Name
Sec
MATH 253
Final Exam
Sections 200,501,502
Solutions
Spring 2008
P. Yasskin
Multiple Choice: (5 points each. No part credit.)
1-13
/65
14
/25
15
/15
Total
/105
1. Find the equation of the plane containing the two lines:
⃗r 1 s = 2 + 3s, −4 − 2s, 3 − s
and
⃗r 2 t = 2 − t, −4 + 2t, 3 + 2t
a. −2x − 5y + 4z = 45
b. −2x − 5y + 4z = 1
c. −2x − 5y + 4z = −3
⃗ s, t = 2 + 3s − t, −4 − 2s + 2t, 3 − s + 2t
d. R
Correct Choice
⃗ s, t = −2 + 3s − t, −5 − 2s + 2t, 4 − s + 2t
e. R
The tangent vectors to the lines and hence the plane are ⃗v 1 = 3, −2, −1 and ⃗v 2 = −1, 2, 2.
The point P = 2, −4, 3 is on both lines and the plane.
So the parametric equation of the plane is
⃗ s, t = P + s ⃗v 1 + t ⃗v 2 = 2 + 3s − t, −4 − 2s + 2t, 3 − s + 2t
R
(The normal equation is −2x − 5y + 4z = 28.)
2. Find the equation of the plane tangent to the graph of the function
fx, y = x 2 + xy + y 2 at the point 2, 3. Then the z-intercept is
a. −38
b. −19
Correct Choice
c. 0
d. 19
e. 38
f = x 2 + xy + y 2
f2, 3 = 19
z = f2, 3 + f x 2, 3x − 2 + f y 2, 3y − 3
f x = 2x + y
f x 2, 3 = 7
= 19 + 7x − 2 + 8y − 3
f y = x + 2y
f y 2, 3 = 8
= 7x + 8y − 19
1
3. Find the arc length of the curve ⃗
rt = ln t, 2t, t 2  between 0, 2, 1 and 1, 2e, e 2 .
Hint:
Look for a perfect square.
a. e 2
Correct Choice
b. 1 + e 2
c. e 2 − 1
d. 2 + e 2
e. e 2 − 2
1 , 2, 2t
|v⃗| = 12 + 4 + 4t 2 = 1t + 2t
t
t
⃗rt = 0, 2, 1 at t = 1
⃗rt = 1, 2e, e 2  at t = e
e
e
e
L = ∫ |v⃗| dt = ∫ 1 + 2t dt = ln t + t 2  1 = ln e + e 2  − ln 1 + 1 = e 2
t
1
1
⃗v =
̂ of the curve ⃗rt = ln t, 2t, t 2  at t = 1.
4. Find the unit binormal B
a.
Plug t = 1 into ⃗v and ⃗
Hint:
a.
b.
c.
d.
e.
⃗v =
2,
3
1,
3
1,
3
2,
3
2,
3
1, 2
3 3
2, 2
3 3
−2 , 2
3 3
2, 1
3 3
−2 , 1
3 3
1 , 2, 2t
t
a| =
|v⃗ × ⃗
Correct Choice
⃗
⃗v = 1, 2, 2
− 12 , 0, 2
At t = 1:
a = −1, 0, 2
t
a = 1 4, −4, 2 = 2 , −2 , 1
16 + 16 + 4 = 6
B̂ = ⃗v × ⃗
6
3 3 3
a|
|v⃗ × ⃗
⃗
a=
⃗v × ⃗
a = 4, −4, 2
2
5. The volume of a square pyramid is V = 1 s 2 h.
3
If the side of the base s 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. increasing at 19 cm 3 / sec
b. increasing at 13 cm 3 / sec
Correct Choice
c. neither increasing nor decreasing
d. decreasing at 13 cm 3 / sec
e. decreasing at 19 cm 3 / sec
dV = ∂V ds + ∂V dh = 2 sh ds + 1 s 2 dh = 2 342 + 1 3 2 −1 = 13
3 dt
3 dt
3
3
dt
∂s dt
∂h dt
This is positive and so increasing.
6. Which of the following is a local minimum of fx, y = sinx cosy?
a. 0, 0
b.
π ,0
2
c. π, π
d.
0, π
2
e. None of the above
f x x, y = cosx cosy
Correct Choice
f y x, y = − sinx siny
Since f x 0, 0 = f x π, π = 1, the points 0, 0 and π, π are not even critical points.
f xx x, y = − sinx cosy
f yy x, y = − sinx cosy
f xy x, y = − cosx siny
Since f xx π , 0 = −1 < 0
f yy π , 0 = −1 < 0
f xy π , 0 = 0
2
2
2
2
π
π
we have
D
, 0 = f xx f yy − f xy = 1
and
,0
is a local maximum.
2
2
Since f xx 0, π = 0
f yy 0, π = 0
f xy 0, π = −1
2
2
2
2
we have
D 0, π = f xx f yy − f xy = −1
and
0, π
is a saddle.
2
2
3
7. Find the equation of the plane tangent to the surface x 2 z 2 + yz 3 = 11 at the point 2, 3, 1.
Then the intersection with the x-axis is at
a. 28, 0, 0
b. 16, 0, 0
c. 14, 0, 0
d. 7, 0, 0
Correct Choice
e. 4, 0, 0
⃗ =⃗
Let f = x 2 z 2 + yz 3 and P = 2, 3, 1. Then ⃗
∇f = 2xz 2 , z 3 , 2x 2 z + 3yz 2  and N
∇f
⃗⋅X = N
⃗⋅P
N
8. Compute
4x + y + 17z = 4 ⋅ 2 + 3 + 17 ⋅ 1 = 28
∫ F⃗ ⋅ ds⃗
P
= 4, 1, 17
If y = z = 0, then x = 7.
⃗ = y, x
for the vector field F
along the curve ⃗rt = t + sin t, t + cos t from ⃗rπ to ⃗r2π.
Hint: Find a scalar potential.
a. 3π 2 + 3π
Correct Choice
b. 3π 2 − 3π
c. 3π 2 + π
d. 3π 2 − π
e. 3π − 3π 2
⃗ = y, x = ⃗
⃗rπ = π, π − 1
⃗r2π = 2π, 2π + 1
∇f where f = xy.
F
∫ F⃗ ⋅ ds⃗ = ∫ ⃗∇f ⋅ ds⃗ = f2π, 2π + 1 − fπ, π − 1 = 2π2π + 1 − ππ − 1 = 3π 2 + 3π
4
9. Find the mass of the solid hemisphere x 2 + y 2 + z 2 ≤ 4 for y ≥ 0
if the density is δ = z 2 .
a. 4 π 2
b.
c.
d.
e.
M=
3
8 π2
3
32π
15
64π
15
128π
15
Correct Choice
π
π
∫∫∫ δ dV = ∫ 0 ∫ 0 ∫ 0 ρ 2 cos 2 ϕ ρ 2 sin ϕ dρ dϕ dθ = π
2
− cos 3 ϕ
3
π
0
r5
5
2
0
= π ⋅ 2 ⋅ 32 = 64π
5
3
15
10. Find the center of mass of the solid hemisphere x 2 + y 2 + z 2 ≤ 4 for y ≥ 0
if the density is δ = z 2 .
a.
b.
c.
d.
e.
0, 5 , 0
8
0, 8 , 0
5
0, 8π , 0
3
0, 3 , 0
8π
0, 3 , 0
4π
M xz =
Correct Choice
π
π
π
π
∫∫∫ yδ dV = ∫ 0 ∫ 0 ∫ 0 ρ sin ϕ sin θρ 2 cos 2 ϕ ρ 2 sin ϕ dρ dϕ dθ = ∫ 0 sin θ dθ ∫ 0 cos 2 ϕ sin 2 ϕ dϕ ∫ 0 ρ 5 dρ
2
2
6
6
sin 2 2ϕ
dϕ r
= 2
4
6 0
3
ȳ = M xz = 8π 15 = 5
By symmetry,
M
3 64π
8
= − cos θ
π
π
∫
0 0
π
∫0
3
1 − cos4ϕ
dϕ = 2
8
3
x̄ = 0
ϕ−
2
sin4ϕ
4
π
0
= 8π
3
z̄ = 0
5
11.
Find the area inside the circle r = 1
but outside the cardioid r = 1 − cos θ.
a.
π
4
b.
π
2
c. 2 − π
4
d. 2 + π
4
Correct Choice
e. 2 − π
2
A=
π/2
π/2
∫∫ 1 dA = ∫ −π/2 ∫ 1−cos θ r dr dθ = ∫ −π/2
= 1
2
1
π/2
∫ −π/2 2 cos θ − cos 2 θ dθ =
sin2θ
= 1 2 sin θ − 1 θ +
2
2
2
12.
Compute
∮ ⃗∇f ⋅ ds⃗
1
2
r2
2
π/2
∫ −π/2
π/2
−π/2
π/2
dθ = 1 ∫
1 − 1 − cos θ 2 dθ
2 −π/2
r=1−cos θ
1 + cos2θ
2 cos θ −
dθ
2
1
= 1 21 − 1
2
2
π
2
− 1 2−1 − 1
2
2
−π
2
= 2− π
4
y
counterclockwise once
2
around the polar curve r = 3 − cos4θ
for the function fx, y = x 2 y.
-2
2
x
a. 2π
-2
b. 4π
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 ⋅ dS⃗ = ∮ F⃗ ⋅ ds⃗
Stokes’ Theorem states
∂C
C
Compute either integral for the cone C given by
z = 2 x2 + y2
for z ≤ 8 oriented up and in,
⃗ = yz, −xz, z.
and the vector field F
⃗ r, θ = r cos θ, r sin θ, 2r
Note: The cone may be parametrized as R
The boundary of the cone is the circle x 2 + y 2 = 16 with z = 8.
a. −768π
b. −256π
Correct Choice
c. 64π
d. 256π
e. 768π
The surface integral:
k̂
̂
î
⃗e r =
cos θ
sin θ,
⃗e θ = −r sin θ, r cos θ,
⃗
⃗ =
∇×F
⃗ = ⃗e r × ⃗e θ = î−2r cos θ − ̂ 2r sin θ + k̂ r cos 2 θ + r sin 2 θ
N
2
= −2r cos θ, −2r sin θ, r
0
î
∂
∂x
̂
∂
∂y
k̂
∂
∂z
yz,
−xz,
z
oriented correctly
= î0 − −x − ̂ 0 − y + k̂ −z − z = x, y, −2z = r cos θ, r sin θ, −4r
∫∫ ⃗∇ × F⃗ ⋅ dS⃗ = ∫∫ ⃗∇ × F⃗ ⋅ N⃗ du dv = ∫ 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
⃗rθ = 4 cos θ, 4 sin θ, 8
The line integral:
⃗ r⃗θ = 32 sin θ, −32 cos θ, 8
F
∮ F⃗ ⋅ ds⃗ = ∫ 0
2π
∂C
⃗ ⋅ ⃗v dθ =
F
⃗v = −4 sin θ, 4 cos θ, 0
∫ 0 −128 sin 2 θ − 128 cos 2 θ dθ = ∫ 0 −128 dθ = −256π
2π
2π
7
Work Out: (Part credit possible. Show all work.)
14.
∫∫∫ ⃗∇ ⋅ F⃗ dV = ∫∫ F⃗ ⋅ dS⃗
(25 points) Verify Gauss’ Theorem
H
∂H
x + y + z ≤ 4 with z ≥ 0
2
for the solid hemisphere
2
2
⃗ = xz 2 , yz 2 , x 2 + y 2 .
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.
Be sure to check and explain the orientations. Use the following steps:
a. Compute the volume integral by successively finding:
⃗
⃗ x, y, z, ⃗
⃗ ρ, θ, ϕ, dV,
∇⋅F
∇⋅F
∫∫∫ ⃗∇ ⋅ F⃗ dV
H
⃗
⃗ = z 2 + z 2 + 0 = 2z 2 = 2ρ 2 cos 2 ϕ
∇⋅F
∫∫∫ ⃗∇ ⋅ F⃗ dV =
H
π/2
∫0 ∫0
2π
= 128π
5
dV = ρ 2 sin ϕ dρ dθ dϕ
5
∫ 0 2ρ cos ϕ ρ sin ϕ dρ dθ dϕ = 2π 2ρ5
2
2
2
−0 − −1
3
3
2
2
0
− cos 3 ϕ
3
π/2
0
= 128π
15
b. Compute the surface integral over the disk by parametrizing the disk and successively finding:
∫∫ F⃗ ⋅ dS⃗
⃗ r, θ, ⃗e r , ⃗e θ , N
⃗, F
⃗ R
⃗ r, θ ,
R
D
⃗ r, θ = r cos θ, r sin θ, 0
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 down (out of the volume). Reverse N
⃗ = 0, 0, −r
We need N
⃗ R
⃗ r, θ
F
= 0, 0, r 2 
∫∫ F⃗ ⋅ dS⃗ = ∫ 0 ∫ 0 −r 3 dr dθ = −2π
2π
D
2
r4
4
2
0
= −8π
8
Recall:
⃗ = xz 2 , yz 2 , x 2 + y 2 
F
c. Compute the surface integral over the hemisphere by parametrizing the surface and successively
finding:
∫∫ F⃗ ⋅ dS⃗
⃗ θ, ϕ, ⃗e θ , ⃗e ϕ , N
⃗, F
⃗ R
⃗ θ, ϕ ,
R
S
⃗ θ, ϕ = 2 sin ϕ cos θ, 2 sin ϕ sin θ, 2 cos ϕ
R
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 + k̂ −4 sin ϕ cos ϕ sin 2 θ − 4 sin ϕ cos ϕ cos 2 θ
N
= −4 sin 2 ϕ cos θ, −4 sin 2 ϕ sin θ, −4 sin ϕ cos ϕ
⃗ to point up (out of the volume). Reverse N
⃗ = 4 sin 2 ϕ cos θ, 4 sin 2 ϕ sin θ, 4 sin ϕ cos ϕ
We need N
⃗ R
⃗ r, θ
F
= 8 sin ϕ cos 2 ϕ cos θ, 8 sin ϕ cos 2 ϕ sin θ, 4 sin 2 ϕ
⃗⋅N
⃗ = 32 sin 3 ϕ cos 2 ϕ cos 2 θ + 32 sin 3 ϕ cos 2 ϕ sin 2 θ + 16 sin 3 ϕ cos ϕ = 32 sin 3 ϕ cos 2 ϕ + 16 sin 3 ϕ cos ϕ
F
π/2
∫∫ F⃗ ⋅ dS⃗ = ∫ 0 ∫ 0
2π
S
= 2π ∫
π/2
0
32 sin 3 ϕ cos 2 ϕ + 16 sin 3 ϕ cos ϕ dϕ dθ
32 sin ϕ1 − cos 2 ϕ cos 2 ϕ dϕ + 2π ∫
u = cos ϕ
First integral:
16 sin 3 ϕ cos ϕ dϕ
0
du = − sin ϕ dϕ
v = sin ϕ
Second integral:
π/2
dv = cos ϕ dϕ
∫∫ F⃗ ⋅ dS⃗ = −2π ∫ 1 321 − u 2 u 2 du + 2π ∫ 0 16v 3 dv = −64π
0
1
u3 − u5
5
3
S
= 64π 1 − 1
5
3
d. Combine
∫∫ F⃗ ⋅ dS⃗
+ 32π 1
4
and
∫∫ F⃗ ⋅ dS⃗ = ∫∫ F⃗ ⋅ dS⃗ + ∫∫ F⃗ ⋅ dS⃗ =
∂H
D
1
4
+ 32π v
4
1
0
= 128π + 8π = 248π
15
15
∫∫ F⃗ ⋅ dS⃗
S
D
0
to get
∫∫ F⃗ ⋅ dS⃗
∂H
128π + 8π − 8π = 128π
15
15
S
which agrees with part (a).
9
15.
(15 points) A rectangular solid sits on the xy-plane with
its top four vertices on the paraboloid z = 9 − 9x 2 − y 2 .
Find the dimensions and volume of the largest such box.
The dimensions are 2x, 2y and z.
V = 2x2yz = 4xyz = 4xy9 − 9x 2 − y 2  = 36xy − 36x 3 y − 4xy 3
V x = 36y − 108x 2 y − 4y 3 = 0
x > 0,
(1)
y>0
and
9 − 27x 2 − y 2 = 0
(1) - (2):
6 − 24x 2 = 0
y 2 = 3 − 3x 2 = 3 − 3 = 9
4
4
9
2
2
z = 9 − 9x − y = 9 −
−
4
The dimensions are 2x =
V y = 36x − 36x 3 − 12xy 2 = 0
z > 0 to give positive, non-zero dimensions.
(2)
3 − 3x 2 − y 2 = 0
x= 1
2
x2 = 1
4
y= 3
2
9 = 9
4
2
1, 2y = 3 and z = 9 .
2
V = 2x2yz = 27
2
10
16. (5 points) (Honors only. Replaces #2.) Find the plane tangent to the parametric surface
⃗ u, v = u + v, u − v, uv at the point R
⃗ 1, 1 = 2, 0, 1.
R
Give both the parametric equation and the normal equation of the tangent plane.
î
̂
⃗e u = 1 1,
⃗e v = 1, −1,
k̂
v
u
⃗ = ⃗e u × ⃗e v = îu + v − ̂ u − v + k̂ −1 − 1 = u + v, v − u, −2
N
⃗ 1, 1 = 2, 0, 1, ⃗e u = 1, 1, 1, ⃗e v = 1, −1, 1, N
⃗ = 2, 0, −2
At P = R
So the parametric equation is
X = P + s ⃗e u + t ⃗e v
x, y, z = 2, 0, 1 + s1, 1, 1 + t1, −1, 1 = 2 + s + t, s − t, 1 + s + t
and the normal equation is
⃗⋅X = N
⃗⋅P
N
2x − 2z = 2 ⋅ 2 − 2 ⋅ 1 = 2
or
x−z = 1
11