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 ͑n͒ ͑nϪ1͒ . . . y ϩ gnϪ1͑x͒y ϩ ϩ g1͑x͒yЈ ϩ g0͑x͒y ϭ f͑x͒

where g0, g1, . . . , gnϪ1 and f are functions of x with a common domain. If f ͑x͒ ϭ 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

x Ϫx Show that both y1 ϭ e and y2 ϭ e are solutions of the second-order linear differential equation yЉ Ϫ y ϭ 0. SOLUTION x x x For the function y1 ϭ e , you have y1Ј ϭ e and y1Љ ϭ e . So, x x y1Љ Ϫ y1 ϭ e Ϫ e ϭ 0 x which means that y1 ϭ e is a solution of the differential equation. Similarly, for Ϫx y2 ϭ e , you have Ϫx Ϫx y2Ј ϭ Ϫ e and y2Љ ϭ e . This implies that

Ϫx Ϫx y2Љ Ϫ y2 ϭ e Ϫ e ϭ 0. Ϫx So,y2 ϭ e 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 CЉ͑Ϫϱ, ϱ͒ of all twice differentiable functions defined on the x Ϫx entire real line, the two solutions y1 ϭ e and y2 ϭ e are linearly independent . This means that the only solution of

C1y1 ϩ C2y2 ϭ 0

that is valid for all x is C1 ϭ C2 ϭ 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 ϭ C1y1 ϩ C2y2. Then x Ϫx y ϭ C1e ϩ C2e x Ϫx yЈ ϭ C1e Ϫ C2e x Ϫx yЉ ϭ C1e ϩ C2e . Substituting into the differential equation yЉ Ϫ y ϭ 0 produces

x Ϫx x Ϫx yЉ Ϫ y ϭ ͑C1e ϩ C2e ͒ Ϫ ͑C1e ϩ C2e ͒ ϭ 0. x Ϫx So,y ϭ C1e ϩ C2e is a solution. The next theorem, which is stated without proof, generalizes these observations. 4.8 Applications of Vector Spaces 213

Solutions of a Linear Homogeneous Differential Equation REMARK The solution Every n th-order linear homogeneous differential equation . . . y ͑n͒ ϩ g ͑x͒y ͑nϪ1͒ ϩ . . . ϩ g ͑x͒yЈ ϩ g ͑x͒y ϭ 0 y ϭ C1y1 ϩ C2 y2 ϩ ϩ Cn yn nϪ1 1 0

is called the general solution of has n linearly independent solutions. Moreover, if ͭy1, y2, . . . , ynͮ is a set of the given differential equation. linearly independent solutions, then every solution is of the form . . . y ϭ C1y1 ϩ C2 y2 ϩ ϩ 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 ͭy1, y2, . . . , ynͮ be a set of functions, each of which has n Ϫ 1 derivatives on an interval I. The determinant

REMARK y1 y2 . . . yn The Wronskian of a set of y Ј y Ј . . . y Ј W y , y , . . . , y ϭ .1 .2 .n functions is named after the ͑ 1 2 n͒ . . . Polish mathematician Josef . . . y͑nϪ1͒ y͑nϪ1͒ . . . y͑nϪ1͒ Maria Wronski (1778–1853). Խ 1 2 n Խ is called the Wronskian of the given set of functions.

Finding the Wronskian of a Set of Functions

a. The Wronskian of the set ͭ1 Ϫ x, 1 ϩ x, 2 Ϫ xͮ is 1 Ϫ x 1 ϩ x 2 Ϫ x W ϭ Ϫ1 1 Ϫ1 ϭ 0. 0 0 0

Խ 2 3Խ b. The Wronskian of the set ͭx, x , x ͮ is x x 2 x 3 W ϭ 1 2x 3x 2 ϭ 2x 3. Խ0 2 6xԽ 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. REMARK The next theorem shows how the Wronskian of a set of functions can be used to This test does not apply to an test for linear independence. arbitrary set of functions. Each of the functions y , y , . . . , 1 2 Wronskian Test for Linear Independence and yn must be a solution of the same linear homogeneous Let ͭy1, y2, . . . , ynͮ be a set of n solutions of an n th-order linear homogeneous differential equation of order n. 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 ϭ 2 is left as an exercise. (See Exercise 40.) 214 Chapter 4 Vector Spaces

Testing a Set of Solutions for Linear Independence Determine whether ͭ1, cos x, sin xͮ is a set of linearly independent solutions of the linear homogeneous differential equation yЈЈЈ ϩ yЈ ϭ 0. SOLUTION Begin by observing that each of the functions is a solution of yЈЈЈ ϩ yЈ ϭ 0. (Try checking this.) Next, testing for linear independence produces the Wronskian of the three functions, as follows. 1 cos x sin x W ϭ 0 Ϫsin x cos x 0 Ϫcos x Ϫsin x ϭ sinԽ 2 x ϩ cos 2 x ϭ 1 Խ Because W is not identically equal to zero, the set ͭ1, cos x, sin xͮ 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 ϭ C1 ϩ C2 cos x ϩ C3 sin x

where C1, C2, and C3 are real numbers.

Testing a Set of Solutions for Linear Independence Determine whether ͭe x, xe x, ͑x ϩ 1͒e xͮ is a set of linearly independent solutions of the linear homogeneous differential equation yЈЈЈ Ϫ 3yЉ ϩ 3yЈ Ϫ y ϭ 0. SOLUTION As in Example 3, begin by verifying that each of the functions is actually a solution of yЈЈЈ Ϫ 3yЉ ϩ 3yЈ Ϫ y ϭ 0. (This verification is left to you.) Testing for linear independence produces the Wronskian of the three functions, as follows. e x xe x ͑x ϩ 1͒e x W ϭ e x ͑x ϩ 1͒e x ͑x ϩ 2͒e x ϭ 0 Խe x ͑x ϩ 2͒e x ͑x ϩ 3͒e xԽ So, the set ͭe x, xe x, ͑x ϩ 1͒e xͮ is linearly dependent.

In Example 4, the Wronskian is used to determine that the set ͭe x, xe x, ͑x ϩ 1͒e xͮ 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, ͑x ϩ 1͒e x ϭ e x ϩ xe x. Try showing that a different set,ͭe x, xe x, x 2e xͮ, forms a linearly independent set of solutions of the differential equation yЈЈЈ Ϫ 3yЉ ϩ 3yЈ Ϫ y ϭ 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 ϩ bxy ϩ cy 2 ϩ dx ϩ ey ϩ f ϭ 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 ͑h, k͒ is the center. For parabolas, the point ͑h, k͒ is the vertex.

Standard Forms of Equations of Conics Circle ͑r ϭ radius ͒: ͑x Ϫ h͒2 ϩ ͑y Ϫ k͒2 ϭ r 2 Ellipse ͑2␣ ϭ major axis length, 2 ␤ ϭ minor axis length ͒:

y 2 2 y 2 2 (x − h) +(y − k) = 1 (x − h) +(y − k) = 1 α 2β 2 β 2 α 2

(h, k) (h, k) 2β 2α

2α 2β x x

Hyperbola ͑2␣ ϭ transverse axis length, 2 ␤ ϭ conjugate axis length ͒:

y (x − h)2 (y − k)2 y (y − k)2 (x − h)2 − = 1 − = 1 α 2 β 2 α 2 β 2

(h, k)

(h, k) 2β 2α

2α 2β x x

Parabola ͑p ϭ directed distance from vertex to focus): y y

p > 0

Focus (h, k + p) p > 0 Vertex Vertex (h , k ) Focus (h, k) (h + p, k ) (x − h)2 = 4 p(y − k) (yk− )2 = 4 pxh (− ) x x 216 Chapter 4 Vector Spaces

Identifying Conic Sections

a. The standard form of x 2 Ϫ 2x ϩ 4y Ϫ 3 ϭ 0 is ͑x Ϫ 1͒2 ϭ 4͑Ϫ1͒͑ y Ϫ 1͒. The graph of this equation is a parabola with the vertex at ͑h, k͒ ϭ ͑1, 1 ͒. The axis of the parabola is vertical. Because p ϭ Ϫ 1, the focus is the point ͑1, 0 ͒. 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 ϩ 4y2 ϩ 6x Ϫ 8y ϩ 9 ϭ 0 is ͑x ϩ 3͒2 ͑y Ϫ 1͒2 ϩ ϭ 1. 4 1 The graph of this equation is an ellipse with its center at ͑h, k͒ ϭ ͑Ϫ3, 1 ͒. The major axis is horizontal, and its length is 2␣ ϭ 4. The length of the minor axis is 2␤ ϭ 2. The vertices of this ellipse occur at ͑Ϫ5, 1 ͒ and ͑Ϫ1, 1 ͒, and the endpoints of the minor axis occur at ͑Ϫ3, 2 ͒ and ͑Ϫ3, 0 ͒, as shown in Figure 4.20(b).

a. y b. y (1, 1) 1 3 (1, 0) x (−3, 2) 2 −2 Focus 2 3 4 −1 (−3, 1) (−5, 1) (−1, 1) 1 −2 (−3, 0) x −3 −5 −4 −3 −2 −1 (x + 3) 2 (y − 1) 2 (x − 1) 2 = 4( −1)( y − 1) + = 1 4 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 xЈ -axis and yЈ -axis. The required rotation angle ␪ (measured counterclockwise) is cot 2 ␪ ϭ ͑a Ϫ c͒͞b. With this rotation, the standard basis for R2, B ϭ ͭ͑1, 0 ͒, ͑0, 1 ͒ͮ is rotated to form the new basis BЈ ϭ ͭ͑cos ␪, sin ␪͒, ͑Ϫsin ␪, cos ␪͒ͮ as shown in Figure 4.21.

y y' (cos θ , sin θ ) (0, 1) (−sin θ , cos θ ) x' θ x (1, 0)

Figure 4.21 To find the coordinates of a point ͑x, y͒ 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 R2

Find the coordinates of a point ͑x, y͒ in R2 relative to the basis BЈ ϭ ͭ͑cos ␪, sin ␪͒, ͑Ϫsin ␪, cos ␪͒ͮ. SOLUTION By Theorem 4.21 you have cos ␪ Ϫsin ␪ 1 0 ͓BЈ B͔ ϭ . ΄ sin ␪ cos ␪ 0 1΅ Because B is the standard basis for R2, P Ϫ1 is represented by ͑BЈ͒Ϫ1. You can use the formula given in Section 2.3 (page 66) for the inverse of a 2 ϫ 2 matrix to find ͑BЈ͒Ϫ1. This results in 1 0 cos ␪ sin ␪ ͓I P Ϫ1͔ ϭ . ΄0 1 Ϫsin ␪ cos ␪΅ By letting ͑xЈ, yЈ͒ be the coordinates of ͑x, y͒ relative to BЈ, you can use the transition matrix P Ϫ1 as follows. cos ␪ sin ␪ x xЈ ϭ ΄Ϫsin ␪ cos ␪΅΄y΅ ΄yЈ΅ The xЈ - and yЈ -coordinates are xЈ ϭ x cos ␪ ϩ y sin ␪ and yЈ ϭ Ϫ x sin ␪ ϩ y cos ␪.

The last two equations in Example 6 give the xЈyЈ -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 xЈyЈ -coordinates. To do this, solve the last two equations in Example 6 for x and y to obtain x ϭ xЈ cos ␪ Ϫ yЈ sin ␪ and y ϭ xЈ sin ␪ ϩ yЈ cos ␪. Substituting these expressions for x and y into the given second-degree equation produces a second-degree equation in xЈ and yЈ that has no xЈyЈ -term.

Rotation of Axes The general second-degree equation ax 2 ϩ bxy ϩ cy 2 ϩ dx ϩ ey ϩ f ϭ 0 can be written in the form aЈ͑xЈ͒2 ϩ cЈ͑yЈ͒2 ϩ dЈxЈ ϩ eЈyЈ ϩ fЈ ϭ 0 by rotating the coordinate axes counterclockwise through the angle ␪, where ␪ a Ϫ c is defined by cot 2 ␪ ϭ . The coefficients of the new equation are obtained b from the substitutions x ϭ xЈ cos ␪ Ϫ y Ј sin ␪ and y ϭ xЈ sin ␪ ϩ y Ј cos ␪.

The proof of the above result is left to you. (See Exercise 82.) LINEAR A satellite dish is an antenna that is designed to transmit ALGEBRA or receive signals of a specific type. A standard satellite APPLIED 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 Ϫ 6xy ϩ 5y2 ϩ 14 Ί2x Ϫ 2Ί2y ϩ 18 ϭ 0 and sketch the graph of the resulting equation in the xЈyЈ -plane. SOLUTION The angle of rotation is given by a Ϫ c 5 Ϫ 5 cot 2 ␪ ϭ ϭ ϭ 0. b Ϫ6 This implies that ␪ ϭ ␲͞4. So, 1 1 sin ␪ ϭ and cos ␪ ϭ . Ί2 Ί2 By substituting

(x' + 3) 2 (y' − 1) 2 1 + = 1 x ϭ xЈ cos ␪ Ϫ yЈ sin ␪ ϭ ͑xЈ Ϫ yЈ͒ 4 1 Ί2 y and y' 2 x' 1 1 y ϭ xЈ sin ␪ ϩ yЈ cos ␪ ϭ ͑xЈ ϩ yЈ͒ (− 2, 0) Ί2 θ = 45 ° x into the original equation and simplifying, you obtain −5 −4 ͑xЈ ͒2 ϩ 4͑yЈ͒2 ϩ 6xЈ Ϫ 8yЈ ϩ 9 ϭ 0. −2 Finally, by completing the square, you find the standard form of this equation to be −3 ͑xЈ ϩ 3͒2 ͑yЈ Ϫ 1͒2 ͑xЈ ϩ 3͒2 ͑yЈ Ϫ 1͒2 (−3 2, −2 2) ϩ ϭ ϩ ϭ 1 −4 22 12 4 1 Figure 4.22 which is the equation of an ellipse, as shown in Figure 4.22.

In Example 7 the new (rotated) basis for R2 is 1 1 1 1 BЈ ϭ , , Ϫ , Ά΂Ί2 Ί2΃ ΂ Ί2 Ί2΃· and the coordinates of the vertices of the ellipse relative to BЈ are Ϫ5 Ϫ1 and . ΄ 1΅ ΄ 1΅ To find the coordinates of the vertices relative to the standard basis B ϭ ͭ͑1, 0 ͒, ͑0, 1 ͒ͮ, use the equations 1 x ϭ ͑xЈ Ϫ yЈ͒ Ί2 and 1 y ϭ ͑xЈ ϩ yЈ͒ Ί2 to obtain ͑Ϫ3Ί2, Ϫ2Ί2͒ and͑ϪΊ2, 0 ͒, as shown in Figure 4.22.