Name
MATH 221
Final
Fall 2009
Section 503
Solutions
1-11
/44
14
/10
12
/15
15
/20
13
/15 Total
/104
P. Yasskin
Multiple Choice: (4 points each. No part credit.)
1. Find the point where the line x = 2t, y = 4 − t, z = −2 + t intersects the plane x − y + z = −2. At
this point x + y + z =
a. −2
b. 0
c. 1
Correct Choice
d. 4
e. 6
x − y + z = 2t − 4 − t + −2 + t = 4t − 6 = −2
x = 2, y = 3, z = −1,
t=1
x+y+z = 4
2. Find the plane tangent to the graph of z = x 2 y 3 at the point 2, 1. The z-intercept is
a. −24
b. −16
Correct Choice
c. −4
d. 0
e. 4
fx, y = x 2 y 3
f2, 1 = 4
f x x, y = 2xy 3
f x 2, 1 = 4
f y x, y = 3x 2 y 2
f y 2, 1 = 12
Tan plane: z = 4 + 4x − 2 + 12y − 1 = 4x + 12y − 16
z-intercept = −16
3. Find the line perpendicular to the cone z 2 − x 2 − y 2 = 0 at the point P = 4, 3, 5. This line
intersects the xy-plane at
a. 3, 4, 0
b. −4, −3, 0
c. 8, 6, 0
Correct Choice
d. −6, −8, 0
e.
4 , 3 ,0
5 5
F = z2 − x2 − y2
⃗
∇F = −2x, −2y, 2z
⃗ =⃗
N
∇F
4,3,5
= −8, −6, 10
⃗ = 4, 3, 5 + t−8, −6, 10 = 4 − 8t, 3 − 6t, 5 + 10t
X = P + tN
Intersects the xy-plane when z = 0 or 5 + 10t = 0 or t = − 1 .
2
So x = 4 − 8 − 1 = 8 and y = 3 − 6 − 1 = 6
2
2
1
4. A box currently has length L = 20 cm which is increasing at 4 cm/sec, width W = 15 cm which
is decreasing at 2 cm/sec, and height H = 12 cm which is increasing at 1 cm/sec. At what rate
is the volume changing?
a. 3 cm 3 /sec
b. 7 cm 3 /sec
c. 540 cm 3 /sec
Correct Choice
d. 1500 cm 3 /sec
e. 3600 cm 3 /sec
dV = WH dL + LH dW + LW dH = 15 ⋅ 12 ⋅ 4 − 20 ⋅ 12 ⋅ 2 + 20 ⋅ 15 ⋅ 1 = 540
dT
dt
dt
dt
V = LWH
5. Duke Skywater is flying the Millenium Eagle through a galactic dust storm. Currently, his position is
P = 30, −20, 10 and his velocity is ⃗v = −4, 3, 12. He measures that currently the dust density is
ρ = 450 and its gradient is ⃗
∇ρ = 2, −2, 1. Find the current rate of change of the dust density as
seen by Duke.
a. −2
Correct Choice
b. 2
c. 110
d. 448
e. 560
⃗
∇⃗v ρ = ⃗v ⋅ ⃗
∇ρ = −4, 3, 12 ⋅ 2, −2, 1 = −8 − 6 + 12 = −2
6. Under the same conditions as in #5, in what unit vector direction should Duke travel to decrease the
dust density as quickly as possible?
a. −2, 2, −1
b. 2, −2, 1
c.
d.
e.
û=
4 ,
13
−4 ,
13
−2 ,
3
−3 , −12
13 13
3 , 12
13 13
2 , −1
3 3
Correct Choice
⃗ρ
−∇
−2, −2, 1
=
=
⃗
4+4+1
∇ρ
−2 , 2 , −1
3 3 3
2
x 2 y 2 + 2x − 4y
. Use the Second Derivative
xy
7. The point 2, −1 is a critical point of the function f =
Test to classify the point.
a. Local Minimum
b. Local Maximum
Correct Choice
c. Inflection Point
d. Saddle Point
e. Test Fails
f = xy + 2y − 4x
f x = y + 42
f y = x − 22
x
y
8
4
f xx = − 3 = −1 < 0
f yy = 3 = −4 < 0
f xy = 1
x
y
8. Compute
∮ F⃗ ⋅ ds⃗
D = 4−1 = 3 > 0
Maximum
counterclockwise around the rectangle 0 ≤ x ≤ π and 0 ≤ y ≤ 2π for
⃗ = y sin x + sin y, x cos y + cos x.
F
HINT: Use the Fundamental Theorem of Calculus for Curves or Green’s Theorem.
a. −16π
b. −8π
Correct Choice
c. 0
d. 8π
e. 16π
⃗ = P, Q
F
By Green’s Theorem:
∂Q
− ∂P dx dy =
∂x
∂y
π
= 4π cos x
= −8π
∮ F⃗ ⋅ ds⃗ = ∫∫
P = y sin x + sin y
Q = x cos y + cos x
π
∫∫cos y − sin x − sin x − cos ydx dy = ∫ 0 ∫ 0 −2 sin x dx dy
2π
0
9. Compute
∫∫ ⃗∇ × F⃗ ⋅ dS⃗
over the sphere x 2 + y 2 + z 2 = 4 with outward normal for
⃗ = xy 2 z, yz 2 x, zx 2 y.
F
HINT: Use Stokes’ Theorem or Gauss’ Theorem.
a. 4π
b. 12π
c. 32 π
3
d. 64 π
3
e. 0
Correct Choice
By Stokes’ Theorem,
∫∫ ⃗∇ × F⃗ ⋅ dS⃗ = ∮ F⃗ ⋅ ds⃗ = 0
By Gauss’ Theorem,
∫∫ ⃗∇ × F⃗ ⋅ dS⃗ = ∫∫∫ ⃗∇ ⋅ ⃗∇ × F⃗ dV = 0
because there is no boundary curve.
⃗ = 0.
because ⃗
∇⋅⃗
∇×F
3
10. Find the area of the piece of the paraboloid z = x 2 + y 2 above the circle x 2 + y 2 ≤ 2.
⃗ r, θ = r cos θ, r sin θ, r 2 .
Note: The paraboloid may be parametrized by R
a. − π
b.
c.
d.
e.
ln 2 2 + 3 − 102 2
32
π ln 2 2 + 3 + 6 2
2
13π
Correct Choice
3
π 5 3/2 − 1
6
π 17 3/2 − 1
6
k̂
̂
î
⃗e r = cos θ
sin θ
2r
⃗e θ = −r sin θ r cos θ 0
⃗ = î−2r 2 cos θ − ̂ 2r 2 sin θ + k̂ r = −2r 2 cos θ, −2r 2 sin θ, r
N
⃗ =
N
A=
4r 4 + r 2 = r 4r 2 + 1
∫∫ 1 dS = ∫ 0 ∫ 0
2π
2
2
r 4r 2 + 1 dr dθ = 2π 4r 2 + 1 3/2
12
11. Use Gauss’ Theorem to compute
0
= π 9 3/2 − π = 26π = 13π
6
6
6
3
∫∫ F⃗ ⋅ dS⃗
outward through the complete surface of the
⃗ = x 2 , xy, xz.
tetrahedron with vertices 0, 0, 0, 2, 0, 0, 0, 3, 0 and 0, 0, 6 for F
Note: The top of the tetrahedron is the plane z = 6 − 3x − 2y.
a. 4
b. 6
c. 9
Correct Choice
d. 12
e. 16
⃗
⃗ = 2x + x + x = 4x
∇⋅F
3− 32 x
∫∫ F⃗ ⋅ dS⃗ = ∫∫∫ ⃗∇ ⋅ F⃗ dV = ∫ 0 ∫ 0
2
∂V
V
∫0
6−3x−2y
4x dz dy dx =
=
∫ 0 24xy − 12x 2 y − 4xy 2  3−y=0 x dx = ∫ 0
=
∫0
=
∫ 0 36x − 36x 2 + 9x 3  dx =
2
2
2
3
2
2
24x 3 −
3
2
2
4x6 − 3x − 2y dy dx
x − 12x 2 3 −
72x − 36x 2 − 36x 2 + 18x 3 − 4x 9 − 9x + 9 x 2
4
18x 2 − 12x 3 + 9 x 4
4
3− 32 x
∫0 ∫0
3
2
x − 4x 3 −
3
2
x
2
dx
dx
2
0
= 12
4
Work Out: (Points indicated. Part credit possible. Show all work.)
12.
(15 points) Compute
∫∫ x dx dy
y
over
D
the "diamond shaped" region D in the
first quadrant bounded by the parabolas
y = 16 − x 2
y = 4 + x2
y = 8 − x2
and
y = x2
x
HINTS: Use the coordinates: u = y + x 2 , v = y − x 2 .
y = v + x2
J=
u = v + 2x 2
∂x, y
∂u, v
x2 = u − v
2
1
1
u−v 2
2
2
1
−1
u−v 2
2
2
=
u−v
2
Integrand: x =
u−v
2
x=
Solve for x and y.
y = v+ u−v
2
1
2
1
2
=
1
1
−−
u−v
u−v
8
8
2
2
=
1
u−v
4
2
Boundaries are:
y = 16 − x 2  y + x 2 = 16  u = 16
y = 8 − x2  y + x2 = 8  u = 8
y = 4 + x2  y − x2 = 4  v = 4
y = x2  y − x2 = 0  v = 0
∫∫ x dx dy = ∫ 0 ∫ 8
4
R
16
u−v
2
y = u+v
2
1
du dv =
u−v
4
2
∫0 ∫8
4
16
1 du dv = 1 u 16 v 4 = 1 84 = 8
8
0
4
4
4
5
13.
y
(15 points) Find the area and y-component
of the centroid (center of mass with ρ = 1)
of the upper half of the cardioid r = 1 + cos θ.
x
In polar coordinates: 0 ≤ r ≤ 1 + cos θ and 0 ≤ θ ≤ π.
1 + cos θ 2
dθ
2
r=0
π
π
= 1 ∫ 1 + 2 cos θ + cos 2 θ dθ = 1 θ + 2 sin θ + 1 θ + sin 2θ
2 0
2
2
4
0
3π
1
1
=
π+
π
=
2
2
4
1+cos θ
π 1+cos θ
π
π 1 + cos θ 3
3
Mx = ∫ ∫
r sin θ r dr dθ = ∫ r
sin θ dθ = ∫
sin θ dθ
3 r=0
3
0 0
0
0
π
3
4
4
1 + cos θ 4
= − ∫ u du = − u = −
= −0 + 2 = 4
3
12
12
12
3
0
A=
π
1+cos θ
∫0 ∫0
1 r dr dθ =
π
∫0
r2
2
1+cos θ
dθ =
π
∫0
ȳ = M x = 4 4 = 16
A
3 3π
9π
14. (10 points) Find 3 positive numbers x, y and z, whose sum is 120 such that fx, y, z = xy 2 z 3
is a maximum.
METHOD 1: Lagrange Multipliers:
⃗
f = xy 2 z 3
∇f = y 2 z 3 , 2xyz 3 , 3xy 2 z 2 
⃗
⃗g
∇f = λ∇

x = 20,

y = 2x,
y = 40,
y 2 z 3 = λ,
z = 3x
2xyz 3 = λ,

x + y + z = 120
⃗
∇g = 1, 1, 1
g = x+y+z
3xy 2 z 2 = λ

x + y + z = x + 2x + 3x = 120

y 2 z 3 = 3xy 2 z 2
6x = 120

x = 20
z = 60
METHOD 2: Eliminate a Variable:
x + y + z = 120
x = 120 − y − z
2 3

f = 120 − y − zy z = 120y 2 z 3 − y 3 z 3 − y 2 z 4
f y = 240yz 3 − 3y 2 z 3 − 2yz 4 = 0
f z = 360y 2 z 2 − 3y 3 z 2 − 4y 2 z 3 = 0

240 − 3y − 2z = 0
360 − 3y − 4z = 0
Subtract:
120 − 2z = 0  z = 60
Substitute back:
240 − 3y − 120 = 0 
Substitute back:
y 2 z 3 = 2xyz 3 ,
y = 40
x = 120 − y − z = 120 − 40 − 60 = 20
6
15.
⃗=
⃗ ⋅ dS
(20 points) Verify Stokes’ Theorem ∫∫ ⃗
∇×F
H
∮ F⃗ ⋅ ds⃗
∂H
⃗ = y, −x, xz + yz
for the vector field F
and the hemisphere z =
9 − x2 − y2
oriented up.
Use the following steps:
a. Parametrize the boundary curve and compute the line integral:
⃗rθ = 3 cos θ, 3 sin θ, 0
⃗vθ = −3 sin θ, 3 cos θ, 0
oriented counterclockwise - OK
⃗ r⃗θ = 3 sin θ, −3 cos θ, 0
F
∮ F⃗ ⋅ ds⃗ = ∫ 0
2π
∫0
2π
⃗ ⋅ ⃗v dθ =
F
∂H
−9 sin 2 θ − 9 cos 2 θ dθ = −9 ∫
2π
0
dθ = −18π
b. Compute the surface integral using the parametrization:
⃗ ϕ, θ =  3 sin ϕ cos θ,
R
⃗e ϕ =  3 cos ϕ cos θ,
⃗e θ = 
3 sin ϕ sin θ,
3 cos ϕ sin θ,
− 3 sin ϕ sin θ,
3 sin ϕ cos θ,
3 cos ϕ 
− 3 sin ϕ 
0 
⃗ = î9 sin 2 ϕ cos θ − ̂ −9 sin 2 ϕ sin θ + k̂ 9 sin ϕ cos ϕ cos 2 θ + 9 sin ϕ cos ϕ sin 2 θ
N
= 9 sin 2 ϕ cos θ, 9 sin 2 ϕ sin θ, 9 sin ϕ cos ϕ
̂
k̂
∂x ∂y
∂z
î
⃗
⃗ =
∇×F
oriented up - OK
= îz − ̂ z + k̂ −1 − 1 = z, −z, −2 = 3 cos ϕ, −3 cos ϕ, −2
y −x xz + yz
π/2
∫∫ ⃗∇ × F⃗ ⋅ dS⃗ = ∫∫ ⃗∇ × F⃗ ⋅ N⃗ dϕ dθ = ∫ 0 ∫ 0
2π
H
27 sin 2 ϕ cos ϕ cos θ − 27 sin 2 ϕ cos ϕ sin θ − 18 sin ϕ cos ϕ dϕ dθ
H
=
π/2
∫0 ∫0
2π
= 2π ∫
π/2
0
−18 sin ϕ cos ϕ dϕ dθ
because ∫
−18 sin ϕ cos ϕ dϕ = −36π
sin 2 ϕ
2
2π
0
cos θ dθ =
∫0
2π
sin θ dθ = 0
π/2
= −18π
0
7