Calculus 3
Exam # 2
October 23, 2012
Solutions
Instructions.
• SHOW ALL YOUR WORK! and write clearly.
NO WORK = NO CREDIT
• Try to write in the form of full sentences. I WILL penalize you for the failure to use full sentences. A sentence
must contain a verb. For example “5” or “x2 + y 2 are NOT sentences. “x2 + y 2 = 5” is a sentence; the symbol
“=” would be read “is equal,” and “is” is a verb.
• Do not use an equal symbol to connect quantities that are not equal, except when setting up equations.
• Use extra paper as necessary. This may be worth repeating: Use extra paper as necessary.
• Anything I don’t understand at once, I will consider wrong.
• Doing exercises 4 and 5 correctly gives you 65 points. You might want to do those first.
• There will be no partial credit for the bonus exercises.
• Try to work carefully; don’t make too many computational mistakes.
The Test.
1. (10 points) Suppose f is a differentiable function of x and y, and g(r, s) = f (2r − s, s2 − 4r). Use the table
∂g
∂g
of values given below to calculate
(1, 2) and
(1, 2).
∂r
∂s
(0, 0)
(1, 2)
f
3
6
g
6
3
fx
4
2
fy
8
5
Note: SHOW WORK! I want to see exactly how you get your answers.
Solution.
We have
∂g
(r, s)
∂r
∂g
(r, s)
∂s
∂f
∂
∂f
∂
(2r − s, s2 − 4r) (2r − s) +
(2r − s, s2 − 4r) (s2 − 4r)
∂x
∂r
∂y
∂r
∂f
∂f
= 2 (2r − s, s2 − 4r) − 4 (2r − s, s2 − 4r),
∂x
∂y
∂f
∂
∂f
∂
2
=
(2r − s, s − 4r) (2r − s) +
(2r − s, s2 − 4r) (s2 − 4r)
∂x
∂s
∂y
∂s
∂f
∂f
= − (2r − s, s2 − 4r) + 2s (2r − s, s2 − 4r).
∂x
∂y
=
When r = 1, s = 2 we have that (2r − s, s2 − 4r) = (0, 0) so that, using the tabled values:
∂g
(1, 2)
∂r
∂g
(1, 2)
∂s
∂f
∂f
(0, 0) − 4 (0, 0) = (2)(4) − (4)(8) = −24,
∂x
∂y
∂f
∂f
= − (0, 0) + 4 (0, 0) = −4 + (4)(8) = 28.
∂x
∂y
=
2
The answer is
∂g
(1, 2) = −24,
∂r
∂g
(1, 2) = 28.
∂s
2
2. (10 points) Find the equation of the tangent plane of the surface x2 + 2y 2 − 3z 2 = 3 at the point (2, −1, 1).
Solution. The plane is to be tangent to a level surface of the function f (x, y, z) = x2 + 2y 2 − 3z 2 . Its
normal is thus the gradient of f at the point in question. We have ∇f (x, y, z) = h2x, 4y, −6zi. Evaluating at
(2, −1, 1), ∇f (1, −1, 1) = h4, −4, −6i. An acceptable answer is
4(x − 2) − 4(y + 1) − 6(z − 1) = 0.
A better answer is obtained expanding and adding constant terms:
2x − 2y − 3z = 3.
3. (5 points) Find the direction in which the function f (x, y, z) = zexy increases most rapidly at the point
(0, 1, 2). What is the maximum rate of increase?
Bonus Question (5 points/ no partial credit): Is there any direction in which the rate of increase is 0?
JUSTIFY! If there is, find at least one such direction, and show it works. If there isn’t, explain why.
4. (40 points) Find all critical points and classify them as local maximum, minimum or saddle point, of the
function
f (x, y) = x4 + y 4 + 2x2 y 2 − 4x2 − 8y 2 .
Solution.
To find the critical points we compute the partial derivatives and set them to 0. We have
fx (x, y) = 4x3 + 4xy 2 − 8x = 4x(x2 + y 2 − 2),
fy (x, y) = 4y 3 + 4x2 y − 16y = 4y(x2 + y 2 − 4).
The critical points are the solutions of
4x(x2 + y 2 − 2) =
4y(x2 + y 2 − 4) =
0
0
The first equation is satisfied if x = 0 or if x2 + y 2 = 4. Suppose first x = 0. The second equation holds if
y = 0, giving one critical point (0, 0); given x = 0 it also holds if y 2 = 4 or y = ±2, giving two more critical
points (0, ±2). Supppose now x 6= 0. It maybe best now to go to the second equation;
√ it holds if y = 0 or
x2 + y 2 = 4. If y√= 0 and x 6= 0, the first equation holds if x62 = 2, thus x = ± 2. We have two more
critical points (± 2, 0). At this point we have exhausted the possibility of either x or y being 0. If x 6= 0,
y 6= 0, the equations could only hold if x2 + y 2 = 2 and x2 + y 2 = 4, which is impossible. There are no more
critical points. The critical points are:
(0, 0),
(0, −2),
(0, 2),
√
(− 2, 0),
√
( 2, 0).
Next comes the classification. The hessian determinant is
D(x, y) =
=
fxx (x, y)fyy (x, y) − (fxy (x, y))2 = (12x2 + 4y 2 − 8)(12y 2 + 4x2 − 16) − (8xy)2
48(x4 + y 4 ) + 160x2 y 2 − 224x2 − 160y 2 + 128.
We now have:
D(0, 0) = 128 > 0,
fxx (0, 0) = −8 < 0,
D(0, −2) = D(0, 2) = 256 > 0, fxx (0, ±2) = 8 > 0,
√
√
D(− 2, 0) = D( 2, 0) = −128 < 0,
(0, 0) is a local maximum.
(0, −2) and (0, 2) are local minima.
√
√
(− 2, 0) and ( 2, 0) are saddle points.
5. (25 points) Find the absolute maximum and the absolute minimum values of the function
f (x, y) = x2 + 4y 2 − 2x
on the set D = {(x, y) : x2 + y 2 ≤ 4}. As a hint, the boundary analysis is quite easy because you can easily
express y in terms of x on the boundary. It also helps to notice that −2 ≤ x ≤ 2 if (x, y) is on the boundary.
3
Solution. We first check if there are any critical points inside the disk D. For this we set the first partial
derivatives to 0:
fx (x, y) = 2x − 2 = 0
fy (x, y) = 8y = 0
The one and only solution is (1, 0) , which is in D. We have f (1, 0) = −1.
Next we have to study what happens at the boundary. At the boundary y 2 = 4 − x2 , so the function assumes
the same values as g(x) = x2 + 4(4 − x2 ) − 2x = −3x2 − 2x + 16. The range of x is −2 ≤ x ≤ 2. To determine
the maximum and minimum values of g we use some Calculus 1 methods. We find g 0 and set it to 0:
g 0 (x) = −6x − 2 = 0
for x = −1/3. Computing g at ±2 and at 1/3:
1
49
=
, g(2) = 0.
3
3
The maximum will thus be the largest value, the minimum the smallest value from the list:
g(−2) = 8,
g(−1/3) = 16 +
−1,
We see that
8,
49/3,
0.
√
49
1
35
the maximum value is
, assumed at (− , ±
),
3
3
3
the minimum value is −1, assumed at (1, 0),
6. (10 points) Find the point on the plane x + 2y + 3z = 1 that is closest to the origin.
Solution 1, using the methods of Chapter 14 We minimize the square of the distance; that is, we
minimize x2 + y 2 + z 2 . We use the equation of the plane to solve for one of the variables; it being easiest to
solve for x = 1 − 2y − 3z. We minimize
f (y, z) = (1 − 2y − 3z)2 + y 2 + z 2 = 5y 2 + 10z 2 + 12yz − 4y − 6z + 1.
Doing the usual,
fy (y, z)
fz
= 10y + 12z − 4 = 0
= 12y + 20z − 6 = 0.
The critical points are y = 1/7, z = 3/14. Then
x=1−
2
9
1
−
=
.
7 14
14
1 1 3
, , ).
14 7 14
Justifying that it actually is the minimum is not too difficult, but will not be required.
The point is (
Solution 2, using earlier methods. The point in question will have the property that the line segment
from it to the origin is perpendicular to the plane; that is, directed by the vector normal to the plane, namely
by h1, 2, 3i. An equation for a line through the origin, perpendicular to the plane is thus x = y/2 = z/3. The
point at which it intersects the plane is the one we are looking for. So, on the line we have y = 2x, z = 3x;
plugging this into the equation of the plane, x + 4x + 9x = 1, thus x = 1/14. Then y = 1/7, z = 3/14,and as
1 1 3
before we see the point is ( , , ).
14 7 14
7. Bonus Question (5 points/ no partial credit):. Let f be defined by

´
³
1

if (x, y) 6= (0, 0),
 xy sin x2 +y
2
f (x, y) =


0
if (x, y) = (0, 0),
Show that both partial derivatives of f exist at (0, 0) but f is not continuous at (0, 0).
Solution. This problem had a very interesting feature, due to a mistake on my part. Not only is the
function in question continuous at 0; it is obviously continuous by the squeeze theorem.
4
sin(x2 + 3y 3 )
must
1 + x2 + y 2
assume an absolute maximum value. That is, there must be a point (x0 , y0 ) such that f (x, y) ≤ f (x0 , y0 ) for
all values of (x, y).
p
Hint: | sin z| ≤ 1 for all z. There are in addition many points where f (x, y) > 0; for example if x = π/2,
y = 0, then f (x, y) = 1/(1 + (π/2)) > 0.
8. Bonus Question (10 points/ no partial credit):. Explain why the function f (x, y) =
Solution.
Anybody wanting to know how to do this exercise can just ask me.