Calculus III
Test 2 Solutions
October 26, 2009
1. Let C be the curve r(t) = het sin(πt) , t − t5 i for 0 ≤ t ≤ 1. Verify that the integral
2
of F(x, y) = h2e2x sin y − ex , e2x cos y + ln(y 2 + 1)i along C is zero by showing that F is
conservative and C is closed.
First note that F is defined on the entire plane, which is simply connected.
Thus to see if it is conservative, it suffices to compare the partials.
For this we have
∂F1
∂F2
= 2e2x cos y =
.
∂y
∂x
To see that the curve is closed we check
r(0) = h1, 0i = r(1).
2. (a) Find a potential function for F(x, y) = h2ey cos(2x) + 2xy, ey sin(2x) + x2 i.
Assuming F is conservative, there is some function f (x, y) so that
fx = 2ey cos(2x) + 2xy
fy = ey sin(2x) + x2 .
From the first equation, it follows that f has the form
f (x, y) = ey sin(2x) + x2 y + g(y).
It follows that
fy = ey sin(2x) + x2 + g ′(y).
Comparing this to the second component of F, we see that g ′ (y) = 0,
so that g(y) = C . Thus the most general potential function is
f (x, y) = ey sin(2x) + x2 y + C.
2
(b) Let C be the curve r(t) = ht cos2 t, et
sin t
i for 0 ≤ t ≤ π , and let F be
Z the vector field
from part (a) above. Use your answer from part (a) to find the value of
F · dr without
directly calculating the integral.
C
We first note that r(0) = h0, 1i and r(π) = hπ, 1i. Then the fundamental
theorem of line integrals says that
Z
F · dr = f (π, 1) − f (0, 1) = π 2 .
C
3. Consider the function f (x, y) = x2 + y 2 + kxy (where k is some unknown constant).
(a) Show that (0, 0) is always a critical point for f , regardless of the value of k .
We compute
fx (0, 0) = (2x + ky)|(0,0) = 0
and
fy (0, 0) = (2y + kx)|(0,0) = 0,
so that both first partials are zero at (0, 0).
point, regardless of k .
Thus (0, 0) is a critical
(b) Classify this critical point for all possible values of k , being sure to indicate for which k
the second derivative test fails.
We compute
fxx (0, 0) = 2
fyy (0, 0) = 2
fxy (0, 0) = k
so that D = fxx fyy − [fxy ]2 = 4 − k 2 . The second derivative test fails
when this is zero, which is when k = ±2. The critical point is a
saddle when D < 0, which occurs when k > 2 or k < −2. The critical
point is a min (since fxx (0, 0) > 0) when D > 0, which occurs when
−2 < k < 2.
4. Write an explicit system of equations that can be used to optimize f (x, y) = x2 y subject
to the constraint x2 + y 3 = 4. Do not solve the system.
Note that ∇f = h2xy, x2i and ∇g = h2x, 3y 2i, so the system is


2xy = λ2x
x2 = λ3y 2

 2
x + y3 = 4
5. Suppose the temperature in the room is given by T (x, y, z) = 3x2 − 4xy + 5yz 2 + xyz .
(a) Find the equation for the plane tangent to the level surface T (x, y, z) = 5 at the point
(0, 1, 1).
We compute the gradient
∇T (0, 1, 1) = h6x − 4y + yz, −4x + 5z 2 + xz, 10yz + xyi(0,1,1) = h−3, 5, 10i.
The plane is then given by
−3(x − 0) + 5(y − 1) + 10(z − 1) = 0.
(b) Find the direction in which one should travel from (0, 1, 1) in order to decrease temperature as fast as possible.
5
4
3
2
1
-5
-4
-3
-2
-1
0
1
2
3
4
5
-1
-2
-3
-4
-5
This is the direction of the negative gradient, which is h3, −5, −10i.
(c) What is the rate of change of temperature at (0, 1, 1) in the direction of v = h1, 1, 1i?
We compute
Dv T (0, 1, 1) = ∇T (0, 1, 1) ·
√
v
1
12
= h−3, 5, 10i · √ h1, 1, 1i = √ = 4 3.
kv k
3
3
6. The following is a contour map for a function f (x, y).
(a) Estimate the value of f at the point (2, −1).
This point lies on the contour curve labeled −2, so f (2, −1) = −2.
(b) Is fx (2, −1) positive, negative, or approximately zero?
Starting at (2, −1) and moving rightward (in the positive x direction),
the next contour curve encountered is labeled −3, so we are decreasing.
Thus fx (2, −1) is negative.
(c) What about fy (2, −1)?
Doing the same as above but moving up (in the positive y direction),
the next contour curve is labeled −1, so we’re increasing. Thus fy (2, −1)
is positive.
(d) Sketch a vector u at (2, −1) with the property that Du (2, −1) = 0.
Any vector tangent to the contour curve at (2, −1) will do.
(e) Which is larger (in absolute value), fy (3, 1) or fy (1, 3)?
Starting at the two points and moving up, it is a longer way to the
next curve when you start at (1, 3) than when you start at (3, 1). This
means that fy (3, 1) is larger (because you’re going up faster).
(f) Sketch the gradient vector for the function at the points (1, 3), (3, 1), and (−1, −1).
Make sure each vector points in the correct direction and that their lengths compare appropriately.
The vectors should be perpendicular to the curves on which they sit.
They all point roughly away from the origin. The one at (−1, −1) is
shorter than the other two.
7. Suppose C is a cylinder that measures L feet along a side and has a circular base of
radius r (see the board). If the cylinder is sheared at an angle of θ from vertical, then its
volume is given by
V = πr 2 L cos θ.
Now suppose the dimensions of C are changing in time, so that r , L, and θ are all functions
of t.
dV
(a) Write out the chain rule for
.
dt
∂V dr ∂V dL ∂V dθ
dV
=
+
+
dt
∂r dt
∂L dt
∂θ dt
(b) Compute
∂V ∂V
∂V
,
, and
.
∂r ∂L
∂θ
∂V
= 2πrL cos θ
∂r
∂V
= πr 2 cos θ
∂L
∂V
= −πr 2 L sin θ
∂θ
(c) When t = 0, you measure that r = 3 in., L = 10 in., and θ = π/6 rad.
You also compute that when t = 0, r is increasing at a rate of
1
a rate of 34 in /sec, and θ is increasing at a rate of 15
rad/sec.
1
5
in/sec, L is decreasing at
Use your answers to parts (a) and (b) to compute the rate at which V is changing when
t = 0. Is the volume of the cylinder increasing or decreasing when t = 0?
From the answers above we have
dV
dr
dL
dθ
= (2πrL cos θ) + (πr 2 cos θ)
+ (−πr 2 L sin θ) .
dt
dt
dt
dt
Plugging in the values given for x, y , and θ , as well as the derivatives
we’re told in this part, we find
dV
(3, 10, π/6) =
dt
√ √ 4
1 1
3 1
3
2π(3)(10)
+ π(9)
−
+ (−π)(9)(10)
2
5
2
3
2 15
√
√
= 6π 3 − 6π 3 − 3π = −3π.
This is negative, so the area is decreasing when t = 0.