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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.