Math 201 – Section 3
September 27, 2004
Homework 4 – Answers
Section 15.1: 38, 40, 53-58.
38. Draw a contour map of f(x, y) = x2 – y2 showing several level curves.
-2
The level curves are the curves of the
form x2 – y2 = k, for all possible
values of k.
-1
1
The easiest case is k = 0. This curve
consists of all points with y = +x or
y = –x.
1
0
2
2
The other curves are easier to plot if
-1
you recognize that they are hyperbolas
with asymptotes consisting of the lines
-2
y = +x and y = –x. Otherwise there is
nothing to do but pick several values of k (+2, +1, –1, –2 are good choices) and
plot several points on each curve.
40. Same for f(x, y) = e y/x.
This one is easier because the function depends
only on y/x, so the places where y/x is constant
are also level curves of the function. These are
lines through the origin—each line corresponds to
one value of y/x, and so one value of the
function itself.
e2
e1
e0=1
e-1/2
There’s a problem at (0, 0), which seems to be
on all the lines. In fact, the function isn’t defined
there (you can’t divide by x=0) and can’t be made
continuous there, so the origin isn’t part of any of the level curves.
In fact, the entire y axis is a problem. You might think of it as the  level curve,
but  isn’t a real number so we should admit that the function isn’t defined
anywhere on the y axis.
1
Matching (to pictures on page 937)
53. z  sin x 2  y 2 … III and B. The level curves are circles.
54. z  x2 y 2e x
55. z 
2
 y2
… II and C. The function is zero along both axes, and goes quickly to
zero whenever r gets large in any direction.
1
… V and F. The level curves are ellipses and the function goes to zero
x  4 y2
whenever x or y is large.
2
56. z = x2 – 3xy2 … VI and A. Requires either close study or eliminating the
alternatives.
57. z = sin x sin y … IV and D. There’s a hump whenever x and y are both odd
multiples of pi.
58. z = sin2 x + (1/4)y2 … I and E. The oscillation in the x direction is the clue.
Section 15.2: 8, 10, 16.
In each case, no limit exists. Level curves work, but other tricks help, too.
8.
10.
x 2  sin 2 y
… Notice that when x=0 and y is close to zero but not zero, the
x , y 0,0 2 x 2  y 2
function value is close to 1. (That’s because the limit of (sin y)/y
is 1.) But when y is zero and x is nonzero, the function value is
1/2. There are points of both kinds as close to (0,0) as you might
like, so there is no limit at (0,0).
lim
6 x3 y
… Divide numerator and denominator by x4 to show that the
x , y 0,0 2 x 4  y 4
6 y
x
function is
— that is, it only depends on (y/x). There
4
y
2
x
are points near the origin with all values of (y/x), so the function
can’t have a limit there.
lim
 
 
16.
xy 4
… If x is zero, the function is zero. But if x = y4, then the function
x , y 0,0 x 2  y 8
equals 1/2. There are points of both kinds as close as you like to
the origin, so the function can’t have a limit there.
(end)
lim
2