# Chapter 9 Notes - Chapter 9 ISM Linear Algebra Chapter 9...

• Notes
• 38

This preview shows page 1 - 6 out of 38 pages.

Chapter 9 ISM: Linear Algebra Chapter 9 9.1 1. x ( t ) = 7 e 5 t , by Fact 9.1.1. 2. x ( t ) = - e · e - 0 . 71 t = - e 1 - 0 . 71 t , by Fact 9.1.1. 3. P ( t ) = 7 e 0 . 03 t , by Fact 9.1.1. 4. This is just an antiderivative problem: y ( t ) = 0 . 8 t 2 2 + C = 0 . 4 t 2 + C , and C = - 0 . 8, so that y ( t ) = 0 . 4 t 2 - 0 . 8. 5. y ( t ) = - 0 . 8 e 0 . 8 t , by Fact 9.1.1. 6. x dx = dt x 2 2 = t + C , and 1 2 = 0 + C , so that x 2 2 = t + 1 2 x 2 = 2 t + 1 x ( t ) = 2 t + 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 ) = . 8. x - 1 / 2 dx = dt 2 x 1 / 2 = t + C , and 2 4 = 0 + C , so that 2 x 1 / 2 = t + 4. x ( t ) = ( t 2 + 2 ) 2 for t ≥ - 4. 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 . 434
ISM: Linear Algebra Section 9.1 10. cos x dx = dt sin x = t + C , and C = 0. x ( t ) = arcsin( t ) for | t | < 1. 11. dx 1+ x 2 = dt arctan( x ) = t + C and C = 0. x ( t ) = tan( t ) for | t | < π 2 . 12. We want e kt = 3 t or e k = 3 or k = ln(3) : dx dt = ln(3) x . 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. 14. a. x ( t ) = e - t 8270 , by Fact 9.1.1 If T is the half-life, then e - T 8270 = 1 2 or - T 8270 = ln ( 1 2 ) or T = - 8270 ln ( 1 2 ) 5732. The half-life is about 5732 years. b. We want to find t such that e - t 8270 = 1 - 0 . 47 = 0 . 53 or - t 8270 = ln(0 . 53) or t = - 8270 ln(0 . 53) 5250. The Iceman died about 5000 years before A.D. 1991, or about 3000 B.C. The Austrian expert was wrong. 15. If P ( t ) = P 0 e k 100 t , then the doubling time T satisfies 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. 16. See Figure 9.1. 17. See Figure 9.2. 18. See Figure 9.3. 19. See Figure 9.4. 20. Ax = 0 - 1 1 0 x 1 x 2 = - x 2 x 1 . See Figure 9.5. It appears that the trajectories will be circles. If we start at 1 0 we will trace out the unit circle x ( t ) = cos( t ) sin( t ) . 435
Chapter 9 ISM: Linear Algebra x 1 x 2 Figure 9.1: for Problem 9.1.16. Figure 9.2: for Problem 9.1.17. Figure 9.3: for Problem 9.1.18. We can verify that dx dt = - sin( t ) cos( t ) equals 0 - 1 1 0 x ( t ) = - sin( t ) cos( t ) , as claimed. 436
ISM: Linear Algebra Section 9.1 Figure 9.4: for Problem 9.1.19. (1, 0) x 2 x 1 Figure 9.5: for Problem 9.1.20. 21. Ax = 0 1 0 0 x 1 x 2 = x 2 0 (see Figure 9.6). x 2 x 1 Figure 9.6: for Problem 9.1.21. 437
Chapter 9 ISM: Linear Algebra The trajectories will be horizontal lines. If we start at p q , then the horizontal velocity will be q , so that x ( t ) = x 1 ( t ) x 2 ( t ) = p + qt q . We can verify that dx dt = q 0 equals 0 1 0 0 x ( t ) = q 0 , as claimed. 22. We are told that dx 1 dt = Ax 1 and dx 2 dt = Ax 2 . Let x ( t ) = x 1 ( t ) + x 2 ( t ). Then dx dt = dx 1 dt + dx 2 dt = Ax 1 + Ax 2 = A ( x 1 + x 2 ) = Ax , as claimed. 23. We are told that dx 1 dt = Ax 1 . Let x ( t ) = kx 1 ( t ). Then dx dt = d dt ( kx 1 ) = k dx 1 dt = kAx 1 = A ( kx 1 ) = Ax , as claimed. 24. We are told that dx dt = Ax . Let c ( t ) = e kt x ( t ). Then dc dt = d dt ( e kt x ) = ( d dt e kt ) x + e kt dx dt = ke kt x + e kt Ax = ( A + kI n )( e kt x ) = ( A + kI n ) c , as claimed. 25. We are told that dx dt = Ax . Let c ( t ) = x ( kt ). Using the chain rule we find that dc dt = d dt ( x ( kt )) = k dx dt | kt = kA ( x ( kt )) = kAc ( t ), as claimed. To get the vector field kAc we scale the vectors of the field Ax by k .
• • • 