π/3 ∫ 0
−π/3
2 0
2 ∫ −π/3
π/3
π/3
= 1 ∫ (1 − 4 cos θ + 4 cos 2 θ) dθ = 1 ∫ (1 − 4 cos θ + 2(1 + cos 2θ)) dθ = 1 3θ − 4 sin θ + sin 2θ
2 −π/3
2 −π/3
2
3
= 3 π − 4 sin π + sin 2 π = π − 2 3 +
= π− 3 3
2
3
3
3
2
Find the angles where r → 0:
8.
− 1 + 2 cos θ = 0
π/3
−π/3
(Non-Honors Only) Find the mass of the hollow solid between the spheres x 2 + y 2 + z 2 = 4
and x 2 + y 2 + z 2 = 9 if the density is δ = 2 12
.
x + y + z2
a. 2π
b.
4π
c.
10π
d.
e.
M=
Correct Choice
π2
3
2π
3
2π
π
3
∫∫∫ δ dV = ∫ 0 ∫ 0 ∫ 2
1 ρ 2 sin ϕ dρ dϕ dθ = 2π(3 − 2) − cos ϕ
ρ2
π
0
= 2π(− −1 − −1) = 4π
3
9.
Duke Skywater is flying the Millenium Eagle through the galaxy. Currently, he is at the point
P = (1, 2, 3) and has velocity ⃗v = (4, 5, 6). He discovers that he is passing through a
dangerous polaron field which has density ρ = x 3 y 2 z. What is the rate of change of the
polaron density as seen by Duke at this instant?
a.
228
b.
144
c.
72
d.
32
e.
12
Correct Choice
⃗ ρ = (3x 2 y 2 z, 2x 3 yz, x 3 y 2 )
∇
⃗ρ
∇
P
= (36, 12, 4)
dρ
⃗ρ
= ⃗v ⋅ ∇
dt
P
= 4 ⋅ 36 + 5 ⋅ 12 + 6 ⋅ 4 = 228
10. Duke Skywater is flying the Millenium Eagle through the galaxy. Currently, he is at the point
P = (1, 2, 3) and has velocity ⃗v = (4, 5, 6). He discovers that he is passing through a
dangerous polaron field which has density ρ = x 3 y 2 z. In what unit vector direction should Duke
travel to decrease the polaron density as fast as possible?
a.
1 (9, 3, 1)
91
b.
(9, −3, 1)
1 (9, −3, 1)
91
c.
d.
e.
(−9, −3, −1)
1 (−9, −3, −1)
91
Correct Choice
⃗ρ
A function decreases fastest in the direction −∇
P
= (−36, −12, −4)
11. (Honors Only) Han Duet is flying the Millenium Eagle through the galaxy along the path
⃗r(t) = (t, t 2 , t 3 ) with velocity ⃗v = dr⃗ . At time t = 2 hours he releases a garbage pod which
dt
travels along his tangent line with constant velocity equal to ⃗v(2), the velocity of the Eagle
when the pod was released. Where is the pod at t = 5 hours, which is 3 hours after its
release?
a.
(3, 9, 27)
b.
(4, 12, 32)
c.
(5, 16, 44)
d.
(6, 20, 56)
e.
(7, 24, 68)
Correct Choice
⃗r(t) = (t, t 2 , t 3 )
⃗v = (1, 2t, 3t 2 )
⃗v(2) = (1, 4, 12)
P = ⃗r(2) = (2, 4, 8)
X(T) = P + Tv(2) = (2, 4, 8) + T(1, 4, 12) = (2 + T, 4 + 4T, 8 + 12T)
t = 2 when T = 0
t = 5 when T = 3
X(3) = (2 + 3, 4 + 4 ⋅ 3, 8 + 12 ⋅ 3) = (5, 16, 44)
4
Work Out: (Points indicated. Part credit possible. Show all work.)
12. (10 points) Find the point (x, y, z) in the first octant on the surface z = 27 + 64
5x
5y
which is closest to the origin.
Minimize the square of the distance to the origin
f = x2 + y2 + z2
subject to the constraint that the point lies on the surface
Method 1:
Eliminate a constraint:
f = x2 + y2 +
27 + 64
5x
5y
2
−64 = 0
5y 2
2
5y 2
Multiply the first equation by 5x
and the second equation by
:
54
128
5y 3
5x 3 = 27 + 64
(2)
= 27 + 64
(1)
5x
5y
5x
5y
27
64
y
x
Equate these to obtain
=
and plug back into (1) to obtain:
3
4
3
5x = 27 + 16 ⋅ 3 = 75
5x
5x
5x
27
Cross multiply:
5 2 x 4 = 75 ⋅ 27 = 3 4 ⋅ 5 2
So x = 3 and y = 4
27
64
9
16
25
and z =
+
=
+
=
= 5.
5x
5y
5
5
5
f x = 2x + 2 27 + 64
5x
5y
Method 2:
=0
Lagrange Multipliers:
⃗ f = (2x, 2y, 2z)
∇
2x = λ 272
5x
−27
5x 2
Minimize
z = 27 + 64 .
5x
5y
⃗g =
∇
27 , 64 , 1
5x 2 5y 2
2z = λ
f y = 2y + 2 27 + 64
5x
5y
The constraint is g = z − 27 − 64 = 0.
5x
5y
⃗ f = λ∇
⃗g
∇
2y = λ 642
Eliminate λ:
2x = 2z 272
2y = 2z 642
5x
5y
5y
3
3
5y
y
Solve for z:
z = 5x =
So x =
or
y = 4x
3
4
3
27
64
3
27
64
5x
Plug into the constraint:
z=
+

= 27 + 16 ⋅ 3 = 75 = 15
x
5x
5y
5x
5x
5x
27
3
Cross multiply:
x 4 = 81
x=3
y=4
z = 5x = 5
27
5
13.
(20 points) Verify Gauss’ Theorem
∫∫∫ ∇⃗ ⋅ F⃗ dV = ∫∫ F⃗ ⋅ dS⃗
∂V
V
⃗ = (xz 2 , yz 2 , −z 3 ) and the solid V
for the vector field F
above the paraboloid P given by z = x 2 + y 2
or parametrized by R(r, θ) = (r cos θ, r sin θ, r 2 ),
below the disk D given by x 2 + y 2 ≤ 4 and z = 4.
Be sure to check and explain the orientations.
Use the following steps:
a.
Compute the volume integral by successively finding:
⃗ (x, y, z), ∇
⃗ (r, θ, z), dV,
⃗ dV
⃗ ⋅F
⃗ ⋅F
⃗ ⋅F
∇
∇
∫∫∫
V
⃗ = z 2 + z 2 − 3z 2 = −z 2
⃗ ⋅F
∇
∫∫∫ ∇⃗ ⋅ F⃗ dV =
V
2π
2
dV = r dr dθ dz
4
∫0 ∫0 ∫r
−z 2 r dz dr dθ = −2π ∫
2
2
8
= −2π 64 r − r
3
2
8
b.
2
r=0
2
0
z3
3
4
z=r 2
r dr = −2π ∫
2
0
64 − r 6
3
3
r dr
= −2π (128 − 32) = −64π
3
Compute the surface integral over the disk by parametrizing the disk and successively
finding:
⃗
⃗ (r, θ), ⃗e r , ⃗e θ , N
⃗, F
⃗ R
⃗ (r, θ) ,
⃗ ⋅ dS
R
F
∫∫
D
⃗ (r, θ) = (r cos θ, r sin θ, 4)
R
î
̂
k̂
⃗e r = (cos θ,
sin θ, 0)
⃗e θ = (−r sin θ, r cos θ, 0)
⃗ = ⃗e r × ⃗e θ = î(0) − ̂ (0) + k̂ (r cos 2 θ + r sin 2 θ) = (0, 0, r)
N
⃗ to point up which it does.
We need N
⃗ = (xz 2 , yz 2 , −z 3 )
F
⃗ R
⃗ (r, θ)
F
= (16r cos θ, 16r sin θ, −64)
∫∫ F⃗ ⋅ dS⃗ = ∫∫ F⃗ ⋅ N⃗ dr dθ = ∫ 0 ∫ 0 −64r dr dθ = −128π
2π
D
D
2
r2
2
2
0
= −256π
6
Recall:
c.
⃗ = (xz 2 , yz 2 , −z 3 ) and P is parametrized by R
⃗ (r, θ) = (r cos θ, r sin θ, r 2 ).
F
Compute the surface integral over the paraboloid P by successively finding:
⃗
⃗, F
⃗ R
⃗ (r, θ) ,
⃗ ⋅ dS
⃗e r , ⃗e θ , N
F
∫∫
P
⃗ (r, θ) = (r cos θ, r sin θ, r 2 )
R
î
k̂
̂
⃗e r = (cos θ,
sin θ, 2r)
⃗e θ = (−r sin θ, r cos θ, 0)
⃗ = ⃗e r × ⃗e θ = î(−2r 2 cos θ) − ̂ (2r 2 sin θ) + k̂ (r cos 2 θ + r sin 2 θ) = (−2r 2 cos θ, −2r 2 sin θ, r)
N
⃗ to point down (out of the volume). Reverse N
⃗ = (2r 2 cos θ, 2r 2 sin θ, −r)
We need N
⃗ = (xz 2 , yz 2 , −z 3 )
F
⃗ R
⃗ (r, θ)
F
= (r 5 cos θ, r 5 sin θ, −r 6 )
⃗⋅N
⃗ = 2r 7 cos 2 θ + 2r 7 sin 2 θ + r 7 = 3r 7
F
∫∫ F⃗ ⋅ dS⃗ = ∫∫ F⃗ ⋅ N⃗ dr dθ = ∫ 0 ∫ 0 3r 7 dr dθ = 2π
2π
d.
P
P
Combine
∫∫ F⃗ ⋅ dS⃗
and
2
∫∫ F⃗ ⋅ dS⃗
D
P
to get
8
3r
8
2
0
= 192π
∫∫ F⃗ ⋅ dS⃗
∂V
∫∫ F⃗ ⋅ dS⃗ = ∫∫ F⃗ ⋅ dS⃗ + ∫∫ F⃗ ⋅ dS⃗ = −256π + 192π = −64π
∂V
D
P
which agrees with part (a).
7
14.
⃗=
⃗ ⋅ dS
⃗ ×F
(20 points) Verify Stokes’ Theorem ∫∫ ∇
C
∮ F⃗ ⋅ ds⃗
∂C
⃗ = (2yz, −2xz, z 2 ) and the portion of the
for the vector field F
cone z =
x2 + y2
between z = 1 and z = 3 oriented up and in.
Notice that the boundary of the piece of cone is two circles.
Be sure to check orientations. Use the following steps:
a.
⃗ (r, θ) = (r cos θ, r sin θ, r).
The cone may be parametrized by R
⃗
⃗, ∇
⃗, ∇
⃗ R
⃗ (r, θ) ,
⃗ ⋅ dS
⃗ ×F
⃗ ×F
⃗ ×F
Successively find: ⃗e r , ⃗e θ , N
∇
∫∫
C
k̂
̂
î
⃗e r = ( cos θ, sin θ, 1)
⃗e θ = (−r sin θ, r cos θ, 0 )
⃗ = ⃗e r × ⃗e θ = î(−r cos θ) − ̂ (r sin θ) + k̂ (r cos 2 θ + r sin 2 θ) = (−r cos θ, −r sin θ, r)
N
⃗ has the correct orientation.
N
⃗ =
⃗ ×F
∇
̂
∂
∂y
î
∂
∂x
2yz −2xz
⃗ R
⃗ (r, θ)
⃗ ×F
∇
k̂
∂
∂z
= î(2x) − ̂ (−2y) + k̂ (−2z − 2z) = (2x, 2y, −4z)
z2
= (2r cos θ, 2r sin θ, −4r)
⃗⋅N
⃗ = −2r 2 cos 2 θ − 2r 2 sin 2 θ − 4r 2 = −6r 2
⃗ ×F
∇
∫∫ ∇⃗ × F⃗ ⋅ dS⃗ = ∫∫ ∇⃗ × F⃗ ⋅ N⃗ dr dθ = ∫ 0 ∫ 1 −6r 2 dr dθ = 2π[−2r 3 ] r=1 = −104π
2π
C
3
3
C
8
⃗ = (2yz, −2xz, z 2 )
Recall F
b.
Parametrize the upper circle U and compute the line integral.
⃗ (r⃗(θ)), ∮ F
⃗ ⋅ ds⃗
Successively find: ⃗r(θ), ⃗v(θ), F
U
⃗r(θ) = (3 cos θ, 3 sin θ, 3)
⃗v(θ) = (−3 sin θ, 3 cos θ, 0)
oriented correctly counterclockwise
⃗ (r⃗(θ)) = (18 sin θ, −18 cos θ, 9)
F
∮ F⃗ ⋅ ds⃗ = ∫ 0
2π
⃗ ⋅ ⃗v dθ =
F
2π
∫0
−54 sin 2 θ − 54 cos 2 θ dθ = − ∫
2π
0
54 dθ = −108π
U
c.
Parametrize the lower circle L and compute the line integral.
⃗ (r⃗(θ)), ∮ F
⃗ ⋅ ds⃗
Successively find: ⃗r(θ), ⃗v(θ), F
L
⃗r(θ) = (cos θ, sin θ, 1)
⃗v(θ) = (− sin θ, cos θ, 0)
oriented counterclockwise
Rev ⃗v(θ) = (sin θ, − cos θ, 0)
⃗ (r⃗(θ)) = (2 sin θ, −2 cos θ, 1)
F
∮ F⃗ ⋅ ds⃗ = ∫ 0
2π
⃗ ⋅ ⃗v dθ =
F
2π
∫0
2 sin 2 θ + 2 cos 2 θ dθ =
2π
∫0
2 dθ = 4π
L
d.
Combine
∮ F⃗ ⋅ ds⃗
and
U
∮ F⃗ ⋅ ds⃗
L
to get
∮ F⃗ ⋅ ds⃗
∂C
∮ F⃗ ⋅ ds⃗ = ∮ F⃗ ⋅ ds⃗_ + ∮ F⃗ ⋅ ds⃗ = −108π + 4π = −104π
∂C
U
L
which agrees with part (a).
9
15. (5 points) Mark the Project that you worked on and then answer the questions in 1 or 2
sentences.
Gauss’ Law and Ampere’s Law
⃗ = 0.
⃗ ⋅E
One of the electric fields produced a charge density of zero: ρ c = 1 ∇
4π
Was there a charge and how did you know? Why was Gauss’ Theorem not violated?
Skimpy Donut
For the minimal donut, what was the relation between a and b?
For the maximal donut, what was the value of b?
Volume Between a Surface and Its Tangent Plane
When minimizing over a square, which tangent point (a, b) minimizes the volume?
Hypervolume of a Hypersphere
The volume enclosed by a sphere of radius R in ℝ n is V n = kπ p R q .
What are the values of p and q when n = 4?
What are the values of p and q when n = 5?
Average Temperatures
What was the shape of the probe used to measure the temperature of the water in the pot?
Which Maple command did you use when Maple was unable to compute the integrals?
Center of Mass of Planet X
In computing the mass of the water, how did you ensure you only integrated where the
land level was below sea level?
Which Maple command did you use when Maple was unable to compute the integrals?
10