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The final exam for math 241: vector calculus, held on may 6, 2008. The exam includes various problems on vector calculus concepts such as finding vectors perpendicular to given vectors, limits, contour diagrams, tangent planes, and the method of lagrange multipliers.
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1
lim
(x,y)→(0,0)
x
3 y
3
x
2
2
exists and determine its value. [Hint: Switch to polar coordinates].
If we set x = r cos(θ), y = r sin(θ), the condition that (x, y) → (0, 0) is that r → 0. Thus, the
limit becomes
lim
r→ 0
r
6
cos
3
(θ) sin
3
(θ)
r
2
= lim
r→ 0
r
4
cos
3
(θ) sin
3
(θ)
In the last time, the reason is that | cos
3 (θ) sin
3 (θ)| ≤ 1 and so the entire expression is bounded
by r
4 , which tends to zero.
contour diagram where ∇f (P ) = 0. Which of these are local extrema, and which are saddle
points?
y
-0.
0
-0.
0.2 0.
x
0
-0.
-0.
-0.6 -0.2 0.
There are five points P where ∇f (P ) = 0. They are (0, 0), and (± 1 / 2 , ± 1 /2) (all four choices
of ± are allowed). At these points, there are no level curves close by, and this indicates that
the function isn’t changing much near those points.
Near any local extremum the function is well-approximated by its quadratic approximation.
The level curves of the quadratic approximation will be ellipses at a local extremum, and will be
hyperbolas at a saddle point. The points (− 1 / 2 , 1 /2), (0, 0) and (1/ 2 , − 1 /2) are local extrema -
the level curves arounds these points look like ellipses. The points (− 1 / 2 , − 1 /2) and (1/ 2 , 1 /2)
will be saddle points since the level curves look like hyperbolas near here.
2
2
2 subject to
x + y + z = 1.
If f (x, y, z) = x
2
2
2
and g(x, y, z) = x + y + z, we have
∇f = λ∇g
(2x, 4 y, 6 z) = λ(1, 1 , 1).
If λ = 0, then x = y = z = 0, and this fails to satisfy x + y + z = 1. Thus, λ 6 = 0 and
λ = 2x = 4y = 6z.
Thus, x = 3z, y = 3/ 2 z and so
1 = x + y + z = 3z + (3/2)z + z = 11z/ 2.
Then, z = 2/11, y = 3/11 and x = 6/11. Thus,
x
2
2
2
=
This is the minimum value.
to explain your reasoning.
Suppose that
F is a vector field defined on an open set U ⊆ R
3
.
(a) (4 points). If
F = ∇f for some function f , then line integrals of
F do not depend on the
path, but only on the endpoints.
True.
(b) (4 points). If ∇ ×
F = 0, then line integrals of
F do not depend on the path, but only on
the endpoints.
False. This is only true under the assumption that U is simply-connected.
(c) (4 points). If U is simply-connected and ∇ ×
F = 0, then line integrals of
F do not depend
on the path, but only on the end points.
True.
(d) (4 points). If the line integrals of
F do not depend on the path, then
F = ∇f for some
function f.
True.
(e) (4 points). If U is star-shaped and ∇ ·
F = 0, then surface integrals of
F only depend on
the boundary.
True. Star-shaped implies that any closed surface can be continuously shrunk to a point in U.
The set U is then a boundary of a region R, and Gauss’s theorem says that
U
F · d
R
F ) dV = 0.
This implies that surface integrals only depend on the boundary.
∗ be the cylindrical region {(u, v, w) : u
2
2 ≤ r
2 , 0 ≤ w ≤ h} and let
D be the conical region {(x, y, z) : x
2
2 ≤
(h−z)
2
h
2
r
2 , 0 ≤ z ≤ h}. The function T : D
∗ → D
given by
T (u, v, w) = (u((h − w)/h), v((h − w)/h), w)
maps D
∗ to D. Compute the Jacobian
∂(x,y,z)
∂(u,v,w)
, and use this to show that the volume of D is
one-third the volume of D
∗
.
The Jacobian is the determinant of the derivative. The derivative is
h−w
h
u
h
h−w
h
v
h
The determinant is
(h−w)
2
h
2
The change of variables formula gives that
D
f (x, y, z) dV =
D
∗
f (x(u, v, w), y(u, v, w), z(u, v, w))
∂(x, y, z)
∂(u, v, w)
dV.
If we set f = 1, we get
volume of D =
D
∗
(h − w)
2
h
2
dV
R
h
0
(h − w)
2
h
2
dz dA
R
h
0
h
2
(h
2
− 2 hw + w
2
) dw dA
R
h −
w
2
h
w
3
3 h
2
h
0
R
[h − h + h/3] dA
R
h
dA
R
h
0
1 dz dA
D
∗
1 dV
volume of D
∗
.
F (x, y, z) = (x/ 3 , y/ 3 , z/3). Let f : [0, 1] × [0, 1] → R be a C
1 function,
and let S be the surface z = f (x, y) (with 0 ≤ x ≤ 1, 0 ≤ y ≤ 1). Show that
S
F · d
1
0
1
0
(−u
∂f
∂u
− v
∂f
∂v
Here we choose the outward pointing normal on S.
We parametrize S by
X(u, v) = (u, v, f (u, v)), 0 ≤ u ≤ 1, 0 ≤ v ≤ 1. Then,
Ts = (1, 0 ,
∂f
∂u
t
∂f
∂v
N (s, t) =
s
t
∂f
∂u
∂f
∂v
Then, we have
S
F · d
1
0
1
0
X(u, v)) ·
N (u, v) du dv
1
0
1
0
u
v
f (u, v)
∂f
∂u
∂f
∂v
du dv
1
0
1
0
−u
∂f
∂u
− v
∂f
∂v
du dv.
