% Problem 6

% Clear variables and figures.

clear
close all

% a)

% Let's use a grid in which x and y go from -1 to 1 with spacing 0.05.

x = -1:0.05:1;
y = -1:0.05:1;
[X,Y] = meshgrid(x,y);
contour(x, y, 3*Y + Y.^3 - X.^3)
title 'Figure 6.1'
xlabel x
ylabel y
pause
print -deps figA6-1

% The graph is shown in Figure 6.1.  The curves are nearly horizontal
% near the origin, but start curving as we move away.  This is because
% when y is close to 0, y^3 is very small compared with 3y, and the
% curves look a lot like level curves of 3y - x^3, or in other words
% y = (1/3)x^3 + C.

% Next let's try x and y from -10 to 10 in increments of 0.2.
x = -10:0.2:10;
y = -10:0.2:10;
[X,Y] = meshgrid(x,y);
contour(x, y, 3*Y + Y.^3 - X.^3, 20)
title 'Figure 6.2'
xlabel x
ylabel y
pause
print -deps figA6-2

% The graph is shown in Figure 6.2.  The curves slope upward, and
% curve around the origin on either side.  The curves are graphs of
% 3y + y^3 - x^3 = C for various C, and when y is large, y^3 is
% large compared with 3y and C.  Thus the curves should tend toward
% the graph of y^3 - x^3 = 0, or in other words y = x, which they do
% in the lower left and upper right corners of the graph.

% b)

contour(x, y, 3*Y + Y.^3 - X.^3, [5 5])
title 'Figure 6.3'
xlabel x
ylabel y
pause
print -deps figA6-3

% The graph is shown in Figure 6.3.  This curve lies close to the line
% y = x throughout much of the region.

% c)

% We need to switch to a grid with positive values of x and y because
% of the logarithms.  Notice that f(1,1) = 0, so we graph the level
% curve with value 0.

x = 0.1:0.1:5;
y = 0.1:0.1:5;
[X,Y] = meshgrid(x,y);
contour(x, y, X.*log(Y) + Y.*log(X), [0 0])
title 'Figure 6.4'
xlabel x
ylabel y
pause
print -deps figA6-4

% The graph is shown in Figure 6.4.

echo off
diary off