Business Calc Homework w answers_Part_19

Business Calc Homework w answers_Part_19 - 91 Section 3.5...

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Unformatted text preview: 91 Section 3.5 (b) Velocity: cos Speed: 0 sin 4 0 m/sec 4 (d) 0 m/sec Acceleration: sin Jerk: sin cos 4 cos 4 2 m/sec2 4 0 m/sec3 4 (c) The body starts at 1, goes to between 2. cos (x h) cos x d cos x lim h dx h→0 cos x cos h sin x sin h cos x lim h h→0 (cos x)(cos h 1) sin x sin h lim h h→0 lim cos x 2 and then oscillates 3 4 Greatest when t Zero when t 4 180 k (e) k 1802 3 4 4 180 180 2 d dx sin x 2 sin x 2 180 1802 180 cos x 3 1803 cos x 2 3 4 d cos x dx 2 k d dx sin x 180 180 180 cos x 2 1802 k cos x 3 d cos x dx 3 25. (a) 2 d dx 2 cos x 2 180 1802 180 sin x 3 1803 26. y [ 360, 360] by [ 0.01, 0.02] The limit is 180 y because this is the conversion factor sin x d csc x csc x cot x dx d ( csc x cot x) dx d d (csc x) (cot x) (cot x) (csc x) dx dx for changing from degrees to radians. (csc x)( csc2 x) (b) csc3 x 27. y [ 360, 360] by [ 0.02, 0.02] sec2 h) sin x h sin x cos h cos x sin h sin x lim h h→0 sin x(cos h 1) cos x sin h lim h h→0 cos h 1 lim sin x lim lim cos x h h→0 h→0 h→0 lim (sin x)(0) 180 sin (x y h→0 (cos x) 180 sin h h→0 h lim (cot x)( csc x cot x) csc x cot2 x d ( tan ) d d (tan ) d This limit is still 0. d sin x dx sin h h sin x d sin x dx 3 k k Greatest (in magnitude) when t (c) lim h→0 180 d cos x dx 180 3 Jerk: 4 lim sin x h→0 2 Greatest (in magnitude) when t Zero when t (sin x) 1 sin x d2 sin x dx 2 Acceleration: Zero when t h→0 (cos x)(0) Speed: cos h h lim h→0 (tan ) d () d tan d ( sec2 tan ) d d d d [(sec )(sec )] (sec2 ) ( ) (tan ) d d d d d (sec ) (sec ) (sec ) (sec ) sec2 d d sec2 2 sec2 (2 tan tan 2 sec2 2)(sec2 ) cos x or, writing in terms of sines and cosines, 2 2 tan cos2 2 cos 2 sin cos3 92 Section 3.6 28. Continuous: 33. Note that g(0) lim g(x) x→0 so b lim g(x) x→0 lim (x x→0 b) lim cos x cos (0) x→0 b. We require lim g(x) x→0 1. The function is continuous if b d sin 2x dx d (2 sin x cos x) dx d 2 (sin x cos x) dx d 2 (sin x) (cos x) dx 1, and g(0), 1. Differentiable: For b 1, the left-hand derivative is 1 and the right-hand derivative is sin (0) 0, so the function is not differentiable. For other values of b, the function is discontinuous at x 0 and there is no left-hand derivative. So, there is no value of b that will make the function differentiable at x 0. 2[(sin x)( sin x) 2 d5 dx 5 d6 dx 6 d7 dx 7 d8 dx 8 sin x cos x sin x cos x cos x 34. d cos 2x dx d [(cos x)(cos x) (sin x)(sin x)] dx d d (cos x) (cos x) (cos x) (cos x) dx dx d d (sin x) (sin x) (sin x) (sin x) dx dx cos x sin x when n 4k 2(2 sin x cos x) cos x 2 sin 2x 35. lim h→0 (cos h h 1) lim h→0 4(249) 3, d 999 cos x dx 999 1) cos2 h 1 1) h→0 h(cos h sin2 h lim 1) h→0 h(cos h sin x. lim h→0 sin h h 0 (1) 2 30. Observe the pattern: d sin x dx 2 d sin x dx 2 d3 sin x dx 3 4 d sin x dx 4 (cos h 1)(cos h h(cos h 1) lim 3 for any whole number k. Since 999 2(sin x)(cos x) 4 sin x cos x sin x Continuing the pattern, we see that dn cos x dx n 2(cos x)( sin x) cos x cos x sin x) 2 cos 2x sin x cos x (cos x)(cos x)] 2 2(cos x 29. Observe the pattern: d cos x dx d2 cos x dx 2 3 d cos x dx 3 4 d cos x dx 4 d dx (cos x) (sin x) sin h lim h→0 cos h 1 0 5 d dx 5 d6 dx 6 d7 dx 7 d8 dx 8 cos x sin x cos x sin x sin x cos x d (A sin x dx d (A cos x dx 36. y sin x sin x y sin x cos x B cos x) A cos x B sin x Solve: sin x B sin x) B cos x y ( A sin x sin x A sin x B cos x) 2A sin x Continuing the pattern, we see that dn sin x dx n At x cos x when n 4k 2 1 for any whole number At x , this gives 0, we have k. Since 725 4(181) 1, d 725 sin x dx 725 32. (a) Using y x, sin (0.12) 0.12. (b) sin (0.12) 0.1197122; The approximation is within 0.0003 of the actual value. 0, so B B cos x) sin x sin x 1 . 2 1, so A 2B 1 and B 2 Thus, A cos x. 31. The line is tangent to the graph of y sin x at (0, 0). Since y (0) cos (0) 1, the line has slope 1 and its equation is y x. 2A sin x 2B cos x (A sin x y 0. 0. s Section 3.6 Chain Rule (pp. 141–149) Quick Review 3.6 1. f (g(x) f (x 2 1) sin (x 2 1) 2. f (g(h(x))) f (g(7x) f ((7x )2 sin [(7x)2 1] sin (49x 2 3. (g h)(x) g(h(x)) 4. (h g)(x) 5. f g(x) h(x) h(g(x)) f x2 1 7x (7x)2 g(7x) h(x 2 sin 1) 1) 1) x2 1 7x 7(x 1 2 49x 2 1) 7x 2 1 7 Section 3.6 6. cos x 2 7. g(cos x) 3 cos2 x g(3 cos2 x) 2 9. g(h(cos x)) g(h( f (x))) dy dx d (x dx f (27x 4) cos 3x 10. cos 2 3x f (g(3x 2)) f (g(h(x))) dy dx d sin (3x dx [cos (3x 2. dy dx 1) 2 f ( 3x 2 f (h(h(x))) 10. 2) dy dx cot x) cot x) d sin (7 dx [cos (7 1 d dx 3 cos (3x 5x) [cos (7 5x)]( 5) 1) 5 cos (7 11. 5x) dy dx dy dx d cos ( dx [ sin ( 3x)]( 4. dy dx d tan (2x dx [sec2 (2x 5. 6. dy dx [ sin ( 3x)] ( 3) ( 5 sin x 3)](2 3x 2) x 3)] (2x 2 2 d 5 cot 5 csc2 x x dx 10 22 2 5 csc ( 2x ) csc2 x x2 dy dx d sin x dx 1 cos x sin x 2 1 cos x 2 2 2 1 sin x cos x 2 2 (1 cos x) sin x cos x cos x 1 sin x cos x 1 (1 2 (1 1 d dx sin x (1 dy dx d cos (sin x) dx dy dx cos x) cos x) d sec (tan x) dx 3 cos2 x sin x 5)3 (2x 5)3(2) 5)4 d dx 5)4 (x 3) (2x d dx 5)3 5)3[8x 5) 3x 2(2x 3x 2(2x 5)4(3x 2) 5)4 5)4 3(2x (2x 5)] 3 5) (14x 15) d (sin3 x tan 4x) dx d d (sin3 x) (tan 4x) (tan 4x) (sin3 x) dx dx d d (sin3 x)(sec2 4x) (4x) (tan 4x)(3 sin2 x) (sin x) dx dx (sin3 x)(sec2 4x)(4) (tan 4x)(3 sin2 x)(cos x) 2 cos x sin x cos x)2 4 sin3 x sec2 4x cos x cos x)2 14. 1 cos x dy dx d (4 dx sec x 1 4 2 sec x 2 d (sin x) dx d dx sec (tan x) tan (tan x) (tan x) sec (tan x) tan (tan x) sec2 x d dx (3 cos2 x) (cos x) 5)4] sec x 2 sec x 2 sec x dy dx d dx d3 [x (2x dx d (x 3) (2x dx x (2x 13. csc2 x) csc x csc x cot x x cos x 2 2 [ sin (sin x)] x) cot x) ( cot x csc x 6 sin (sin x) cos x 8. (csc x x) (sin x) x 2(2x 2 sin x (1 cos x)2 7. 6 8x 3(2x d (2x 1) dx 2 x cos x)(cos x) (sin x)( sin x) (1 cos x)2 1 sin x 2 1 cos x d sin x dx (1 (1 x 3) sin x d sin x cos x dx 1 cos x 1 dy dx x 3) 3x 2) sec2 (2x (2 dx (x 3)(4)(2x 12. d dx [sec2 (2x 2d (x 3)(4)(2x 3x) 3 sin ( 3x) x 3) (x cos3 x) 5x) d dx 3x) cot x)2 d (sin 5 x dx 5 sin 3. dx 1 (csc x)(cot x csc x) (csc x cot x)2 1) d 5x)] (7 dx 3d 1 1 d (csc x dx (csc x 1)] (3x x) 2x f (h(3x 2)) 2 [cos (3x 1)](3) 3 x) Section 3.6 Exercises 1. 2(x (csc x f (3(3x 2)2) 2 2 x) 2(x 3(cos x 2) 3( cos x 2)2 2) h(g(cos x)) h(g( f (x))) 8. 3 cos x 6 h( cos x 9. cos 27x 4 g( f (x)) 93 3 sin2 x cos x tan 4x tan x) d (sec x tan x dx (sec x tan x tan x sec x tan x sec x sec x tan x tan x tan x) sec2 x) 94 15. Section 3.6 dy dx 3 d dx 19. 2x 1 dy dx d (1 dx cos2 7x)3 ( 2x 2x ( d dx d dx 1)(0) 2x d (2x 1 dx 1 3 2 2x 2x 1 3 1) 20. dy dx 3 16. dy dx 1) 2x 1) 1 21. d x ) (x) dx d x( dx 1 21 21 x) x (1 x 2)( 1 (1 x2) 2 x sin x 2) 3 sin x2 (2x) 22. x2 2 (1 d (1 dx x2 1 x 1 3t x) 2 ds dt 2 3t 2 d [t cos ( dt d (t) [cos ( dt 23. d sin (3x dx 2) 2 sin (3x 2) cos (3x 2) (3x 2 sin (3x 2) cos (3x 6 sin (3x 2) cos (3x 2) 3 sin (6x 4) ds dt 2)(3) d (1 dx cos 2x)2 2) d dx 2) 4t)] d dt d dt cos ( 4t) (t) 4t)] ( 4t) 4t)]( 4) cos ( cos 2x) (1 d dx d dt 4 3 4 3 4 cos ( cos ( 4t)(1) 4t) 4t) 4 4 sin 3t cos 5t 3 5 d 4 d (cos 3t) (3t) ( sin 5t) (5t) dt 5 dt 4 (cos 3t)(3) ( sin 5t)(5) 5 4 cos 3t ds dt d sin dt 3 cos 2 3 cos 2 25. dr d d tan (2 d cos 2x) cos 2x)( sin 2x) (2x) 2(1 4t) 24. d dx 2(1 2(1 cos 2x)( sin 2x)(2) 4(1 4t)] sin 2a. 2) 2 sin (3x dy dx 3t t[ sin ( x 2) 17. The last step here uses the identity 2 sin a cos a 18. d dt 2 t[ sin ( 3/2 d sin2 (3x dx sec2 5x 3t ( 3) 2 4t sin ( dy dx 1/2 3t 2 1 x 1 x2 d cos 2 dt x) x )(1) 1 tan 5x 22 2 1 ds dt 5 2 or (tan 5x) sin x2 (1 ( 1 d tan 5x 2 tan 5x dx 1 d (sec2 5x) (5x) dx 2 tan 5x 1 2 tan 5x) 2 tan 5x 5 sec2 5x 2 (1 d ( dx 2 1 cos2 7x)2 cos 7x sin 7x (sec 5x)(5) 3/2 x d dx d dx 42(1 1 3(2x cos2 7x)2(2 cos 7x)( sin 7x)(7) d dx 1 2x (2x cos2 7x)2(2 cos 7x)( sin 7x) (7x) 1 cos2 7x) (2) 2x 2 1) cos2 7x)2(2 cos 7x) (cos 7x) 3(1 1)2 1) (3) cos2 7x)2 (1 3(1 3( 3(1 3(1 d dx ( 2x sec2 (2 sin 5t 3 7 t cos t 2 4 7 d3 d7 t t sin t t 4 dt 4 dt 2 3 7 7 t sin t 2 4 4 ) )( 1) sec2 (2 sec2 (2 ) d (2 d ) ) cos 2x)(sin 2x) 26. dr d d (sec 2 tan 2 ) d d d (sec 2 ) (tan 2 ) (tan 2 ) (sec 2 ) d d d d (sec 2 )(sec2 2 ) (2 ) (tan 2 )(sec 2 tan 2 ) (2 ) d d 2 sec3 2 2 sec 2 tan2 2 Section 3.6 27. dr d d d 1 sin sin 1 2 cos dr d 2 (1 sec ) 2 ( sec ) d (sec ) d sec 1 (2 ) ( g (x) 1 (2 ) (sec y 2 (4) sec 2 sec 2) d tan x sec2 x dx d d sec2 x (2 sec x) (sec x) dx dx 10 36. f (u) 3 csc (3x y 3[2 csc (3x 18 csc (3x 32. y y d 9 tan dx x 3 sec2 3 d 3 sec2 dx x 6 sec 3 x 2 sec2 3 (1) csc2 ud u 10 du 10 5 x 2 f (5)g (1) 5 2 2 5 2 4 2 (cos u) d 2(cos u) 3 cos u du 2 sin u cos3 u 1) 1 1) 37. f (u) d dx 1)] (3x 9 sec2 xdx 3 dx 3 3 2 sec x x tan 3 3 x tan 3 x 3 d x sec dx 3 dx dx 3 (u 2 1) 1) cot (3x 1) sec 1 g ( 1) 2 2 1 () 3 5 1) cot (3x x 3 f 2 1)][ csc (3x x 3 x) x)2 (1 f (g(1))g (1) d [u du 1)] 1) cot (3x 3[2 csc (3x (1 d ( x) dx 1 1 1 ( f g) f (g( ))g 4 4 4 1 f g 4 4 1)] [ csc (3x 2 1) (3x d 1)] csc (3x dx 3[2 csc (3x dx 1 ( 1) 1) d [ 3 csc2(3x dx 2d 2 d dx csc2 (3x 5 2 1 u2 x) 1 2 g (x) 2 csc2 x cot x 1) 1 2 u f (g( 1))g ( 1) csc2 1 ( 2 csc x)( csc x cot x) 2 (5) 1 1 d cot x csc2 x dx d d ( csc2 x) ( 2 csc x) (csc x) dx dx d cot (3x dx 1 4 10 2 sec x tan x 31. y 2 x) ( f g) (1) 2 y x) ) u d cot 10 du u csc2 10 10 d g (x) (5 x) dx (2 sec x)(sec x tan x) 30. y f (1)g (1) (1 1 35. f (u) sec sec 2 u ( f g) ( 1) d (2 ) d tan ) sec )(tan ) sec ( tan 29. y d (1 du d (1 dx 34. f (u) sec ) d ( d f (g(1))g (1) sin ) sin d (2 d (2 ) ( f g) (1) sin 2 28. g (x) sin ( cos d5 (u 1) 5u 4 du 1 d ( x) dx 2x 33. f (u) sin 2 d d (sin ) (sin ) ( ) d d 1 2 d ( sin ) d 1)](3) g (x) 2u d du u 2 1 (u 2 1) d (2u) du (u 2 (2u) 1) d2 (u du 1) 2 1)(2) (2u)(2u) 2u 2 2 (u 2 1)2 (u 2 1)2 d (10x 2 x 1) 20x 1 dx ( f g) (0) f (g(0))g (0) f (1)g (0) (0)(1) 0 95 ...
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This document was uploaded on 10/31/2011 for the course MAC 2311 at University of Florida.

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