Chapter 1 Solutions
Exercise 1.1 f (x, y) = x: a level curve f (x, y) =constant is a vertical line x = c. Curves for c = −1, 0, 1/2, 3 and
4 are graphed below:
Level sets x=c for c=-1,0,1/2,3, and 4
2
1.5
1
0.5
c=-1
c=1/2
c=4
y
0
c=0
c=3
-0.5
-1
-1.5
-2
-1
0
1
2
3
4
x
Here is a surface plot of graph(f ):
z=x
2
1.5
1
0.5
0
z
-0.5
-1
-1.5
-2 2
1.5
1
y
0.5
0
-0.5
-1
-1.5
-2 -2
The Matlab m-file used for this problem is
% level sets and graph(f) for f: (x,y)-->x
clear
clf
x=-2:0.1:2;
y=x;
-1.5
-1
-0.5
0
0.5
x
1
1.5
2
% level curves f(x,y)=c for c=-1.0,1/2,3,4
c=[-1,0,0.5,3,4];
% generate level curves all at once
xx=c’*ones(size(x));
plot(xx,y,’k’);
% put in axes, curve labels and title
xlabel(’x’), ylabel(’y’)
text(-0.9,0.25,’c=-1’)
text(-0.25,-0.25,’c=0’)
text(0.2,0.25,’c=1/2’)
text(2.75,-0.25,’c=3’)
text(3.75,0.25,’c=4’)
title(’Level sets x=c for c=-1,0,1/2,3, and 4’)
print -deps contour1.eps
pause
close
% print plot to .eps file
% hit a key to continue
% close plot
% surface plot of the graph of f
[X,Y]=meshgrid(x,y);
mesh(X,Y,X)
% plot labels and title
xlabel(’x’), ylabel(’y’), zlabel(’z’)
title(’z=x’)
print -deps surface1.eps
pause
close
2
Exercise 1.3 f (x, y) = x2 − y 2 : a level curve f (x, y) =constant is a hyperbola x2 − y 2 = c if c 6= 0. Vertices are on
the y-axis if c < 0 and on the x-axis if c > 0. If c = 0 the level curve is the graph of the equation |x| = |y|, that is,
the set of all points for which y = ±x.
x2-y2=c for c=-1,0,1/2,3,4
2
c=-1
1.5
1
c=3
c=3
c=0
y
0.5
0
c=1/2
-0.5
c=1/2
c=4
c=4
c=0
-1
-1.5
c=-1
-2
-2
-1
0
1
2
x
A surface plot of graph(f ):
z=x2-y2
4
3
2
1
0
z
-1
-2
-3
-4 2
1.5
2
1
1.5
0.5
1
0
y
0.5
0
-0.5
-0.5
-1
-1
-1.5
x
-1.5
-2-2
3
The Matlab m-file used for this problem is:
% level sets and graph(f) for f: (x,y)-->x^2-y^2
clear
clf
x=-2:0.1:2;
y=-2:0.1:2;
% level curves f(x,y)=c for c=-1.0,1/2,3,4
% superimpose level curves
hold on
% label x & y axes and give plot a title
xlabel(’x’)
ylabel(’y’)
title(’x^2-y^2=c for c=-1,0,1/2,3,4’)
% c=-1
plot(x,[sqrt(x.^2+1); -sqrt(x.^2+1)],’k’);
text(-1.25,1.8,’c=-1’)
text(-1.25,-1.8,’c=-1’)
% c=0
plot(x,[x;-x],’k’)
text(-0.35,0.55,’c=0’)
text(-0.35,-0.55,’c=0’)
% c=1/2
plot([sqrt(y.^2+0.5); -sqrt(y.^2+0.5)],y,’k’)
text(-1.1,0,’c=1/2’)
text(0.75,0,’c=1/2’)
% c=3
plot([sqrt(y.^2+3); -sqrt(y.^2+3)],y,’k’)
text(-1.8,0.65,’c=3’)
text(1.6,0.65,’c=3’)
% c=4
plot([sqrt(y.^2+4); -sqrt(y.^2+4)],y,’k’)
text(-2.4,-0.5,’c=4’)
text(2.2,-0.5,’c=4’)
pause
% print plot to .eps file
print -dps contour3.eps
% release plot
hold off
% surface plot of graph(f)
% close above plot
close
[X,Y]=meshgrid(x,y);
Z=X.^2-Y.^2;
4
mesh(X,Y,Z)
% set axis labels and title
xlabel(’x’)
ylabel(’y’)
zlabel(’z’)
title(’z=x^2-y^2’)
% print surface plot
print -dps surface3.eps
pause
close
5
Exercise 1.4 f (x, y) = 3R8 − 8R6 + 6R4 where R2 = x2 + y 2 : the values of f depend only on how far the point
(x, y) is from the origin (0, 0). So level sets f (x, y) =constant are concentric circles centered at (0, 0).
f(x,y)=c
1
c=1.04287843
0.5
c=67/256
y
c=1411/4,294,967,296
0
. c=0
c=1
-0.5
-1
-1
-0.5
0
0.5
1
x
Here is a surface plot of graph(f ):
graph(f)
350
300
250
200
z
150
100
50
01
1
0.5
0.5
0
0
y
-0.5
-0.5
-1-1
6
x
It’s hard to see detail in the surface plot because f grows so rapidly for R > 1. The values of f near the corners of
the grid are much larger than values of f within and on the unit circle. Since graph(f ) has radial symmetry with the
z-axis as the axis of the symmetry then we can look at a profile of the surface by slicing it with any plane containing
the z-axis.
f(R)=3R8-8R6+6R4
1.4
1.2
1
f(R)
0.8
0.6
0.4
0.2
0
-1
-0.5
0
0.5
1
R
The profile of graph(f ) above esposes detail of f near the unit circle R = 1. The last two plots show graph(f ) in a
neighborhood of the point (−1, 0, 1), a point on the unit circle, and some corresponding level curves.
graph(f) in a neighborhood of (-1,0,1)]
1.015
1.01
1.005
z
1
0.995
0.1
-0.97
0.05
-0.98
-0.99
0
-1
y
-1.01
-0.05
x
-1.02
-1.03
-0.1
7
40
35
30
25
20
15
10
5
2
4
6
8
10
12
Here is a Matlab m-file used for this problem:
% level sets and graph(f) for f: (x,y)-->3r^8-8r^6+6r^4
% where r=x^2+y^2
clear
clf
%
%
%
%
f has "radial" symmetry, i.e., its value at (x,y) depends only on
how far (x,y) is from (0,0).
Level sets are nested circles with center (0,0): if f(r)=c then
circle x^2+y^2=r^2 is level set.
% superimpose level sets
hold on
% label x & y axes and title
xlabel(’x’)
ylabel(’y’)
title(’f(x,y)=c’)
% c=0 (circle
text(0,0,’.’)
text(0.05,0,’c=0’)
% c=67/256 (circle of radius 1/2)
x=linspace(-0.5,0.5);
plot(x,[sqrt(0.25-x.^2);-sqrt(0.25-x.^2)],’k’)
text(0.4,0.4,’c=67/256’)
% c=1411/(2^32) (circle of radius 1/4)
x=linspace(-0.25,0.25);
8
plot(x,[sqrt(0.0625-x.^2);-sqrt(0.0625-x.^2)],’k’)
text(-0.25,0.3,’c=1411/4,294,967,296’)
% c=1 (unit circle)
x=linspace(-1,1);
plot(x,[sqrt(1-x.^2);-sqrt(1-x.^2)],’k’)
text(0.9,-0.05,’c=1’)
% c=1.04287843 (circle of radius 1.1)
x=linspace(-1.1,1.1);
plot(x,[sqrt((1.1)^2-x.^2);-sqrt((1.1)^2-x.^2)],’k’)
text(0.8,0.8,’c=1.04287843’)
print -dps contour4a.eps
% release plot
hold off
pause
% surface plot of graph(f)
% close previous plot
close
x=-1:0.1:1;
[X,Y]=meshgrid(x); % for square grids meshgrid() needs only one argument
R=X.^2+Y.^2;
Z=R.^4.*(3*R.^4-8*R.^2+6);
mesh(X,Y,Z)
% label axes and title
xlabel(’x’)
ylabel(’y’)
zlabel(’z’)
title(’graph(f)’)
print -dps surface4a.eps
pause
% z-axis is axis of symmetry: a slice through graph(f) using a plane
% containing the z-axis gives following profile
close
R=-1.2:0.01:1.2;
Z=R.^4.*(3*R.^4-8*R.^2+6);
plot(R,Z)
% label axes and title
xlabel(’R’)
ylabel(’f(R)’)
title(’f(R)=3R^8-8R^6+6R^4’)
9
print -dps profile4.eps
pause
close
% above profile shows some detail for R near 1 that are not seen on
% previous surface plot of graph(f). So plot a portion of surface in
% a neighborhood he point of (-1,0,1).
x=-1.03:0.005:-0.97;
y=-0.1:0.005:0.1;
[X,Y]=meshgrid(x,y);
R=X.^2+Y.^2;
Z=R.^4.*(3*R.^4-8*R.^2+6);
mesh(X,Y,Z)
% label axes and title
xlabel(’x’)
ylabel(’y’)
zlabel(’z’)
title(’graph(f) in a neighborhood of (-1,0,1)]’)
print -dps surface4b.eps
pause
% now look at some contour lines corresponding to this part of surface
contour(Z,’k’)
print -dps contour4b.eps
pause
close
10
Exercise 1.5
(a) n = 0: in this case f (x) = x and level sets of f corresponding to c = −1, 0, 1, 2 are {−1} {0}, {1}, and {2},
respectively. These level sets are superimposed on the number line below.
•
−1
−2
•
0
•
1
•
2
(b) n = 1: f (x, y) = y and level sets are horizontal lines y = c.
level sets y=c for c=-1,0,1 & 2 (horizontal lines)
2
c=2
1.5
c=1
y
1
0.5
c=0
0
-0.5
c=-1
-1
-2
-1.5
-1
-0.5
0
0.5
1
1.5
x
(c) n = 2: f (x, y, z) = z and level sets are stacked planes parallel to the xy-plane.
level sets for f(x,y,z)=z
2
1.5
1
0.5
z
0
-0.5
-1 2
1.5
2
1
1.5
0.5
1
0
y
0.5
0
-0.5
-0.5
-1
-1
-1.5
-1.5
-2-2
11
x
2
Exercise 1.7
(a) n = 0: f (x) = x and level sets of f corresponding to c = −1, 0, 1, 2 are {−1} {0}, {1} {2}, just like the previous
problem.
•
−1
−2
•
0
•
1
•
2
(b) n = 1: f (x, y) = x − y 2 and level sets are nested parabolas along the x-axis opening to the right.
level sets for f(x,y)=x-y2
2
1.5
1
y
0.5
0
-0.5
-1
-1.5
-2
-1
0
1
2
3
4
5
6
x
(c) n = 2: f (x, y) = x − y 2 − z 2 and level sets are nested paraboloids opening along the positive x-axis. The first
surface plot is the level set z − y 2 − z 2 = −1. The subsequent plot contains all four level sets.
2
1.5
1
0.5
0
-0.5
-1
-1.5
-2 2
1.5
7
1
6
0.5
5
0
4
3
-0.5
2
-1
1
-1.5
0
-2-1
12
level sets for f(x,y,z)=x-y2-z2
2
1.5
1
0.5
0
z
-0.5
-1
-1.5
-2 2
1.5
10
1
8
0.5
0
y
6
-0.5
4
-1
2
-1.5
-2
x
0
Here is the Matlab m-file I used on this problem:
clear
clf
% n=0 -- draw by hand
% n=1: f(x,y)=x-y^2. So level sets are parabolas x=y^2+c.
% Plot them all at once
y=-2:0.1:2;
c=[-1,0,1,2];
% put all level curves on one plot
a=ones(size(y));
b=ones(size(c));
x=c’*a+b’*y.^2;
plot(x,y,’k’)
title(’& level sets for f(x,y)=x-y^2’)
xlabel(’& x’), ylabel(’& y’)
print -dps contour7n1.eps
pause
% n=2: f(x,y,z)=x-y^2-z^2. So level sets are stacked paraboloids x=c+y^2+z^2.
% plot each one separately and superimpose on one plot.
close
[Y,Z]=meshgrid(y);
X=c(1)+Y.^2+Z.^2;
surf(X,Y,Z)
print -dps surface7c_1.eps % save one surface
pause
13
hold
X=c(2)+Y.^2+Z.^2;
surf(X,Y,Z)
X=c(3)+Y.^2+Z.^2;
surf(X,Y,Z)
X=c(4)+Y.^2+Z.^2;
surf(X,Y,Z)
title(’& level sets for f(x,y,z)=x-y^2-z^2’)
xlabel(’& x’), ylabel(’& y’), zlabel(’& z’)
print -dps surface7c_all.eps % all plots superimposed
hold off
pause
close
14
Exercise 1.10 The graph of any function f : Rn → R is a level set for some function F : Rn+1 → R.
Here is a proof: graph(f ) = {(x1 , . . . , xn , xn+1 ) | xn+1 = f (x1 , . . . , xn )} so if F is defined by F (x1 , . . . , xn , xn+1 ) =
xn+1 − f (x1 , . . . , xn ) then F −1 (0) = graph(f ).
15