chapter09-student solutions manual

Linear Algebra with Applications (3rd Edition)

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SSM: Linear Algebra Section 9.1 Chapter 9 9.1 1. x ( t ) = 7 e 5 t , by Fact 9.1.1. 3. P ( t ) = 7 e 0 . 03 t , by Fact 9.1.1. 5. y ( t ) = - 0 . 8 e 0 . 8 t , by Fact 9.1.1. 7. x - 2 dx = dt - x - 1 = t + C - 1 x = t + C , and - 1 = 0 + C , so that - 1 x = t - 1 x ( t ) = 1 1 - t ; note that lim x 1 - x ( t ) = . 9. x - k dx = dt 1 1 - k x 1 - k = t + C , and 1 1 - k = C , so that 1 1 - k x 1 - k = t + 1 1 - k x 1 - k = (1 - k ) t + 1 x ( t ) = ((1 - k ) t + 1) 1 / 1 - k . 11. dx 1+ x 2 = dt arctan( x ) = t + C and C = 0. x ( t ) = tan( t ) for | t | < π 2 . 13. a. The debt in millions is 0 . 45(1 . 06) 212 104 , 245, or about 100 billion dollars. b. The debt in millions is 0 . 45 e 0 . 06 · 212 150 , 466, or about 150 billion dollars. 15. If P ( t ) = P 0 e k 100 t , then the doubling time T satis±es the equation P ( T ) = P 0 e k 100 T = 2 P 0 or e k 100 T = 2 or k 100 T = ln(2) or T = 100 k ln(2) 69 k since ln(2) 0 . 69. 17. See Figure 9.1. 217
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Chapter 9 SSM: Linear Algebra Figure 9.1: for Problem 9.1.17. Figure 9.2: for Problem 9.1.19. x 2 x 1 Figure 9.3: for Problem 9.1.21. 19. See Figure 9.2. 21. A~x = ± 0 1 0 0 ² ± x 1 x 2 ² = ± x 2 0 ² (see Figure 9.3). 218
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SSM: Linear Algebra Section 9.1 The trajectories will be horizontal lines. If we start at ± p q ² , then the horizontal velocity will be q , so that ~x ( t ) = ± x 1 ( t ) x 2 ( t ) ² = ± p + qt q ² . We can verify that d~x dt = ± q 0 ² equals ± 0 1 0 0 ² ~x ( t ) = ± q 0 ² , as claimed. 23. We are told that d~x 1 dt = A~x 1 . Let ~x ( t ) = k~x 1 ( t ). Then d~x dt = d dt ( k~x 1 ) = k d~x 1 dt = kA~x 1 = A ( k~x 1 ) = A~x , as claimed. 25. We are told that d~x dt = A~x . Let ~ c ( t ) = ~x ( kt ). Using the chain rule we Fnd that d~ c dt = d dt ( ~x ( kt )) = k d~x dt | kt = kA ( ~x ( kt )) = kA~ c ( t ), as claimed. To get the vector Feld kA~ c we scale the vectors of the Feld A~x by k . 27. Use ±act 9.1.3. The eigenvalues of A = ± - 4 3 2 - 3 ² are λ 1 = - 6 and λ 2 = - 1, with associated eigenvec- tors ~v 1 = ± - 3 2 ² and ~v 2 = ± 1 1 ² . The coordinates of ~x (0) = ± 1 0 ² with respect to ~v 1 and ~v 2 are c 1 = - 1 5 and c 2 = 2 5 . By ±act 9.1.3 the solution is ~x ( t ) = - 1 5 e - 6 t ± - 3 2 ² + 2 5 e - t ± 1 1 ² . 29. λ 1 = 0, λ 2 = 5; ~v 1 = ± - 2 1 ² , ~v 2 = ± 1 2 ² ; c 1 = - 2, c 2 = 1, so that ~x ( t ) = - 2 ± - 2 1 ² + e 5 t ± 1 2 ² = ± 4 - 2 ² + e 5 t ± 1 2 ² . 31. λ 1 = 1, λ 2 = 6, λ 3 = 0; ~v 1 = 1 - 2 1 . Since ~x (0) = ~v 1 we need not Fnd ~v 2 and ~v 3 . c 1 = 1 , c 2 = c 3 = 0, so that ~x ( t ) = e t 1 - 2 1 . In Exercises 32 to 35, Fnd the eigenvalues and eigenspaces. Then determine the direction of the ²ow along the eigenspaces (outward if λ > 0 and inward if λ < 0). Use ±igure 11 of Section 9.1 as a guide to sketch the other trajectories.
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This homework help was uploaded on 01/23/2008 for the course MATH 253B taught by Professor Bretsch during the Spring '08 term at Colby.

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chapter09-student solutions manual - SSM: Linear Algebra...

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