Name
MATH 253
Final
Section 501,503
Solutions
Fall 2009
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 = 1 + 2t, y = 8 − 3t, z = 2 − 2t intersects the plane x − y + z = 1.
At this point x + y + z =
a. 9
b. 5
Correct Choice
c. 2
d. 1
e. 0
x − y + z = 1 + 2t − 8 − 3t + 2 − 2t = 3t − 5 = 1
x = 5, y = 2, z = −2,
x+y+z = 5
t=2
2. Find the plane tangent to the graph of z = cosx + 2y at the point
π , π . The z-intercept is
6 6
a. 0
b. π
6
c. π
3
π
d.
2
e. π
Correct Choice
fx, y = cosx + 2y
f π , π = cos π
6 6
2
f x x, y = − sinx + 2y
f y x, y = −2 sinx + 2y
=0
f x π , π = − sin π = −1
f y π , π = −2 sin π = −2
6 6
2
6 6
2
π
π
π
π
π
−2 y−
= −x − 2y +
+
= −x − 2y +
Tan plane: z = −1 x −
6
6
6
3
2
π
z-intercept =
2
3. Find the plane tangent to the surface xz + yz = 5 at the point P = 6, 1, 3. The z-intercept is
a. 0, 0, 0
Correct Choice
b. 0, 0, −5
c. 0, 0, 5
d. 0, 0, −10
e. 0, 0, 10
1 ,− z ,− x + 1
⃗ =⃗
N
∇F
=
z
y
6,1,3
y2
z2
1 x − 3y + 1 z = 1 6 − 31 + 1 3 = 0
⃗⋅X = N
⃗⋅P
N
3
3
3
3
Intersects the z-axis when x = y = 0. So z = 0.
F = xz + yz
⃗
∇F =
1 , −3, − 6 + 1
3
9
=
1 , −3, 1
3
3
1
4. A circuit has two resistors R 1 = 200 Ω and R 2 = 300 Ω in parallel. The net resistance R
1 = 1 + 1 . If R 1 is increasing at 2 Ω/sec and R 2 is decreasing at 9 Ω/sec at
R1
R2
R
what rate is R changing?
satisfies
a.
b.
c.
d.
e.
9 Ω/sec
50
18 Ω/sec
25
− 9 Ω/sec
50
− 9 Ω/sec
25
− 18 Ω/sec
25
Correct Choice
1 = 1 + 1 = 300 + 200 = 1
R = 120
R
200
300
200 ⋅ 300
120
dR = R 2 dR 1 + R 2 dR 2 = 120 2 2 − 120 2 9 = − 18
− 12 dR = − 12 dR 1 − 12 dR 2
25
dt
dt
dt
dt
200 2
300 2
R
R 21 dt
R 22 dt
R1
R2
5. Ham Duet is flying the Millenium Eagle through a galactic dust storm. Currently, his position is
P = 10, −20, 30 and his velocity is ⃗v = 4, −12, 3. He measures that currently the dust density is
ρ = 500 and its gradient is ⃗
∇ρ = −2, 1, 2. Find the current rate of change of the dust density as
seen by Ham.
a. 514
b. 486
c. 28
d. 14
e. −14
Correct Choice
⃗
∇ρ = 4, −12, 3 ⋅ −2, 1, 2 = −8 − 12 + 6 = −14
∇⃗v ρ = ⃗v ⋅ ⃗
6. Under the same conditions as in #5, in what unit vector direction should Ham travel to decrease the
dust density as quickly as possible?
a. −2, 1, 2
b. 2, −1, −2
c.
d.
e.
û=
2 , −1 , −2
3 3 3
4 , −12 , 3
13 13 13
−4 , 12 , −3
13 13 13
Correct Choice
⃗ρ
−∇
−−2, 1, 2
=
=
⃗
4+1+4
∇ρ
2 , −1 , −2
3 3 3
2
16
7. The point 1, −2 is a critical point of the function f = x 2 y 2 + 8
x − y . Use the Second Derivative
Test to classify the point.
a. Local Minimum
Correct Choice
b. Local Maximum
c. Inflection Point
d. Saddle Point
e. Test Fails
f = x 2 y 2 + 8x − 16
f x = 2xy 2 − 82
f y = 2x 2 y + 162
y
x
y
16
32
2
2
f xx = 2y + 3 = 24 > 0
f yy = 2x − 3 = 6 > 0
f xy = 4xy = −8
x
y
D = 144 − 64 = 80 > 0
Minimum
8. Compute
∮ F⃗ ⋅ ds⃗
⃗ = x 4 − y 3 , y 4 + x 3 .
counterclockwise around the circle x 2 + y 2 = 4 for F
HINT: Use the Fundamental Theorem of Calculus for Curves or Green’s Theorem.
a. 0
b. 8π
c. 16π
Correct Choice
d. 24π
e. 32π
⃗ = P, Q
F
By Green’s Theorem:
∮ F⃗ ⋅ ds⃗ = ∫∫
∂Q
− ∂P
∂x
∂y
dx dy =
P = x4 − y3
Q = y4 + x3
∫∫3x 2 − −3y 2 dx dy = ∫ 0 ∫ 0 3r 2 r dr dθ = 2π
2π
2
3r 4
4
2
0
= 24π
9. The surface of an apple A may be given in spherical coordinates by ρ = 1 − cos ϕ and may be
parametrized by
Compute
Rφ, θ = 1 − cos ϕ sin ϕ cos θ, 1 − cos ϕ sin ϕ sin θ, 1 − cos ϕ cos ϕ.
∫∫ ⃗∇ × F⃗ ⋅ dS⃗
⃗ = xyz 2 , yzx 2 , zxy 2 .
over the apple with outward normal for F
HINT: Use Stokes’ Theorem or Gauss’ Theorem.
a. 0
Correct Choice
b. 4π
c. 12π
d. 32 π
3
e. 64 π
3
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 mass of the spiral ⃗
rθ = θ cos θ, θ sin θ for 0 ≤ θ ≤ 6π if the linear density is
ρ=
x2 + y2 .
a. 1 ln 6π +
b.
c.
d.
e.
1 + 36π 2 + 3π 1 + 36π 2
2
1 ln 6π + 1 + 6π − 3π 1 + 6π
2
1 ln 6π + 1 + 6π + 3π 1 + 6π
2
1 1 + 36π 2  3/2 − 1
Correct Choice
3
3
1 1 + 6π 3/2 − 1
3
3
⃗v = cos θ − θ sin θ, sin θ + θ cos θ
|v⃗| =
cos θ − θ sin θ 2 + sin θ + θ cos θ 2 =
ρ=
θ 2 cos 2 θ + θ 2 sin 2 θ = θ
M=
∫ ρ ds = ∫ 0
6π
θ 1 + θ 2 dθ =
1 + θ2
1 1 + θ 2  3/2
3
6π
0
= 1 1 + 36π 2  3/2 − 1
3
3
∮ F⃗ ⋅ ds⃗
around the triangle with vertices A = 2, 0, 0,
⃗ = y, z, x.
B = 0, 3, 0 and C = 0, 0, 6, traversed from A to B to C to A for F
11. Use Stokes’ Theorem to compute
⃗ x, y = x, y, 6 − 3x − 2y.
Note: The plane of the triangle may be parametrized as R
a. −24
b. −18
Correct Choice
c. 12
d. 18
e. 24
k̂
̂
î
⃗ = î3 − ̂ −2 + k̂ 1 = 3, 2, 1
N
⃗e x = 1 0 −3
⃗e y = 0 1 −2
⃗
⃗ =
∇×F
î
̂
k̂
∂x ∂y ∂z
y
z
= î−1 − ̂ 1 + k̂ −1 = −1, −1, −1
x
⃗
⃗⋅N
⃗ = −3 − 2 − 1 = −6
∇×F
∮ F⃗ ⋅ ds⃗ =
∂T
∫∫ ⃗∇ × F⃗ ⋅ dS⃗ =
T
3− 32 x
∫0 ∫0
2
−6 dy dx = −6 Area = −6 1 2 ⋅ 3
2
= −18
4
Work Out: (Points indicated. Part credit possible. Show all work.)
12.
(15 points) Compute
∫∫ y 2 dx dy
over
y
D
the "diamond shaped" region D in the
first quadrant bounded by the hyperbolas
y = 1x
and
y = 4x
and the lines
y=x
and
y = 2x
x
y
HINT: Use the coordinates u = xy, v = x .
y = xv
u = x2v
x 2 = uv
Solve for x and y.
u = u 1/2 v −1/2
v
x=
u v
v
y=
y=
uv = u 1/2 v 1/2
1 u −1/2 v −1/2 1 u −1/2 v 1/2
2
2
J=
=
= 1 −− 1 = 1
4v
4v
2v
1
1
1/2 −3/2
1/2 −1/2
− u v
u v
2
2
Integrand: y 2 = uv
Boundaries are:
y = 1x  xy = 1  u = 1
y = 4x  xy = 4  u = 4
y
y
y=x x =1v=1
y = 2x  x = 2  v = 2
∂x, y
∂u, v
∫∫ y 2 dx dy = ∫ 1 ∫ 1 uv 2v1 du dv =
2
D
4
1
2
∫ 1 ∫ 1 u du dv =
2
4
1 v 2 u 2
2 1 2
4
1
= 15
4
5
13.
(15 points) Find the volume and z-component of the centroid
(center of mass with ρ = 1) of the solid between the surfaces
z = x 2 + y 2 
3/2
z = 8.
and
In cylindrical coordinates: r 3 ≤ z ≤ 8.
V=
∫0 ∫0 ∫r
2π
2
r dz dr dθ = 2π ∫
3
8
= 2π 16 − 32
5
M xy =
∫0 ∫0 ∫r
2π
2
2
rz
0
8
zr dz dr dθ = 2π ∫
3
2
0
2
5
dr = 2π ∫ 8r − r 4  dr = 2π 4r 2 − r
5
0
= 96π
5
z=r 3
= 32π 1 − 2
5
8
So 0 ≤ r ≤ 2.
2
rz
2
2
0
8
2
8
dr = π ∫ 64r − r 7  dr = π 32r 2 − r
8
0
z=r 3
2
0
= π128 − 32 = 96π
M xy
z̄ =
= 96π 5 = 5
V
96π
14. (10 points) Find the point in the first octant on the graph of xy 2 z 4 = 32 which is closest to the
origin.
HINTS: What is the square of the distance from a point to the origin? Lagrange multipliers are
easier.
Minimize f = x 2 + y 2 + z 2 subject to g = xy 2 z 4 = 32.
Method 1: Lagrange multipliers:
⃗
∇f = 2x, 2y, 2z
⃗
⃗g
∇f = λ∇
λ=

⃗
∇g = y 2 z 4 , 2xyz 4 , 4xy 2 z 3 
2x = λy 2 z 4 ,
2x = 1 =
1
2 4
4
y z
xz
2xy 2 z 2
32 = xy 2 z 4 = x
2x
2
2y = λ2xyz 4 ,

2z = λ4xy 2 z 3
2x 2 = y 2 ,
2x 4 = 32x 7

4x 2 = z 2
x=1
y=

2
y=
2 x,
z = 2x
z=2
Method 2: Eliminate a variable:
10
f = 24 8 + y 2 + z 2
y z
x = 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

z=
2y


y6
y 6 z 8 = 2 11
2y
8
= 2 11
y 4 z 10 = 2 12

y 14 = 2 7
5
x = 32
= 2 4 =1
2 4
y z
2⋅2
6
15.
(20 points) Verify Gauss’ Theorem
∫∫∫ ⃗∇ ⋅ F⃗ dV = ∫∫ F⃗ ⋅ dS⃗
∂V
V
⃗ = xy , yx , z  and the volume
for the vector field F
2
above the cone z =
x2 + y2
2
3
and below the plane z = 2.
Use the following steps:
a. Compute the volume integral:
⃗
⃗ = y 2 + x 2 + 3z 2 = r 2 + 3z 2
∇⋅F
∫∫∫ ⃗∇ ⋅ F⃗ dV = ∫ 0 ∫ 0 ∫ r r 2 + 3z 2 r dz dr dθ = 2π ∫ 0
2π
2
2
2
r3z + z3r
2
z=r
dr = 2π ∫ 2r 3 + 8r − 2r 4  dr
V
4
5
= 2π r + 4r 2 − 2r
5
2
2
0
2
0
= 2 5 π 5 + 10 − 8
10
= 25π 1 + 1 − 4
5
2
4
= 7 ⋅ 2 π = 112 π
5
5
b. Compute the surface integral over the disk using the parametrization
⃗ r, θ =  r cos θ ,
R
r sin θ ,
2 :
⃗e r = 
cos θ ,
sin θ , 0 
⃗e θ =  − r sin θ , r cos θ , 0 
⃗ = î0 − ̂ 0 + k̂ r cos 2 θ + r sin 2 θ = 0, 0, r
N
⃗ R
⃗ r, θ = xy 2 , yx 2 , z 3  = r 3 cos θ sin 2 θ, r 3 sin θ cos 2 θ, 8
F
∫∫ F⃗ ⋅ dS⃗ = ∫∫ F⃗ ⋅ N⃗ dr dθ = ∫ 0 ∫ 0 8r dr dθ = 2π
2π
2
4r 2
2
0
= 32π
C
D
c. Compute the surface integral over the cone using the parametrization
⃗ r, θ =  r cos θ ,
R
r sin θ ,
r :
⃗e r = 
cos θ ,
sin θ , 1 
⃗e θ =  − r sin θ , r cos θ , 0 
⃗ = î−r cos θ − ̂ r sin θ + k̂ r cos 2 θ + r sin 2 θ = −r cos θ, −r sin θ, r
N
⃗ = r cos θ, r sin θ, −r
Reverse orientation:
N
⃗ R
⃗ r, θ
F
= xy 2 , yx 2 , z 3  = r 3 cos θ sin 2 θ, r 3 sin θ cos 2 θ, r 3 
∫∫ F⃗ ⋅ dS⃗ = ∫∫ F⃗ ⋅ N⃗ dr dθ = ∫ 0 ∫ 0 r 4 cos 2 θ sin 2 θ + r 4 sin 2 θ cos 2 θ − r 4  dr dθ
2π
C
2
C
=
∫ 0 r 4 dr ∫ 0 2 sin 2 θ cos 2 θ − 1 dθ =
2
= 32
5
2π
∫0
2π
1 − cos 4θ − 1 dθ = 32
5
4
r 5 2 ∫ 2π sin 2 2θ − 1 dθ =
5 0 0
2
2π − 2π = − 48 π
5
4
d. Compute the surface integral over the total boundary:
∫∫ F⃗ ⋅ dS⃗ = ∫∫ F⃗ ⋅ dS⃗ + ∫∫ F⃗ ⋅ dS⃗ = 32π −
∂V
D
48 π = 112 π
5
5
C
7