Download 4.8 Applications of Vector Spaces

212
Chapter 4
Vector Spaces
4.8 Applications of Vector Spaces
Use the Wronskian to test a set of solutions of a linear
homogeneous differential equation for linear independence.
Identify and sketch the graph of a conic section and perform a
rotation of axes.
LINEAR DIFFERENTIAL EQUATIONS (CALCULUS)
A linear differential equation of order n is of the form
y snd 1 gn21sxdy sn21d 1 . . . 1 g1sxdy9 1 g0sxdy 5 f sxd
where g0, g1, . . . , gn21 and f are functions of x with a common domain. If f sxd 5 0,
then the equation is homogeneous. Otherwise it is nonhomogeneous. A function y is
called a solution of the linear differential equation if the equation is satisfied when y
and its first n derivatives are substituted into the equation.
A Second-Order Linear Differential Equation
Show that both y1 5 e x and y2 5 e2x are solutions of the second-order linear
differential equation y0 2 y 5 0.
SOLUTION
For the function y1 5 ex, you have y19 5 e x and y10 5 e x. So,
y10 2 y1 5 e x 2 e x 5 0
which means that y1 5 e x is a solution of the differential equation. Similarly, for
y2 5 e2x, you have
y29 5 2e2x
and y20 5 e2x.
This implies that
y20 2 y2 5 e2x 2 e2x 5 0.
So, y2 5 e2x is also a solution of the linear differential equation.
There are two important observations you can make about Example 1. The first is
that in the vector space C 0 s2 `, `d of all twice differentiable functions defined on the
entire real line, the two solutions y1 5 e x and y2 5 e2x are linearly independent. This
means that the only solution of
C1y1 1 C2 y2 5 0
that is valid for all x is C1 5 C2 5 0. The second observation is that every linear
combination of y1 and y2 is also a solution of the linear differential equation. To see this,
let y 5 C1y1 1 C2 y2. Then
y 5 C1ex 1 C2e2x
y9 5 C1e x 2 C2e2x
y 0 5 C1e x 1 C2e2x.
Substituting into the differential equation y0 2 y 5 0 produces
y0 2 y 5 sC1e x 1 C2e2xd 2 sC1e x 1 C2e2xd 5 0.
So, y 5 C1e x 1 C2e2x is a solution.
The next theorem, which is stated without proof, generalizes these observations.
4.8
REMARK
The solution
y 5 C1y1 1 C2 y2 1 . . . 1 Cn yn
is called the general solution of
the given differential equation.
Applications of Vector Spaces
213
Solutions of a Linear Homogeneous Differential Equation
Every nth-order linear homogeneous differential equation
y snd 1 gn21sxdy sn21d 1 . . . 1 g1sxdy9 1 g0sxdy 5 0
has n linearly independent solutions. Moreover, if Hy1, y2, . . . , ynJ is a set of
linearly independent solutions, then every solution is of the form
y 5 C1y1 1 C2 y2 1 . . . 1 Cn yn
where C1, C2, . . . , Cn are real numbers.
In light of the preceding theorem, you can see the importance of being able to
determine whether a set of solutions is linearly independent. Before describing a way
of testing for linear independence, you are given the following preliminary definition.
Definition of the Wronskian of a Set of Functions
Let H y1, y2, . . . , ynJ be a set of functions, each of which has n 2 1 derivatives
on an interval I. The determinant
REMARK
The Wronskian of a set of
functions is named after the
Polish mathematician Josef
Maria Wronski (1778–1853).
Ws y1, y2, . . . , yn d 5
|
y1
y19
.
.
.
y s1n21d
y2
y29
.
.
.
. . .
. . .
y s2n21d . . .
yn
yn9
.
.
.
|
y snn21d
is called the Wronskian of the given set of functions.
Finding the Wronskian of a Set of Functions
a. The Wronskian of the set H1 2 x, 1 1 x, 2 2 xJ is
W5
|
|
| |
12x
21
0
11x
1
0
22x
21 5 0.
0
b. The Wronskian of the set Hx, x 2, x 3J is
x
W5 1
0
REMARK
This test does not apply to an
arbitrary set of functions. Each
of the functions y1, y2, . . . ,
and yn must be a solution of
the same linear homogeneous
differential equation of order n.
x2
2x
2
x3
3x 2 5 2x 3.
6x
The Wronskian in part (a) of Example 2 is said to be identically equal to zero,
because it is zero for any value of x. The Wronskian in part (b) is not identically equal
to zero because values of x exist for which this Wronskian is nonzero.
The next theorem shows how the Wronskian of a set of functions can be used to
test for linear independence.
Wronskian Test for Linear Independence
Let H y1, y2, . . . , ynJ be a set of n solutions of an nth-order linear homogeneous
differential equation. This set is linearly independent if and only if the
Wronskian is not identically equal to zero.
The proof of this theorem for the case where n 5 2 is left as an exercise. (See Exercise 40.)
214
Chapter 4
Vector Spaces
Testing a Set of Solutions
for Linear Independence
Determine whether H1, cos x, sin xJ is a set of linearly independent solutions of the
linear homogeneous differential equation
y999 1 y9 5 0.
SOLUTION
Begin by observing that each of the functions is a solution of y999 1 y9 5 0. (Try
checking this.) Next, testing for linear independence produces the Wronskian of the
three functions, as follows.
|
1
cos x
sin x
W 5 0 2sin x
cos x
0 2cos x 2sin x
5 sin2 x 1 cos2 x 5 1
|
Because W is not identically equal to zero, the set
H1, cos x, sin xJ
is linearly independent. Moreover, because this set consists of three linearly
independent solutions of a third-order linear homogeneous differential equation, the
general solution is
y 5 C1 1 C2 cos x 1 C3 sin x
where C1, C2, and C3 are real numbers.
Testing a Set of Solutions
for Linear Independence
x
x
Determine whether He , xe , sx 1 1de xJ is a set of linearly independent solutions of the
linear homogeneous differential equation
y999 2 3y0 1 3y9 2 y 5 0.
SOLUTION
As in Example 3, begin by verifying that each of the functions is actually a solution
of y999 2 3y0 1 3y9 2 y 5 0. (This verification is left to you.) Testing for linear
independence produces the Wronskian of the three functions, as follows.
|
ex
W 5 ex
ex
So, the set H
e x,
xe x
sx 1 1de x
sx 1 2de x
xe x,
|
sx 1 1de x
sx 1 2de x 5 0
sx 1 3de x
sx 1 1d J is linearly dependent.
ex
In Example 4, the Wronskian is used to determine that the set
He x, xe x, sx 1 1de xJ
is linearly dependent. Another way to determine the linear dependence of this set is to
observe that the third function is a linear combination of the first two. That is,
sx 1 1de x 5 e x 1 xe x.
Try showing that a different set, He x, xe x, x 2e xJ, forms a linearly independent set of
solutions of the differential equation
y999 2 3y0 1 3y9 2 y 5 0.
4.8
Applications of Vector Spaces
215
CONIC SECTIONS AND ROTATION
Every conic section in the xy-plane has an equation that can be written in the form
ax 2 1 bxy 1 cy 2 1 dx 1 ey 1 f 5 0.
Identifying the graph of this equation is fairly simple as long as b, the coefficient of the
xy-term, is zero. When b is zero, the conic axes are parallel to the coordinate axes, and
the identification is accomplished by writing the equation in standard (completed
square) form. The standard forms of the equations of the four basic conics are given
in the following summary. For circles, ellipses, and hyperbolas, the point sh, kd is the
center. For parabolas, the point sh, kd is the vertex.
Standard Forms of Equations of Conics
Circle sr 5 radiusd: sx 2 hd2 1 s y 2 kd2 5 r 2
Ellipse s2a 5 major axis length, 2b 5 minor axis lengthd:
y
(x − h) 2
α2
+
y
(y − k) 2
β
=1
2
(x − h) 2
β
(h, k)
2
+
(y − k) 2
=1
α2
(h, k)
2β
2α
2α
2β
x
x
Hyperbola s2a 5 transverse axis length, 2b 5 conjugate axis lengthd:
y
(x − h) 2
α2
−
(y − k) 2
β2
y
=1
(y − k) 2
α2
−
(x − h) 2
β
2
=1
(h, k)
(h, k)
2β
2α
2α
2β
x
x
Parabola s p 5 directed distance from vertex to focus):
y
y
p>0
Focus
(h, k + p)
p>0
Vertex
(h , k )
Vertex
(h, k)
(x − h) 2 = 4p(y − k)
x
Focus
(h + p, k)
( y − k ) 2 = 4 p (x − h )
x
216
Chapter 4
Vector Spaces
Identifying Conic Sections
a. The standard form of x 2 2 2x 1 4y 2 3 5 0 is
sx 2 1d2 5 4s21ds y 2 1d.
The graph of this equation is a parabola with the vertex at sh, kd 5 s1, 1d. The axis
of the parabola is vertical. Because p 5 21, the focus is the point s1, 0d. Finally,
because the focus lies below the vertex, the parabola opens downward, as shown
in Figure 4.20(a).
b. The standard form of x 2 1 4y 2 1 6x 2 8y 1 9 5 0 is
sx 1 3d2 s y 2 1d2
1
5 1.
4
1
The graph of this equation is an ellipse with its center at sh, kd 5 s23, 1d. The major
axis is horizontal, and its length is 2a 5 4. The length of the minor axis is 2b 5 2.
The vertices of this ellipse occur at s25, 1d and s21, 1d, and the endpoints of the
minor axis occur at s23, 2d and s23, 0d, as shown in Figure 4.20(b).
y
a.
y
b.
(1, 1)
1
3
(1, 0)
(−3, 2)
x
−2
Focus 2
3
2
4
(−3, 1)
(− 5, 1)
(− 1, 1)
(− 3, 0)
−1
−2
1
x
−3
−5
−4
−3
−2
−1
(x + 3)2 (y − 1)2
=1
+
1
4
(x − 1)2 = 4(−1)( y − 1)
Figure 4.20
The equations of the conics in Example 5 have no xy-term. Consequently, the axes
of the graphs of these conics are parallel to the coordinate axes. For second-degree
equations that have an xy-term, the axes of the graphs of the corresponding conics are
not parallel to the coordinate axes. In such cases it is helpful to rotate the standard
axes to form a new x9-axis and y9-axis. The required rotation angle u (measured
counterclockwise) is cot 2u 5 sa 2 cdyb. With this rotation, the standard basis for R 2,
B 5 Hs1, 0d, s0, 1dJ
is rotated to form the new basis
B9 5 Hscos u, sin ud, s2sin u, cos udJ
as shown in Figure 4.21.
y'
(− sin θ , cos θ )
y
(cos θ , sin θ )
(0, 1)
x'
θ
(1, 0)
x
If b is nonzero, it is helpful for identifying
the graph of a conic section by rotating the
coordinate axes such that the conic section
is parallel to the new axes
Figure 4.21
To find the coordinates of a point sx, yd relative to this new basis, you can use a
transition matrix, as demonstrated in Example 6.
4.8
Applications of Vector Spaces
217
A Transition Matrix for Rotation in R 2
Find the coordinates of a point sx, yd in R2 relative to the basis
B9 5 Hscos u, sin ud, s2sin u, cos udJ.
target basis
SOLUTION
By Theorem 4.21 you have
fB9 Bg 5
3 sin u
cos u
original basis
2sin u
cos u
1
0
4
0
.
1
Because B is the standard basis for R2, P 21 is represented by sB9 d21. You can use the
formula given in Section 2.3 (page 66) for the inverse of a 2 3 2 matrix to find sB9 d21.
This results in
fI
3
cos u
2sin u
1 0
g5
0 1
P21
[B'|B] → [I | P^(-1)] vs.
[B'| I] → [I | B'^(-1)]
4
sin u
.
cos u
By letting sx9, y9d be the coordinates of sx, yd relative to B9, you can use the transition
matrix P 21 as follows.
32sin u
cos u
sin u
cos u
4 3y4 5 3y94
x
x9
P^(-1) * [v]_B = [v]_B'
The x9- and y9-coordinates are x9 5 x cos u 1 y sin u and y9 5 2x sin u 1 y cos u .
The last two equations in Example 6 give the x9y9-coordinates in terms of the
xy-coordinates. To perform a rotation of axes for a general second-degree equation, it is
helpful to express the xy-coordinates in terms of the x9y9-coordinates. To do this, solve
the last two equations in Example 6 for x and y to obtain
x 5 x9 cos u 2 y9 sin u
and
y 5 x9 sin u 1 y9 cos u.
Substituting these expressions for x and y into the given second-degree equation
produces a second-degree equation in x9 and y9 that has no x9y9-term.
Rotation of Axes
The general second-degree equation ax 2 1 bxy 1 cy 2 1 dx 1 ey 1 f 5 0 can
be written in the form
a9sx9 d2 1 c9s y9 d2 1 d9x9 1 e9y9 1 f 9 5 0
by rotating the coordinate axes counterclockwise through the angle u, where u
a2c
is defined by cot 2u 5
. The coefficients of the new equation are obtained
b
from the substitutions
x 5 x9 cos u 2 y9 sin u
and
y 5 x9 sin u 1 y9 cos u.
The proof of the above result is left to you. (See Exercise 82.)
LINEAR
ALGEBRA
APPLIED
A satellite dish is an antenna that is designed to transmit
or receive signals of a specific type. A standard satellite
dish consists of a bowl-shaped surface and a feed horn
that is aimed toward the surface. The bowl-shaped surface
is typically in the shape of an elliptic paraboloid. (See
Section 7.4.) The cross section of the surface is typically
in the shape of a rotated parabola.
Chris H. Galbraith/Shutterstock.com
218
Chapter 4
Vector Spaces
Example 7 demonstrates how to identify the graph of a second-degree equation by
rotating the coordinate axes.
Rotation of a Conic Section
Perform a rotation of axes to eliminate the xy-term in
5x 2 2 6xy 1 5y 2 1 14!2 x 2 2!2y 1 18 5 0
and sketch the graph of the resulting equation in the x9y9-plane.
SOLUTION
The angle of rotation is given by
cot 2u 5
a2c 525
5
5 0.
b
26
This implies that u 5 py4. So,
sin u 5
1
!2
and cos u 5
1
!2
.
By substituting
(x' + 3)2 (y' − 1)2
=1
+
1
4
x 5 x9 cos u 2 y9 sin u 5
y
y'
(−
−5
2
2, 0)
sx9 2 y9d
and
x'
y 5 x9 sin u 1 y9 cos u 5
1
θ = 45°x
−4
1
!2
1
sx9 1 y9d
!2
into the original equation and simplifying, you obtain
sx9 d2 1 4s y9 d2 1 6x9 2 8y9 1 9 5 0.
−2
Finally, by completing the square, you find the standard form of this equation to be
−3
(− 3 2, − 2 2)
Figure 4.22
sx9 1 3d2 s y9 2 1d2 sx9 1 3d2 s y9 2 1d2
1
5
1
51
22
12
4
1
−4
which is the equation of an ellipse, as shown in Figure 4.22.
In Example 7 the new (rotated) basis for R2 is
B9 5
51 12, 122, 12 12, 1226
!
!
!
!
and the coordinates of the vertices of the ellipse relative to B9 are
32514
and
32114.
Two vertex points on the major axis
(長軸) in the new basis
To find the coordinates of the vertices relative to the standard basis
B 5 Hs1, 0d, s0, 1dJ, use the equations
x5
1
sx9 2 y9 d
!2
y5
1
sx9 1 y9 d
!2
and
Two vertex points on the major
axis in the original basis
to obtain s23!2, 22!2 d and s2 !2, 0d, as shown in Figure 4.22.