Name
1-10
/50
Fall 2006
11
/15
P. Yasskin
12
/15
13
/15
14
/15
Total
/110
ID
MATH 251
Final Exam
Sections 507
Solutions
Multiple Choice: (5 points each. No part credit.)
1.
For the curve ⃗r(t) = (t cos t, t sin t), which of the following is false?
a.
The velocity is ⃗v = (cos t − t sin t, sin t + t cos t)
b.
The speed is |v⃗| =
c.
1 + t2
The acceleration is ⃗
a = (−2 sin t − t cos t, 2 cos t − t sin t)
d.
The arclength between t = 0 and t = 1 is L =
e.
The tangential acceleration is a T =
1
∫0 t
1 + t 2 dt
Correct Choice
t
1 + t2
⃗v = (cos t − t sin t, sin t + t cos t)
|v⃗| =
(cos t − t sin t) 2 + (sin t + t cos t) 2 =
cos 2 t + t 2 cos 2 t + sin 2 t + t 2 sin 2 t =
1 + t2
⃗
a = (−2 sin t − t cos t, 2 cos t − t sin t)
L=
1
1
∫ 0 |v⃗| dt = ∫ 0
1 + t 2 dt
d|v⃗|
2t
=
or
dt
2 1 + t2
aT = ⃗
a ⋅ T̂ = (−2 sin t − t cos t, 2 cos t − t sin t) ⋅
aT =
=
2.
1
(cos t − t sin t, sin t + t cos t)
1 + t2
1
t
[(−2 sin t − t cos t)(cos t − t sin t) + (2 cos t − t sin t)(sin t + t cos t)] =
1 + t2
1 + t2
Find the plane tangent to the surface x 2 z 2 + y 4 = 5 at the point (2, 1, 1).
a.
2x + y + z = 6
b.
2x + y + z = 5
c.
x + y + 2z = 5
d.
x − y + 2z = 3
e.
x − y + 2z = 6
Correct Choice
⃗ f = (2xz 2 , 4y 3 , 2x 2 z)
∇
f = x2z2 + y4
P = (2, 1, 1)
⃗⋅X = N
⃗⋅P
N
4x + 4y + 8z = 8 + 4 + 8 = 20
⃗ =∇
⃗f
N
P
= (4, 4, 8)
x + y + 2z = 5
1
3.
Let L =
lim
(x,y)→(0,0)
x 2 + xy 2
x2 + y4
a.
L exists and L = 1 by looking at the paths y = mx.
b.
L does not exist by looking at the paths y = x and y =
c.
L does not exist by looking at the paths y =
d.
L does not exist by looking at the paths x = my 2 .
e.
L does not exist by looking at the paths x = y
and y = − x .
x
3
x.
Correct Choice
and x = −y 3 .
2
2 4
2 3
Along y = mx, we have L = lim x 2 + m 4 x 4 = lim 1 + m 4 x 2 = 1, which proves nothing.
x→0 x + m x
x→0 1 + m x
2
2
Along y = ± x , we have L = lim x 2 + x 2 = 1, which proves nothing.
x→0 x + x
m 2 y 4 + my 4
m 2 + m , which depends on m and
Along x = my 2 , we have L = lim
=
lim
y→0 m 4 y 4 + y 4
y→0 m 4 + 1
proves the limit does not exist.
y6 ± y5
y2 ± y
Along x = ± y 3 , we have L = lim 6
=
lim
= 0, which (by itself) proves nothing.
y→0 y + y 4
y→0 y 2 + 1
4.
The point (1, −3) is a critical point of the function f = xy 2 − 3x 3 + 6y. It is a
a.
local minimum.
b.
local maximum.
c.
saddle point.
d.
inflection point.
e.
The Second Derivative Test fails.
f x = y 2 − 9x 2
Correct Choice
f y = 2xy + 6
f xx (1, −3) = −18
f yy (1, −3) = 2
f xx = −18x
f yy = 2x
f xy (1, −3) = −6
f xy = 2y
D = f xx f yy − f xy = −36 − 36 = −72
2
saddle point
5.
The dimensions of a closed rectangular box are measured as 70 cm, 50 cm and 40 cm with
a possible error of 0. 2 cm in each dimension. Use differentials to estimate the maximum error
in the calculated surface area of the box.
a.
8
b.
16
c.
32
d.
64
e.
128
Correct Choice
A = 2xy + 2xz + 2yz
dS = ∂S dx + ∂S dy + ∂S dz = 2(y + z) dx + 2(x + z) dy + 2(x + y) dz
∂x
∂y
∂z
dA = 2(90)(. 2) + 2(110)(. 2) + 2(120)(. 2) =. 4(320) = 128
2
6.
Consider the quarter of the cylinder x 2 + y 2 ≤ 4 with x ≥ 0, y ≥ 0 and 0 ≤ z ≤ 8.
Find the total mass of the quarter cylinder if the density is ρ = e x
a.
2π(e 4 − 1)
b.
8π(e 4 − 1)
c.
2πe 4
d.
8πe 4
e.
4
M = ∫∫∫ ρ dV =
7.
2 +y 2
.
Correct Choice
8
π/2
2
∫ 0 ∫ 0 ∫ 0 e r r dr dθ dz = (8)
2
π
2
2
2
er
2
= 2π(e 4 − 1)
0
Consider the quarter of the cylinder x 2 + y 2 ≤ 4 with x ≥ 0, y ≥ 0 and 0 ≤ z ≤ 8.
Find the z- component of the center of mass of the quarter cylinder if the density is ρ = e x
a.
2π(e 4 − 1)
b.
8π(e 4 − 1)
c.
2πe 4
d.
8πe 4
e.
4
8.
8
π/2
2
∫0 ∫0 ∫0
ze r r dr dθ dz =
2
8π(e 4 − 1)
z-mom
=
=4
M
2π(e 4 − 1)
Compute the line integral
from (3, 0) to (−3, 0).
a.
−9π
b.
−3π
c.
π
d.
3π
e.
9π
.
Correct Choice
z-mom = ∫∫∫ zρ dV =
z̄ =
2 +y 2
z2
2
8
0
π
2
2
er
2
2
= 8π(e 4 − 1)
0
OR by symmetry, z̄ must be half way up.
∫ y dx − x dy
counterclockwise around the semicircle x 2 + y 2 = 9
(HINT: Parametrize the curve.)
Correct Choice
⃗r(θ) = (3 cos θ, 3 sin θ)
⃗v = (−3 sin θ, 3 cos θ) Oriented correctly.
⃗ = (y, −x) = (3 sin θ, −3 cos θ)
⃗ ⋅ ⃗v = −9 sin 2 θ − 9 cos 2 θ = −9
F
F
⃗ ⋅ ds⃗ = F
⃗ ⋅ ⃗v dθ = π −9 dθ = −9π
y dx − x dy = F
∫
∫
∫
∫0
3
9.
⃗ ⋅ ds⃗ for the vector field F
⃗ = (y, x) along the curve
Compute the line integral ∫ F
2
2
⃗r(t) = e cos t , e sin t
for 0 ≤ t ≤ π .
(HINT: Find a scalar potential.)
d.
e − 1e
1 −e
e
2
e
2e
e.
0
a.
b.
c.
⃗ =∇
⃗f
F
Correct Choice
By the F.T.C.C.
10.
A = ⃗r(0) = (e cos 0 , e sin 0 ) = (e, 1)
for f = xy
B = ⃗r( π ) = (e cos π , e sin π ) = (e −1 , 1)
∫ A F⃗ ⋅ ds⃗ = ∫ A ∇⃗ f ⋅ ds⃗ = f(B) − f(A) = f(e −1 , 1) − f(e, 1) = (e −1 ⋅ 1) − (e ⋅ 1) =
B
B
1 −e
e
Consider the parabolic surface P given by
z = x 2 + y 2 for z ≤ 4 with normal pointing up and in,
the disk D given by x 2 + y 2 ≤ 4 and z = 4 with
normal pointing up, and the volume V between them.
⃗ we have
Given that for a certain vector field F
⃗=3
⃗ dV = 14
⃗ ⋅ dS
⃗ ⋅F
∇
and
F
∫∫∫
∫∫
V
D
compute
∫∫ F⃗ ⋅ dS⃗.
P
a.
17
b.
11
c.
8
d.
−11
e.
−17
Correct Choice
∫∫∫ ∇⃗ ⋅ F⃗ dV = ∫∫ F⃗ ⋅ dS⃗ − ∫∫ F⃗ ⋅ dS⃗
By Gauss’ Theorem:
V
D
P
The minus sign reverses the orientation of P to point outward. Thus
∫∫ F⃗ ⋅ dS⃗ = ∫∫ F⃗ ⋅ dS⃗ − ∫∫∫ ∇⃗ ⋅ F dV = 3 − 14 = −11
P
D
V
4
Work Out: (15 points each. Part credit possible.)
11.
Find the dimensions and volume of the largest box
which sits on the xy-plane and whose upper vertices
are on the elliptic paraboloid z + 2x 2 + 3y 2 = 12.
You do not need to show it is a maximum.
You MUST use the Method of Lagrange multipliers.
Half credit for the Method of Elminating the Constraint.
Maximize V = LWH = (2x)(2y)z = 4xyz subject to the constraint g = z + 2x 2 + 3y 2 = 12.
⃗ V = (4yz, 4xz, 4xy)
∇
⃗ g = (4x, 6y, 1)
∇
⃗ V = λ∇
⃗g
∇
4yz = λ4x
λ = 4xy


4xz = λ6y
4yz = 16x 2 y
4xy = λ
4xz = 24xy 2
Since V ≠ 0, we can assume x ≠ 0 and y ≠ 0 and z ≠ 0.
So z = 4x 2
z = 6y 2
2x 2 = 3y 2
The constraint becomes:
x=
3
2
y=
2 x=1
3
The dimensions are:
The volume is:
4x 2 + 2x 2 + 2x 2 = 12 or 8x 2 = 12
z = 4x 2 = 6
L = 2x =
V = LWH =
6
W = 2y = 2
H=z=6
6 (2)(6) = 12 6
5
12.
The hemisphere H given by
x 2 + y 2 + (z − 2) 2 = 9 for z ≥ 2
has center (0, 0, 2) and radius 3. Verify Stokes’ Theorem
⃗=
⃗ ⋅ dS
⃗ ×F
∫∫ ∇
H
∮ F⃗ ⋅ ds⃗
∂H
for this hemisphere H with normal pointing up and out
⃗ = (yz, −xz, z).
and the vector field F
Be sure to check and explain the orientations. Use the following steps:
a.
The hemisphere may be parametrized by
⃗ (θ, ϕ) = (3 sin ϕ cos θ, 3 sin ϕ sin θ, 2 + 3 cos ϕ)
R
Compute the surface integral by successively finding:
⃗, ∇
⃗, ∇
⃗ R
⃗ (θ, ϕ) ,
⃗ ×F
⃗ ×F
⃗e θ , ⃗e ϕ , N
∫∫ ∇⃗ × F⃗ ⋅ dS⃗
H
̂
î
k̂
⃗e θ = (−3 sin ϕ sin θ, 3 sin ϕ cos θ,
0 )
⃗e ϕ = (3 cos ϕ cos θ, 3 cos ϕ sin θ, −3 sin ϕ)
⃗ = ⃗e θ × ⃗e ϕ = î(−9 sin 2 ϕ cos θ) − ̂ (9 sin 2 ϕ sin θ) + k̂ (−9 sin ϕ cos ϕ sin 2 θ − 9 sin ϕ cos ϕ cos 2 θ)
N
= (−9 sin 2 ϕ cos θ, −9 sin 2 ϕ sin θ, −9 sin ϕ cos ϕ)
⃗ points down and in. Reverse it:
⃗ = (9 sin 2 ϕ cos θ, 9 sin 2 ϕ sin θ, 9 sin ϕ cos ϕ)
N
N
⃗ =
⃗ ×F
∇
î
∂
∂x
̂
∂
∂y
k̂
∂
∂z
yz,
−xz,
z
⃗ R
⃗ (θ, ϕ)
⃗ ×F
∇
= î(0 − −x) − ̂ (0 − y) + k̂ (−z − z) = (x, y, −2z)
= (3 sin ϕ cos θ, 3 sin ϕ sin θ, −2(2 + 3 cos ϕ))
⃗⋅N
⃗ = 27 sin 3 ϕ cos 2 θ + 27 sin 3 ϕ sin 2 θ − 18 sin ϕ cos ϕ(2 + 3 cos ϕ)
⃗ ×F
∇
= 27 sin 3 ϕ − 36 sin ϕ cos ϕ − 54 sin ϕ cos 2 ϕ
⃗= ∇
⃗ ⋅ dS
⃗⋅N
⃗ dθ dϕ = π/2 2π(27 sin 3 ϕ − 36 sin ϕ cos ϕ − 54 sin ϕ cos 2 ϕ) dθ dϕ
⃗ ×F
⃗ ×F
∇
∫∫
∫∫
H
H
= 2π ∫
π/2
0
(27(1 − cos 2 ϕ) sin ϕ − 36 sin ϕ cos ϕ − 54 sin ϕ cos 2 ϕ) dϕ
= 2π −27 cos ϕ −
= −36π
∫0 ∫0
cos 3 ϕ
3
+ 18 cos 2 ϕ + 18 cos 3 ϕ
π/2
0
Let u = cos ϕ.
= −2π −27 1 − 1
3
+ 18 + 18
Problem Continued
6
b.
Parametrize the boundary circle ∂H and compute the line integral by successively
finding:
⃗ (r⃗(θ)), ∮ F
⃗ ⋅ ds⃗.
⃗ = (yz, −xz, z)
⃗r(θ), ⃗v(θ), F
Recall:
F
∂H
⃗r(θ) = (3 cos θ, 3 sin θ, 2)
⃗v(θ) = (−3 sin θ, 3 cos θ, 0)
By the right hand rule the upper curve must be traversed counterclockwise which ⃗v
does.
⃗ (r⃗(θ)) = (6 sin θ, −6 cos θ, 2)
F
∮ F⃗ ⋅ ds⃗ = ∫ 0
2π
⃗ ⋅ ⃗v dθ =
F
∂C
2π
∫0
−18 sin 2 θ − 18 cos 2 θ dθ =
2π
∫0
−18 dθ = −36π
They agree!
13.
The spider web at the right is the graph
of the hyperbolic paraboloid z = xy .
It may be parametrized as
⃗ (r, θ) = (r cos θ, r sin θ, r 2 sin θ cos θ).
R
Find the area of the web for r ≤
⃗ r = (cos θ,
R
sin θ,
3.
2r sin θ cos θ)
⃗ θ = (−r sin θ, r cos θ, r 2 (cos 2 θ − sin 2 θ))
R
⃗ = i(r 2 sin θ(cos 2 θ − sin 2 θ) − 2r 2 sin θ cos 2 θ) − j(r 2 cos θ(cos 2 θ − sin 2 θ) − −2r 2 sin 2 θ cos θ)
N
+ k(r cos 2 θ − −r sin 2 θ) = (−r 2 sin θ cos 2 θ − r 2 sin 3 θ, −r 2 cos 3 θ − r 2 sin 2 θ cos θ, r)
=(−r 2 sin θ, −r 2 cos θ, r)
⃗ =
N
A=
2π
r 4 sin 2 θ + r 4 cos 2 θ + r 2 r 4 + r 2 = r r 2 + 1
∫0 ∫0
3
⃗ dr dθ =
N
2π
∫0 ∫0
= 2π (4) 3/2 − (1) 3/2
3
3
r(r + 1)
2
1/2
dr dθ = 2π
3/2
2
2 (r + 1)
3
2
3
0
= 2π [8 − 1] = 14π
3
3
7
14.
Green’s Theorem states:
∫∫
R
∂Q
− ∂P
∂x
∂y
dx dy =
∮ P dx + Q dy
∂R
Verify Green’s Theorem for the functions
P = −x 2 y
and
Q = xy 2
on the region inside the circle x 2 + y 2 = 16.
Use the following steps:
a.
∂Q
− ∂P .
∂x
∂y
∂Q
Then compute ∫∫
− ∂P
∂x
∂y
Compute
dx dy by converting to polar coordinates.
R
∂Q
− ∂P = y 2 − −x 2 = x 2 + y 2 = r 2
dx dy = r dr dθ
∂y
∂x
∂Q
∂P dx dy = 2π 4 r 2 r dr dθ = 2π r 4 4 = 128π
−
∫0 ∫0
∫∫ ∂x ∂y
4 r=0
R
b.
Parametrize the boundary circle.
Compute P, Q, dx and dy on the boundary curve.
Then compute
∮ P dx + Q dy
around the boundary.
∂R
The boundary circle x 2 + y 2 = 16 may be parametrized by ⃗r(θ) = (4 cos θ, 4 sin θ).
P = −x 2 y = −64 cos 2 θ sin θ
dx = −4 sin θ dθ
Q = xy 2 = 64 cos θ sin 2 θ
dy = 4 cos θ dθ
P dx + Q dy = 256 cos 2 θ sin 2 θ dθ + 256 cos 2 θ sin 2 θ dθ = 512 cos 2 θ sin 2 θ dθ
2π
2π
2π
dθ
∮ P dx + Q dy = ∫ 0 512 cos 2 θ sin 2 θ dθ = 128 ∫ 0 sin 2 (2θ) dθ = 128 ∫ 0 1 − cos(4θ)
2
∂R
= 64 θ −
sin(4θ)
4
2π
= 128π
0
The answers are the same!
8