Name
ID
MATH 251
Final Exam
Sections 509
Spring 2007
Solutions
P. Yasskin
1-10
/60 12
/15
11
/20 13
/15
Total
Multiple Choice: (6 points each. No part credit.)
/110
1. Consider the triangle with vertices A = 2, 4, B = 1, 1 and C = 0, 3.
Find the angle at B.
a. 30°
Correct Choice
b. 45°
c. 60°
d. 120°
e. 135°
BA = A − B = 1, 3
BA =
cos θ =
BC = C − B = −1, 2
12 + 32 =
BA ⋅ BC
BA BC
BC =
10
=
5
5 10
=
12 + 22 =
1
2
2. For the "twisted cubic" curve ⃗
rt =
BA ⋅ BC = −1 + 6 = 5
5
θ = 45°
t, t 2 , 2 t 3
3
find the tangential acceleration a T = T̂ ⋅ ⃗
a.
a. 4t + 8t 2
b.
1
4t + 8t 2
c. 4t
Correct Choice
4t
1 + 2t 2
1
e.
1 + 2t 2
d.
⃗v = 1, 2t, 2t 2 
⃗
a = 0, 2, 4t
OR
|v⃗| =
1 + 4t 2 + 4t 4 = 1 + 2t 2
a T = T̂ ⋅ ⃗
a=
1 1, 2t, 2t 2 
T̂ = ⃗v =
|v⃗|
1 + 2t 2
1 1, 2t, 2t 2  ⋅ 0, 2, 4t =
1 4t + 8t 2  = 4t
1 + 2t 2
1 + 2t 2
a T = d |v⃗| = d 1 + 2t 2  = 4t
dt
dt
1
3. Find the linear approximation to fx, y = x 2 + 4y 3 + 1 at x, y = 2, 1.
Use it to estimate f2. 1, 1. 1.
a. 19. 6
b. 19. 2
Correct Choice
c. 16. 2
d. 12. 8
e. 5. 87
f2, 1 = 16
fx, y = x 2 + 4y 3 + 1
3
f x x, y = 2xy + 1
f x 2, 1 = 8
2
2
f y x, y = x + 43y
f y 2, 1 = 24
f tan x, y = f2, 1 + f x 2, 1x − 2 + f y 2, 1y − 1 = 16 + 8x − 2 + 24y − 1
f tan 2. 1, 1. 1 = 16 + 82. 1 − 2 + 241. 1 − 1 = 16 +. 8 + 2. 4 = 19. 2
4. Find the equation of the plane tangent to the surface 3x sin z + y cos z = 4 at x, y, z =
2, 2, π .
4
What is the z-intercept?
a. π
2
b. 8 + π
c. 4 + π
2
π
d. 2 +
4
π
e. 1 +
8
Correct Choice
fx, y, z = 3x sin z + y cos z
⃗
∇f = 3 sin z, cos z, 3x cos z − y sin z
P=
2, 2, π
4
⃗ =⃗
N
∇f =
3 , 1 ,2
2
2
π
3
1
3
1
⃗
⃗
N⋅X = N⋅P
x+
y + 2z =
2 +
2 +2
= 4+ π
4
2
2
2
2
2
π
π
z-intercept:
x=0
y=0
2z = 4 +
z = 2+
2
4
P
2
5. Find the arc length of the curve ⃗
rt = 2t 2 , t 3  between t = 0 and t = 1.
a. 31
b.
c.
d.
e.
27
61
27
91
27
31
9
61
9
⃗v = 4t, 3t 2 
L=
6.
Correct Choice
|v⃗| =
16t 2 + 9t 4 = t 16 + 9t 2
∫ ds = ∫|v⃗| dt = ∫ 0 t
1
3/2
16 + 9t 2 dt = 1 16 + 9t 2 
27
= 1 125 − 64 = 61
27
27
1
0
= 1 25 3/2 − 16 3/2 
27
Find the mass of the region inside the
upper half of the limaçon r = 2 − cos θ
if the surface density is ρ = y.
a.
5
3
b.
10
3
c.
13
3
d.
15
3
e.
20
3
M=
Correct Choice
π
2−cos θ
∫∫ ρ dA = ∫∫ y dA = ∫ 0 ∫ 0
r sin θ r dr dθ =
π
∫0
r3
3
2−cos θ
r=0
sin θ dθ = 1
3
π
∫ 0 2 − cos θ 3 sin θ dθ
u = 2 − cos θ
du = sin θ dθ
3
3
4
M = 1 ∫ u 3 du = u
= 81 − 1 = 20
3 1
12 1
12
3
3
Find the mass of the solid between the
7.
cones z = r and z = 6 − r
if the volume mass density is ρ = z.
Correct Choice
a. 54π
b. 108π
c. 216π
d.
297 π
8
e.
297 π
4
M=
6−r
∫∫∫ ρ dV = ∫∫ z dV = ∫ 0 ∫ 0 ∫ r z r dz dr dθ = 2π ∫ 0 z2 z=r r dr = π ∫ 0 6 − r 2 − r 2
3
3
= π ∫ 36 − 12r r dr = π ∫ 36r − 12r 2  dr = π18r 2 − 4r 3  30 = π162 − 108 = 54π
0
0
2π
∫ F⃗ ⋅ ds⃗
8. Compute
3
6−r
3
2
3
r dr
⃗ = 2x + y + z, 2y + x + z, 2z + x + y along the curve
for F
⃗r
⃗rt = t cos t, t sin t, t e t/π  between t = 0 and t = π.
HINT: Find a scalar potential and use the Fundamental Theorem of Calculus for Curves.
a. π 2 1 + e 2 − e
Correct Choice
b. π 2 1 + e 2 − 2e
c. π 2 1 + e 2 + e
d. π 2 1 + e 2 + 2e
e. π 2 1 + e 2 
⃗ =⃗
F
∇f for f = x 2 + y 2 + z 2 + xy + xz + yz
A = ⃗r0 = 0, 0, 0
B = ⃗rπ = −π, 0, πe
∫ F⃗ ⋅ ds⃗ = ∫ ⃗∇f ⋅ ds⃗ = fB − fA = π 2 + π 2 e 2 − π 2 e
⃗r
⃗r
4
9. Compute
∮ F⃗ ⋅ ds⃗
⃗ = −x 2 y + x 3 − y 3 , xy 2 + x 3 − y 3  counterclockwise around the circle x 2 + y 2 = 9.
for F
C
HINT: Use Green’s Theorem.
a. 36π
b. 72π
c. 144π
Correct Choice
d. 162π
e. 324π
P = −x 2 y + x 3 − y 3
Q = xy 2 + x 3 − y 3
∂Q
− ∂P = 4x 2 + 4y 2 = 4r 2
∂x
∂y
∂Q
− ∂P dA =
∂x
∂y
∮ F⃗ ⋅ ds⃗ = ∫∫
C
10. Compute
∫∫ F⃗ ⋅ dS⃗
∂Q
= y 2 + 3x 2
∂x
∂P = −x 2 − 3y 2
∂y
∫ 0 ∫ 0 4r 2 r dr dθ = 2πr 4  30 = 162π
2π
3
over the complete boundary of the cylinder x 2 + y 2 ≤ 4 for 0 ≤ z ≤ 5
∂C
⃗ = x 3 + xy 2 , x 2 y + y 3 , x 2 z + y 2 z.
for the vector field F
HINT: Use Gauss’ Theorem.
a. 80 π
3
b. 40π
c. 100π
d. 400 π
3
e. 200π
Correct Choice
⃗ = 3x 2 + y 2  + x 2 + 3y 2  + x 2 + y 2  = 5x 2 + 5y 2
∇⋅F
∫∫ F⃗ ⋅ dS⃗ = ∫∫∫ ∇ ⋅ F⃗ dV = ∫ 0 ∫ 0 ∫ 0 5r 2  r dr dθ dz = 252π
5
∂C
2π
2
r4
4
2
0
= 200π
5
Work Out: (Points indicated. Part credit possible. Show all work.)
11.
⃗=
⃗ ⋅ dS
20 points Verify Stokes’ Theorem ∫∫ ⃗
∇×F
C
∮ F⃗ ⋅ ds⃗
∂C
⃗ = yz 2 , −xz 2 , z 3  and the
for the vector field F
cylinder x 2 + y 2 = 9 for 1 ≤ z ≤ 2 oriented out.
Be sure to check and explain the orientations.
Use the following steps:
⃗ θ, z = 3 cos θ, 3 sin θ, z.
a. The cylindrical surface may be parametrized by R
Compute the surface integral:
⃗ , check orientation, ⃗
⃗, ⃗
⃗ R
⃗ θ, z ,
Successively find: ⃗e θ , ⃗e z , N
∇×F
∇×F
∫∫ ⃗∇ × F⃗ ⋅ dS⃗
C
k̂
̂
î
⃗e θ = −3 sin θ, 3 cos θ, 0
⃗e z =
 0,
0,
1
⃗ = ⃗e θ × ⃗e z = î3 cos θ − ̂ −3 sin θ + k̂ 0 = 3 cos θ, 3 sin θ, 0
N
⃗ has the correct orientation.
N
⃗
⃗ =
∇×F
î
∂
∂x
̂
∂
∂y
yz 2 −xz 2
⃗
⃗ R
⃗ θ, z
∇×F
k̂
∂
∂z
= î0 − −2xz − ̂ 0 − 2yz + k̂ −z 2 − z 2  = 2xz, 2yz, −2z 2 
z3
= 6z cos θ, 6z sin θ, −2z 2 
⃗
⃗⋅N
⃗ = 18z cos 2 θ + 18z sin 2 θ = 18z
∇×F
∫∫ ⃗∇ × F⃗ ⋅ dS⃗ = ∫∫ ⃗∇ × F⃗ ⋅ N⃗ dθ dz = ∫ 1 ∫ 0
2
C
2π
18z dθ dz = 2π9z 2  2z=1 = 54π
C
6
b. Let U be the upper circle. Parametrize U and compute the line integral.
⃗ r⃗θ,
Successively find: ⃗rθ, ⃗vθ, check orientation, F
∮ F⃗ ⋅ ds⃗.
U
⃗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 clockwise but ⃗v points counterclockwise.
So reverse ⃗v:
⃗vθ = 3 sin θ, −3 cos θ, 0
⃗ r⃗θ = yz 2 , −xz 2 , z 3  = 12 sin θ, −12 cos θ, 8
F
∮ F⃗ ⋅ ds⃗ = ∫ 0
2π
⃗ ⋅ ⃗v dθ =
F
∂C
∫0
2π
36 sin 2 θ + 36 cos 2 θ dθ =
∫0
2π
36 dθ = 72π
c. Let L be the lower circle. Parametrize L and compute the line integral.
⃗ r⃗θ,
Successively find: ⃗rθ, ⃗vθ, check orientation, F
∮ F⃗ ⋅ ds⃗.
L
⃗rθ = 3 cos θ, 3 sin θ, 1
⃗vθ = −3 sin θ, 3 cos θ, 0
By the right hand rule the lower curve must be traversed counterclockwise and ⃗v is
counterclockwise.
⃗ r⃗θ = yz 2 , −xz 2 , z 3  = 3 sin θ, −3 cos θ, 1
F
∮ F⃗ ⋅ ds⃗ = ∫ 0
2π
⃗ ⋅ ⃗v dθ =
F
∂C
d. Combine
∮ F⃗ ⋅ ds⃗
and
U
∫0
2π
−9 sin 2 θ − 9 cos 2 θ dθ = − ∫
∮ F⃗ ⋅ ds⃗
L
to get
2π
0
9 dθ = −18π
∮ F⃗ ⋅ ds⃗.
∂C
∮ F⃗ ⋅ ds⃗ = ∮ F⃗ ⋅ ds⃗ + ∮ F⃗ ⋅ ds⃗ = 72π − 18π = 54π
∂C
U
L
which agrees with part (a).
7
12. (15 points) Find 3 numbers a, b and c whose sum is 12 for which ab + 2ac + 3bc is a
maximum.
Maximize f = ab + 2ac + 3bc subject to the constraint a + b + c = 12.
c = 12 − a − b
Solve the constraint:
Substitute into the function:
f = ab + 2a12 − a − b + 3b12 − a − b = 24a + 36b − 2a 2 − 4ab − 3b 2
Set the partial derivatives equal to zero and solve:
f a = 24 − 4a − 4b = 0
f b = 36 − 4a − 6b = 0
4a + 4b = 24
4a + 6b = 36
2b = 12
a+b = 6
b=6
c = 12 − a − b = 12 − 0 − 6 = 6
a = 6−b = 0
a = 0, b = 6, c = 6
13. (15 points) Find the mass and center of mass of the conical surface z =
x2 + y2
for z ≤ 2
⃗ r, θ = r cos θ, r sin θ, r.
with density ρ = x 2 + y 2 . The cone may be parametrize as R
k̂
̂
î
⃗e r = cos θ,
sin θ, 1
⃗e θ = −r sin θ, r cos θ, 0
⃗ = ⃗e θ × ⃗e z = î−r cos θ − ̂ r sin θ + k̂ r cos 2 θ + r sin 2 θ = −r cos θ, −r sin θ, r
N
⃗ =
N
M=
r 2 cos 2 θ + r 2 sin 2 θ + r 2 = r 2
∫∫ ρ dS = ∫ ∫ r 2
⃗ dr dθ =
N
ρ = r2
∫0 ∫0 r3
2π
2
2 dr dθ = 2π 2
r4
4
2
0
= 8π 2
x̄ = ȳ = 0 by symmetry.
z-mom = M xy =
z̄ =
∫∫ z ρ dS = ∫ ∫ r 3
⃗ dr dθ =
N
∫0 ∫0 r4
2π
2
2 dr dθ = 2π 2
r5
5
2
0
=
64π 2
5
M xy
z-mom
64π 2
1
=
=
= 8
M
M
5
5
8π 2
8