Name
Sec
MATH 253
Final Exam
Sections 501-503,200
1-10
/50 12
/25
11
/15 13
/15
Fall 2008
Solutions
P. Yasskin
Multiple Choice: (5 points each. No part credit.)
/105
Total
1. Find a parametric equation of the line tangent to the curve ⃗
r(θ) = (2 sin θ, 2 cos θ, θ)
at the point (0, −2, π).
a. X(t) = (−2, −2t, 1 + πt)
b. X(t) = (0, −2t − 2, π + t)
c. X(t) = (−2t, −2, π + t)
correct choice
d. X(t) = (−2t − 2, 0, π + t)
e. X(t) = (0, −2t − 2, 1 + πt)
⃗r(θ) = (2 sin θ, 2 cos θ, θ) = (0, −2, π) at θ = π.
⃗v(θ) = (2 cos θ, −2 sin θ, 1)
⃗v(π) = (−2, 0, 1)
X(t) = (0, −2, π) + t(−2, 0, 1) = (−2t, −2, π + t)
2. The density of the fog is given by ρ = 30 − x 2 − y 2 − z. If an airplane is at the position
(x, y, z) =
2 , 2, 4 , in what unit vector direction should the airplane initially travel to get out of the
fog as quickly as possible?
a.
−2 2 −4 −1
,
,
5
5 5
b.
2 2 4 1
, ,
5
5 5
c.
− 2
, −2 , −2
10
10
10
correct choice
d.
2
,
10
2 ,
10
2
10
e.
− 2
,
10
2 ,
10
2
10
⃗ ρ = (−2x, −2y, −1)
∇
⃗ρ
∇
2 , 2, 4
= −2 2 , −4, −1
⃗ ρ 2 , 2, 4
To decrease the density, the airplane should travel in the direction − ∇
2 2 4 1
⃗ ρ = 8 + 16 + 1 = 5, the unit vector direction is
Since −∇
, ,
.
5
5 5
= 2 2 , 4, 1 .
1
3. Find an equation of the plane tangent to the graph of the function z = x 2 y + xy 2
at the point (2, 1).
a. z = 5x + 8y + 6
b. z = −5x − 8y + 6
c. z = −5x − 8y + 24
d. z = 5x + 8y − 12
correct choice
e. z = 5x + 8y − 6
f(x, y) = x 2 y + xy 2
f x (x, y) = 2xy + y 2
f y (x, y) = x 2 + 2xy
f(2, 1) = 6
f x (2, 1) = 5
f y (2, 1) = 8
z = f(2, 1) + f x (2, 1)(x − 2) + f y (2, 1)(y − 1) = 6 + 5(x − 2) + 8(y − 1) = 5x + 8y − 12
4. Find an equation of the plane tangent to the level surface x 2 y 2 + x 2 z 2 + y 2 z 2 = 49
at the point (1, 2, 3).
a. 13x + 20y + 15z = 98
correct choice
b. 13x + 10y + 5z = 48
c. 13x − 10y + 5z = 8
d. 13x − 20y + 15z = 18
e. 39x + 20y + 5z = 94
P = (1, 2, 3)
f = x2y2 + x2z2 + y2z2
⃗ =∇
⃗ f = (2xy 2 + 2xz 2 , 2yx 2 + 2yz 2 , 2zx 2 + 2zy 2 )
⃗ f(P) = (26, 40, 30)
∇
N
⃗⋅X = N
⃗⋅P
N
26x + 40y + 30z = 26 ⋅ 1 + 40 ⋅ 2 + 30 ⋅ 3 = 196
13x + 20y + 15z = 98
⃗ (r, θ) = (r cos θ, r sin θ, θ)
5. Find the equation of the plane tangent to the parametric surface R
at the point where (r, θ) = (2, π).
a. −x = −2y + z
b. −x + 2y − z = −2 + π
c. −x − 2y − z = −2 + π
d. −y + 2z = 2π
e. y + 2z = 2π
correct choice
⃗ (2, π) = (2 cos π, 2 sin π, π) = (−2, 0, π)
P=R
⃗e r = ( cos θ, sin θ, 0)
⃗e r (2, π) = ( cos π, sin π, 0) = (−1, 0, 0)
⃗e θ = (−r sin θ, r cos θ, 1)
⃗e θ (2, π) = (−2 sin π, 2 cos π, 1) = (0, −2, 1)
⃗ = ⃗e r × ⃗e θ =
N
î
̂
k̂
−1
0
0
0
−2 1
= î(0) − ̂ (−1) + k̂ (2) = (0, 1, 2)
⃗⋅X = N
⃗⋅P
N
y + 2z = 2π
2
6. A satellite is travelling from East to West directly above the equator. In what direction does the
binormal B̂ point?
a. North
b. South
correct choice
c. Up
d. Down
e. West
T̂ points West, N̂ points Down. So B̂ = T̂ × N̂ points South by the right hand rule.
7. Compute
Q
∫ P 2x dx + 2y dy + 2z dz
along the straight line from P = (1, −2, 2) to Q = (3, −4, 12).
HINT: Use the Fundamental Theorem of Calculus for Curves.
a. −10
b.
10
c. 10
d. 108
e. 160
correct choice
⃗ = (2x, 2y, 2z). Then F
⃗ =∇
⃗ f where f = x 2 + y 2 + z 2 . So by the FTCC,
Let F
∫ P 2x dx + 2y dy + 2z dz = ∫ P F⃗ ds⃗ = ∫ P ∇⃗ f ds⃗ = f(Q) − f(P) = (9 + 16 + 144) − (1 + 4 + 4) = 160
Q
8. Compute
Q
∮ 2x dx + 2xy dy
Q
counterclockwise around the boundary of the rectangle 2 ≤ x ≤ 4,
1 ≤ y ≤ 4.
HINT: Use Green’s Theorem.
a. 6
b. 18
c. 30
correct choice
d. 36
e. 72
By Green’s Theorem:
∮ 2x dx + 2xy dy = ∫ 1 ∫ 2 ∂ x (2xy) − ∂ y (2x) dx dy = ∫ 1 ∫ 2 2y dx dy = ∫ 2 1 dx ∫ 1 2y dy = 2
4
4
4
4
4
4
y2
4
1
= 30
3
9.
∫∫ F⃗ ⋅ dS⃗
Compute
over the complete boundary of the solid
∂V
above the paraboloid z = x 2 + y 2 and below the plane z = 4
⃗ = (xy 2 , yx 2 , z 2 ).
with outward normal, for the vector field F
HINT: Use Gauss’ Theorem.
a. − 128 π
3
b. 40π
c. 72π
d.
160 π
3
e.
896 π
15
correct choice
⃗ = y 2 + x 2 + 2z = r 2 + 2z in cylindrical coordinates.
⃗ ⋅F
∇
⃗=
⃗ ⋅ dS
⃗ dV = 2π 2 4 (r 2 + 2z)r dz dr dθ = 2π
⃗ ⋅F
F
∇
∫∫
∫0 ∫0 ∫r
∫∫∫
∂V
V
6
= 2π r 4 + 8r 2 − r
3
10.
2
0
2
= 2π 16 + 32 − 64
3
2
∫0
r3z + z2r
= 32π 3 − 4
3
4
z=r
2
dr = 2π ∫ (4r 3 + 16r) − (2r 5 ) dr
2
0
= 5 ⋅ 32π = 160 π
3
3
Find the area of one petal of the 4 leaf rose r = sin(4θ).
The petal in the first quadrant.is shown.
a.
π
16
b.
π
8
c.
π
4
d.
π
2
correct choice
e. π
r = sin(4θ) = 0
A=
∫∫ 1 dA =
= 1
2
π/4
∫0
at
π/4
sin(4θ)
∫0 ∫0
4θ = π
r dr dθ =
θ= π
4
or
π/4
∫0
r2
2
sin(4θ)
0
1 − cos(8θ)
sin(8θ)
dθ = 1 θ −
2
4
8
dθ = 1
2
π/4
0
π/4
∫0
sin 2 (4θ) dθ
= π
16
4
Work Out: (Points indicated. Part credit possible. Show all work.)
11. (15 points) Find the point in the first octant on the graph of 4x 4 y 2 z = 1 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 = 4x 4 y 2 z = 1.
Method 1: Lagrange multipliers:
⃗ f = (2x, 2y, 2z)
∇
⃗ f = λ∇
⃗g
∇
λ=

⃗ g = (16x 3 y 2 z, 8x 4 yz, 4x 4 y 2 )
∇
2x = λ16x 3 y 2 z,
2y = λ8x 4 yz,
2y
2x
=
= 2z
3 2
16x y z
8x 4 yz
4x 4 y 2

1 = 4x 4 y 2 z = 4(4z 2 ) 2 (2z 2 )z = 128z 7

2z = λ4x 4 y 2
x 2 = 4z 2 ,
y 2 = 2z 2

z= 1
2
(x, y, z) =
1,
x = 2z,
y=
2z
2 1
,
2 2
Method 2: Eliminate a variable:
z=
1
4x 4 y 2
f x = 2x −
equate:

y=
f = x2 + y2 +
1
=0
2x 9 y 4
x 2 = 2y 2
1
2
1
16x 8 y 4
f y = 2y −

(x, y, z) =
x=
1
=0
4x 8 y 5
2y

4x 10 y 4 = 1
substitute back:
8x 8 y 6 = 1
4(2y 2 ) 5 y 4 = 1

y 14 = 17
2
1, 1 , 1
2 2
5
12.
∫∫ ∇⃗ × F⃗ ⋅ dS⃗ = ∮ F⃗ ⋅ ds⃗
(25 points) Verify Stokes’ Theorem
∂C
C
for the cone C given by z 2 = x 2 + y 2 for z ≤ 2
⃗ = (yz 2 , −xz 2 , z 3 ).
oriented down and out, and the vector field F
Be sure to check and explain the orientations.
Use the following steps:
⃗ (r, θ) = (r cos θ, r sin θ, r)
a. Note: The cone may be parametrized as R
Compute the surface integral by successively finding:
⃗, ∇
⃗, ∇
⃗ R
⃗ (r, θ) ,
⃗ ×F
⃗ ×F
⃗e r , ⃗e θ , N
∫∫ ∇⃗ × F⃗ ⋅ dS⃗
C
k̂
̂
î
⃗e r =
(cos θ
sin θ,
⃗e θ = (−r sin θ, r cos θ,
1)
0)
⃗ = ⃗e r × ⃗e θ = 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 N
⃗ =
⃗ ×F
∇
k̂
∂
∂z
̂
∂
∂y
î
∂
∂x
yz 2 −xz 2
⃗ R
⃗ (r, θ)
⃗ ×F
∇
= î(0 − −2xz) − ̂ (0 − 2yz) + k̂ (−z 2 − z 2 ) = (2xz, 2yz, −2z 2 )
z3
= (2r 2 cos θ, 2r 2 sin θ, −2r 2 )
∫∫ ∇⃗ × F⃗ ⋅ dS⃗ = ∫∫ ∇⃗ × F⃗ ⋅ N⃗ dr dθ = ∫ 0 ∫ 0 (2r 3 cos 2 θ + 2r 3 sin 2 θ + 2r 3 ) dr dθ
2π
C
2
C
=
2π
2
∫ 0 ∫ 0 (4r 3 ) dr dθ = 2π[r 4 ] 0 = 32π
2
b. Compute the line integral by parametrizing the boundary curve and successively finding:
⃗ (r⃗(θ)),
⃗r(θ), ⃗v, F
∮ F⃗ ⋅ ds⃗
∂C
⃗r(θ) = (2 cos θ, 2 sin θ, 2)
⃗v = (−2 sin θ, 2 cos θ, 0)
Reverse ⃗v = (2 sin θ, −2 cos θ, 0)
⃗ (r⃗(θ)) = (yz 2 , −xz 2 , z 3 ) = (8 sin θ, −8 cos θ, 8)
F
∮ F⃗ ⋅ ds⃗ = ∫ 0
2π
∂C
⃗ ⋅ ⃗v dθ =
F
2π
2π
∫ 0 (16 sin 2 θ + 16 cos 2 θ) dθ = ∫ 0
16 dθ = 32π
6
13. (15 points) Find the mass and center of mass of the
octant if the density is δ = x 2 + y 2 + z 2 .
M=
∫∫∫ δ dV =
π/2
π/2
∫0 ∫0
∫ 0 ρ 2 ρ 2 sin ϕ dρ dϕ dθ = π2 − cos ϕ
2
1
8
π/2
0
of the sphere x 2 + y 2 + z 2 ≤ 4 in the first
ρ5
5
2
0
= 16 π
5
By symmetry, x̄ = ȳ = z̄ . So
M xy =
∫∫∫ zδ dV =
x̄ = ȳ = z̄ =
π/2
π/2
∫0 ∫0
∫ 0 ρ cos(ϕ) ρ 2 ρ 2 sin(ϕ) dρ dϕ dθ = π2
2
sin 2 ϕ
2
π/2
0
ρ6
6
2
0
= 8π
3
M xy
= 8π 5 = 5
M
3 16π
6
7