Name
1-10
/50
Fall 2006
11
/15
P. Yasskin
12
/15
13
/15
14
/15
Total
/110
ID
MATH 253
Final Exam
Sections 201,202
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 line perpendicular to the surface x 2 z 2 + y 4 = 5 at the point (2, 1, 1).
a.
(x, y, z) = (1 + t, 1 + t, 2 + 2t)
b.
(x, y, z) = (1 + 2t, 1 + t, 2 + t)
c.
(x, y, z) = (2 + t, 1 + t, 1 + 2t)
d.
(x, y, z) = (1 + 2t, 1 + t, 2 + 2t)
e.
(x, y, z) = (2 + 2t, 1 + t, 1 + 1t)
f = x2z2 + y4
X = P + tv⃗
P = (2, 1, 1)
Correct Choice
⃗ f = (2xz 2 , 4y 3 , 2x 2 z)
∇
⃗f
∇
P
= (4, 4, 8)
⃗v = (1, 1, 2)
(x, y, z) = (2, 1, 1) + t(1, 1, 2) = (2 + t, 1 + t, 1 + 2t)
1
3.
Let L =
x 2 +y 2
−1
x2 + y2
e
lim
(x,y)→(0,0)
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 = −x.
c.
L does not exist by looking at polar coordinates.
d.
L exists and L = 0 by looking at polar coordinates.
e.
L exists and L = 1 by looking at polar coordinates.
1+m 2 x 2
e
− 1 l’H
Along y = mx, we have L = lim e
= lim
2
2
x→0 (1 + m )x
x→0
Correct Choice
(1 + m 2 )2x
= 1,
(1 + m 2 )2x
1+m 2 x 2
− 1 which proves nothing.
for all m including 1 and
r
l’H
In polar coordinates, L = lim e 2− 1 = lim e 2r = 1, which proves the limit exists and = 1.
r→0
r→0
2r
r
r2
4.
2
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 xx = −18x
f yy (1, −3) = 2
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.
Compute the line integral
⃗r(t) = e sin
a.
b.
c.
d.
e.
t2
, e cos
t2
e 2 + e − 1e − 12
e
1 + 1 − e − e2
e
e2
e 2 − e + 1e − 12
e
1 − 1 + e − e2
e
e2
0
⃗ =∇
⃗f
F
∫ F⃗ ⋅ ds⃗
for 0 ≤ t ≤
⃗ = (y, x + 2y) along the curve
for the vector field F
π.
(HINT: Find a scalar potential.)
Correct Choice
for f = xy + y 2
A = ⃗r(0) = (e sin 0 , e cos 0 ) = (1, e)
B = ⃗r( π ) = (e sin π , e cos π ) = (1, e −1 )
By the F.T.C.C.
∫ A F⃗ ⋅ ds⃗ = ∫ A ∇⃗ f ⋅ ds⃗ = f(B) − f(A) = f(1, e −1 ) − f(1, e) = (e −1 + e −2 ) − (e + e 2 ) =
B
B
1 + 1 − e − e2
e
e2
2
6.
∫ y dx − x dy
Compute the line integral
along the curve y = x 2 from (−3, 9) to (0, 0).
HINT: The curve may be parametrized as r(t) = (t, t 2 ).
a.
−9
b.
−3
c.
1
d.
3
e.
9
Correct Choice
⃗v = (1, 2t) Orientation OK.
r(t) = (t, t 2 )
⃗ = (y, −x) = (t 2 , −t)
⃗ ⋅ ⃗v = t 2 − 2t 2 = −t 2
F
F
∫ y dx − x dy = ∫ F⃗ ⋅ ds⃗ = ∫ F⃗ ⋅ ⃗v dt =
7.
0
∫ −3
−t 2 dθ =
3
−t
3
0
−3
= 0 − − −27
3
= −9
Consider the quarter cylinder surface x 2 + y 2 = 4 with x ≥ 0, y ≥ 0 and 0 ≤ z ≤ 8.
Find the total mass of the quarter cylinder surface if the density is ρ = x.
⃗ (θ, h) = (2 cos θ, 2 sin θ, h).
The surface may be parametrized by R
a.
32
b.
32π
c.
8
d.
8π
e.
64π
Correct Choice
⃗e θ = (−2 sin θ, 2 cos θ,
0 )
⃗e h = (
1 )
0 ,
M = ∫∫ ρ dS =
8.
0 ,
8
π/2
∫0 ∫0
⃗ dθ dh =
x N
⃗ = (2 cos θ, 2 sin θ, 0)
N
⃗ = 4 cos 2 θ + 4 sin 2 θ = 2
N
8
π/2
∫0 ∫0
2 cos θ 2 dθ dh = 4(8) sin θ
π/2
0
= 32
Consider the quarter cylinder surface x 2 + y 2 = 4 with x ≥ 0, y ≥ 0 and 0 ≤ z ≤ 8.
Find the y-component of the center of mass of the quarter cylinder if the density is ρ = x.
c.
4
π
π
4
32
d.
2
e.
1
a.
b.
Correct Choice
y-mom = ∫∫ yρ dS =
ȳ =
8
π/2
∫0 ∫0
⃗ dθ dh =
yx N
8
π/2
∫0 ∫0
2
4 sin θ cos θ 2 dθ dh = 8(8) sin θ
2
π/2
= 32
0
y-mom
= 32 = 1
M
32
3
9.
∮ x 2 y dx − xy 2 dy
Compute the line integral
counterclockwise around the circle x 2 + y 2 = 16.
(HINT: Use a theorem.)
a.
−128π
b.
−64π
c.
0
d.
64π
e.
128π
Correct Choice
Use Green’s Theorem:
− ∂P
∮ P dx + Q dy = ∫∫ ∂Q
∂x
∂y
∂R
with P = x 2 y and Q = −xy 2 .
dx dy
R
∂Q
− ∂P = −y 2 − x 2 = −r 2
dx dy = r dr dθ
∂y
∂x
4
2π 4
∂Q
dx dy = − ∫ ∫ r 2 r dr dθ = −2π r
∫∫ ∂x − ∂P
0
0
∂y
4
4
r=0
R
10.
= −128π
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
⃗=4
⃗ dV = 13
⃗ ⋅ dS
⃗ ⋅F
∇
and
F
∫∫∫
∫∫
V
D
compute
∫∫ F⃗ ⋅ dS⃗.
P
a.
−17
b.
−9
c.
5
d.
9
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 = 4 − 13 = −9
P
D
V
4
Work Out: (15 points each. Part credit possible.)
11. Find the point in the first octant on the graph of xy 2 z 4 = 32 which is closest to the origin.
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.
Minimize f = x 2 + y 2 + z 2 subject to g = xy 2 z 4 = 32.
Method 1: Lagrange Multipliers:
⃗ f = (2x, 2y, 2z)
∇
⃗ f = λ∇
⃗g
∇

⃗ g = (y 2 z 4 , 2xyz 4 , 4xy 2 z 3 )
∇
2x = λy 2 z 4 ,
2y = λ2xyz 4 ,
2z = λ4xy 2 z 3
x, y and z cannot be 0 to satisfy the constraint.
1
λ = 2x
= 14 =
 2x 2 = y 2 ,
4x 2 = z 2
y2z4
xz
2xy 2 z 2
32 = xy 2 z 4 = x
2x
2
(2x) 4 = 32x 7

x=1
y=

2
y=
2 x,
z = 2x
z=2
Method 2: Eliminate the Constraint:
x=
10
f = 24 8 + y 2 + z 2
y z
32
y2z4
12
f y = − 25 8 + 2y = 0
y z

2=

y=
2
y 4 z 10
= z2
6 8
y z
y
2
z=2
13
f z = − 24 9 + 2z = 0
y z

x=
z=
2y


y6
y 6 z 8 = 2 11
2y
8
= 2 11
y 4 z 10 = 2 12

y 14 = 2 7
32 = 2 5 = 1
2 ⋅ 24
y2z4
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
⃗ (r, θ)
⃗ ×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
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
b.
∂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.
y
1 dx dy over the
x2
diamond shaped region bounded by the curves
Compute
y=
∫∫
8
6
x , y = 3 x , y = x and y = 3x.
y2
HINT: Let u = x
10
y
and v = x .
4
2
0
0
2
4
6
8
10
x
We solve for x and y so we can compute the Jacobian:
2
u = y x =y
v
x y
J=
∂x
∂u
∂y
∂u
y
x = v = u2
v
∂x
∂v
∂y
∂v
=
1
v2
1
v
So x = u2
v
−2u
v3
−u
v2
=
The boundaries are: y 2 = x or u = 1.
y = x or v = 1.
The integrand is:
∫∫
1 dx dy =
x2
3
9
∫1 ∫1
1 = v4
x2
u2
v 4 ⋅ u du dv =
u2 v4
y = uv
−u − −2u
v4
v4
= u4
v
y 2 = 9x or u = 9.
y = 3x or v = 3.
So
3
9
∫ 1 dv ∫ 1
1 du = v
u
3
1
ln|u|
9
1
= [3 − 1][ln 9 − ln 1] = 2 ln 9
7
14.
The surface of a football may be approximated
in cylindrical coordinates by
r = sin z
for 0 ≤ z ≤ π
∫∫∫ ∇⃗ ⋅ F⃗ dV = ∫∫ F⃗ ⋅ dS⃗
Verify Gauss’ Theorem
∂V
V
for the volume inside the football and the vector field
⃗ = (2x, 2y, x 2 + y 2 )
F
Use the following steps:
a.
⃗ in rectangular coordinates
⃗ ⋅F
Compute the volume integral by computing ∇
and then
∫∫∫ ∇⃗ ⋅ F⃗ dV
in cylindrical coordinates.
V
⃗ = 2+2+0 = 4
⃗ ⋅F
∇
π
∫∫∫ ∇⃗ ⋅ F⃗ dV = ∫ 0 ∫ 0 ∫ 0
2π
sin z
4r dr dz dθ = 2π ∫
π
= 2π ∫ 1 − cos 2z dz = 2π z − sin 2z
0
2
b.
π
0
2r 2
π
0
sin z
r=0
π
dz = 2π ∫ 2 sin 2 z dz
0
= 2π 2
⃗ (θ, h) = (sin h cos θ, sin h sin θ, h).
The surface of the football may be parametrized by R
Compute the surface integral by successively finding
⃗, F
⃗ R
⃗ (θ, h) , F
⃗⋅N
⃗ , and
⃗e θ , ⃗e h , N
∫∫ F⃗ ⋅ dS⃗.
⃗e θ = (− sin h sin θ, sin h cos θ, 0)
⃗e h = (cos h cos θ, cos h sin θ, 1)
⃗ = ⃗e θ × ⃗e h = î(sin h cos θ) − ̂ (− sin h sin θ) + k̂ (− sin h cos h sin 2 θ − sin h cos h cos 2 θ)
N
= (sin h cos θ, sin h sin θ, − sin h cos h)
⃗ R
⃗ (θ, h)
F
= (2 sin h cos θ, 2 sin h sin θ, sin 2 h)
⃗⋅N
⃗ = 2 sin 2 h cos 2 θ + 2 sin 2 h sin 2 θ − sin 3 h cos h = 2 sin 2 h − sin 3 h cos h
F
π
π
∫∫ F⃗ ⋅ dS⃗ = ∫ 0 ∫ 0 F⃗ ⋅ N⃗ dh dθ = ∫ 0 ∫ 0 (2 sin 2 h − sin 3 h cos h) dh dθ
4
π
= 2π ∫ (1 − cos 2h − sin 3 h cos h) dh = 2π h − sin 2h − sin h
0
2
4
2π
2π
π
= 2π 2
0
8