FDWKCalcSM_ch6

# FDWKCalcSM_ch6 - 264 Section 6.1 Chapter 6 Differential...

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264 Section 6.1 Chapter 6 Differential Equations and Mathematical Modeling Section 6.1 Slope Fields and Euler’s Method (pp. 321–330) Exploration 1 Seeing the Slopes 1. Since dy dx = 0 represents a line with a slope of 0, we should expect to see intervals with no change in y . We see this at odd multiples of π / . 2 2. Since y is the dependent variable, I t will have no effect on the value of dy dx x = cos . 3. The graph of dy dx will look the same at all values of y . 4. When x dy dx x = = = 0 1 , cos . This can be seen on the graph near the origin. At that point, the change in y and change in x are the same. 5. When x dy dx x = = = − π , cos . 1 This can be seen in the graph at x = π . At this point, the change in y is negative of the change in x . 6. This is true because each point on the graph has a negative of itself. Quick Review 6.1 1. Yes. d dx e e x x = 2. Yes. d dx e e x x 4 4 4 = 3. No. d dx x e xe x e x x x ( ) 2 2 2 = + 4. Yes. d dx e xe x x 2 2 2 = 5. No. d dx e xe x x ( ) 2 2 5 2 + = 6. Yes. d dx x x x 2 1 2 2 2 1 2 = = ( ) 7. Yes. d dx x x x sec sec tan = 8. No. d dx x x = − 1 2 9. y x x C = + + = + + = − 3 4 2 2 3(1) 4(1) 5 2 C C 10. y x x C = + 2 3 sin cos 4 2 sin(0) 3 cos(0) 7 = + = − C C 11. y e x C x = + + 2 sec 5 0 3 2 0 = + + = e C C ( ) sec( ) 12. y x x = + + tan ln(2 1) 1 C π π = + + = tan (1) ln(2(1) 1) 3 4 1 C C Section 6.1 Exercises 1. dy x x dx = ( sec ) 5 4 2 y x x C = + 5 tan 2. dy x x e dx x = (sec tan ) y x e C x = + sec 3. dy x e x dx x = + (sin ) 8 3 y x e x C x = − + + + cos 2 4 4. dy x x dx x x C = = + + 1 1 1 2 ln 5. dy x dx x C x x = + + = + + 5 5 1 1 5 2 1 ln tan 6. dy x x dx x x C = = + 1 1 1 2 2 1 sin 7. dy t t dt t C = = + ( cos( )) sin( ) 3 3 3 8. dy te dt t = cos sin = + e C t sin 9. dy x x dx = (sec ( )( )) 2 5 4 5 = + tan x C 5 10. dy u udu = 4 3 (sin ) cos = + (sin ) u C 4 11. dy x dx x C = = − + 3 3 sin cos 2 3 0 5 = − + = cos( ) , C C y x = − + 3 5 cos 12. dy e x dx e x C x x = = + 2 2 cos sin 3 2 0 1 2 1 0 = + = = + e C C y e x x sin( ) , sin 13. du x x dx x x x C = + = + + ( ) 7 3 5 5 6 2 7 3 1 1 1 5 4 5 4 7 3 7 3 = + + = − = + C C u x x x ,

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Section 6.1 265 14. dA x x x dx x x x x C = + + = + + + ( ) 10 5 2 4 4 9 4 10 4 2 6 1 1 1 4 1 1 4 1 10 4 2 10 4 2 = + + + = = + + + ( ) , C C A x x x x 15. dy x x dx x x x c = + = + + + 1 3 12 12 2 4 1 3 3 1 1 12 1 11 12 1 1 3 1 3 = + + + = − = + + ( ) , C C y x x x 1 0 ( ) x > 16. dy x x dx x x c = = + 5 3 2 5 2 3 2 sec tan / 7 5 0 0 7 5 7 3 2 3 2 = + = = + tan( ) ( ) , tan / / C C y x x 17. dy t dt t C t t = + + = + + 1 1 2 2 2 2 1 ln tan 3 0 2 2 2 1 0 1 = + + = = + + tan ( ) , tan C C y t C t 18. dx t t dt t t t C = + = + + + 1 1 6 6 2 1 ln 0 1 1 6 1 7 6 1 1 = + + + = − = + + ln( ) ( ) , ln C C x t t t 7 0 ( ) t > 19. dv t t e t dt t e t C t t = + + ( ) = + + + 4 6 4 3 2 sec tan sec 5 4 0 3 0 0 0 2 = + + + = sec( ) ( ) , e C C V t e t t t = + + 4 3 2 2 sec π < < π 2 20. ds t t dt t t C = = + ( ) 3 2 3 2 0 1 1 0 3 2 3 2 = + = = ( ) ( ) , C C s t t 21.
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