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PRACTICE FINAL EXAM: SOLUTIONS
- Complete the following problems. You may use any result from class you like, but if you cite a theorem
be sure to verify the hypotheses are satisfied.
- In order to receive full credit, please show all of your work and justify your answers. You do not need
to simplify your answers unless specifically instructed to do so.
- If you need extra room, use the back sides of each page. If you must use extra paper, make sure to
write your name on it and attach it to this exam. Do not unstaple or detach pages from this exam.
- Please circle one of the following:
- Keep Both Midterms, and keep my Final Exam Worth 40% of my overall grade.
- Drop Only Midterm 1, and make my Final Exam Worth 60% of my overall grade.
- Drop Only Midterm 2, and make my Final Exam Worth 60% of my overall grade.
- Drop Both Midterms, and make my Final Exam Worth 80% of my overall grade.
(If no selection is made, I will go with Choice 1)
- Please sign the following:
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The following boxes are strictly for grading purposes. Please do not mark.
1 8 pts
2 6 pts
3 7 pts
4 8 pts
5 8 pts
6 13 pts
7 10 pts
Total 60 pts
(1) Consider the curve in R
3 parameterized by
r(t) = 〈 3 t, 2 cos(2t), 2 sin(2t)〉.
(a) (2 pts) Compute T(t), the unit tangent vector to the curve at time t.
Solution: We know T =
r ′
‖r′‖
. Since r
′ (t) = 〈 3 , −4 sin(2t), 4 cos(2t)〉, we have
‖r
′ ‖ =
2
- (−4 sin(2t)) 2
- (4 cos(2t)) 2
9 + 16 sin
2 (2t) + 16 cos 2 (2t)
Thus,
T(t) =
sin(2t),
cos(2t)
(b) (2 pts) Compute N(t), the principal unit normal vector to the curve at time t.
Solution: We know N =
T ′
‖T′‖
. Since T ′ (t) =
8 5
cos(2t), −
8 5
sin(2t)
, we have
‖T
′ ‖ =
2
cos(2t)
sin(2t)
cos 2 (2t) +
sin
2 (2t)
Thus,
N(t) = 〈 0 , − cos(2t), − sin(2t)〉.
(2) Suppose the temperature in a room is given by the function T (x, y, z) = x
2 − 2 yz + xz
3 .
(a) (2 pts) Compute ∇T (x, y, z).
Solution: We have
∇T =
∂T
∂x
∂T
∂y
∂T
∂z
= 〈 2 x + z
3 , − 2 z, − 2 y + 3xz
2 〉.
(b) (2 pts) A fly is hovering in the air at the point (1, 1 , 0). At that point, in which unit direction does
the temperature increase most rapidly?
Solution: The gradient always points in the direction of most rapid increase. By part (a), we have
∇T (1, 1 , 0) = 〈 2 , 0 , − 2 〉. The unit vector pointing in this direction is
1 √ 2
1 √ 2
(c) (2 pts) What is the rate of temperature increase in the unit direction pointing toward (5, 1 , 3)?
Solution: The vector from the point (1, 1 , 0) to the point (5, 1 , 3) is (4, 0 , 3). The unit vector in this
direction is u =
4 5
3 5
. The directional derivative of T in the direction u at the point (1, 1 , 0) is
DuT = ∇T (1, 1 , 0) · u = 〈 2 , 0 , − 2 〉 ·
(3) Consider the function f (x, y) = x
3 − 3 x
2
2 − 8 y + 2.
(a) (3 pts) Find all the stationary points of f ; i.e., the points where ∇f = 0.
Solution: We have ∇f = 〈 3 x
2 − 6 x, 2 y − 8 〉, and so ∇f = 0 exactly when
3 x
2 − 6 x = 0
2 y − 8 = 0.
The first equation implies x = 0 or x = 2, while the second equation implies y = 4. So, the
stationary points are (0, 4), (2, 4). §
(b) (4 pts) Use the Second Derivative Test to determine the nature of each stationary point.
Solution: We compute the Hessian matrix of f :
Hf (x, y) =
[
6 x − 6 0
0 2
]
So,
Hf (0, 4) =
[
]
which gives the sequence of principal minors {− 6 , − 12 }. Thus, (0, 4) is a saddle point.
On the other hand, for the point (2, 4) we have
Hf (2, 4) =
[
]
which gives the sequence of principal minors { 6 , 12 }. Thus, (2, 4) is a local minimum. §
(5) Let S be the solid square in R
2 with vertices (0, 0), (0, 2), (2, 0), (2, 2). Let F(x, y) = 〈x
2 y, xy
2 〉.
(a) (4 pts) Compute the flux of F through ∂S; i.e., compute
∂S
F · n ds.
(Hint: Use Green’s Theorem.)
Solution: By Green’s Theorem, ∮
∂S
F · n ds =
S
div F dA
S
∂x
(x
2 y) +
∂y
(xy
2 ) dA
S
2 xy + 2xy dA
2
0
2
0
xy dx dy
0
[
x
2 y
]
x=
x=
dy
2
0
2 y dy
= 4[y
2 ]
2 0
(b) (4 pts) Compute the circulation of F around ∂S; i.e., compute
∂S F · T ds.
(Hint: Use Green’s Theorem.)
Solution: By Green’s Theorem, ∮
∂S
F · T ds =
S
∂x
(xy
2 ) −
∂y
(x
2 y) dA
0
0
y
2 − x
2 dx dy
2
0
[
xy
2 −
x
3
]x=
x=
dy
0
2 y
2 −
dy
[
y
3 −
y
] 2
0
= 0.
(6) Let G be the paraboloid z = x
2
2 with circle x
2
2 = 1, z = 1 as its boundary. Let F(x, y, z) =
〈y, −x, yz〉.
(a) (3 pts) Compute
∂G
F · T ds directly (without appealing to Stokes’s theorem).
Solution: Parameterize the boundary circle by
x = cos(t)
y = sin(t)
z = 1,
where t ranges from t = 0 to t = 2π. Then ∮
∂G
F · T ds =
∂G
y dx − x dy + yz dz
∫ (^2) π
0
(sin(t))(− sin(t) dt) − (cos(t))(cos(t) dt) + (sin(t))(1)(0 dt)
2 π
0
(sin
2 (t) + cos
2 (t)) dt
∫ (^2) π
0
1 dt
= − 2 π.
(b) (2 pts) Find n, the upward unit normal to the surface.
Solution: The surface G is given by the equation −x 2 − y 2
- z = 0. The gradient of this equation
is 〈− 2 x, − 2 y, 1 〉, which gives a normal vector to the surface (and this vector points upward). The
upward unit normal is therefore
n =
4 x 2
〈− 2 x, − 2 y, 1 〉.
(e) (3 pts) Compute
G (curl F) · n dS directly (without appealing to Stokes’s theorem).
Solution: Combining our results from parts (b)-(d), above, we have ∫ ∫
G
(curl F) · n dS =
G
〈z, 0 , − 2 〉 ·
4 x 2
〈− 2 x, − 2 y, 1 〉 dS
S
− 2 xz − 2 √ 4 x 2
dS
1
0
2 π
0
− 2 u
3 cos(v) − 2 √ 4 u 2
· u
4 u 2
0
∫ (^2) π
0
(− 2 u
4 cos(v) − 2 u) dv du
0
[− 2 u
4 sin(v) − 2 uv]
v=2π v= du
= − 4 π
1
0
u du
= − 2 π.
Note that this agrees with the answer from part (a), as predicted by Stokes’s theorem. §
(7) (10 pts) Let B be the parabolic solid 0 ≤ z ≤ 4 − x
2 − y
2
. Let F(x, y, z) = 〈x
2 , y
2 , z
2 〉.
Compute the flux of F through ∂B; i.e, compute
∂B
F · n dS.
(Hint: First use Gauss’s Divergence Theorem, and then use cylindrical coordinates, integrating θ first.)
Solution: By Gauss’s Divergence Theorem, ∫ ∫
∂B
F · n dS =
B
div F dV
B
∂x
(x
2 ) +
∂y
(y
2 ) +
∂z
(z
2 ) dV
B
x + y + z dV.
Using cylindrical coordinates, we then have
B
x + y + z dV = 2
2
0
4 −r 2
0
2 π
0
(r cos(θ) + r sin(θ) + z) · r dθ dz dr
0
∫ (^4) −r^2
0
r
2
∫ (^2) π
0
cos(θ) dθ + r
2
∫ (^2) π
0
sin(θ) dθ + rz
∫ (^2) π
0
dθ
0
∫ (^4) −r^2
0
2 πrz dz dr
2
0
[πrz
2 ]
z=4−r 2
z=0 dr
= 2π
0
r(4 − r
2 )
2 dr.
For this final integral, we can either expand the square and compute the integral directly, or we can
make the substitution u = 4 − r
2 , which gives
= −π
4
u
2 du
= −π
[
u
3
] 0
4
= −π
64 π
