Business Calc Homework w answers_Part_47

Business Calc Homework w answers_Part_47 - Chapter 5 Review...

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Unformatted text preview: Chapter 5 Review 49. y 1 x 1 x2 2x y 2 b (b) Let h S2n a 2n h [y 30 1 y(1) 1 1 y (1) 1 dt t 1 2 1 3 0 1 1 [h(y0 3 2 h( y2i 2t dt x t2 4 (x 2 4 1 1) x2 4 1 [2.5 2(2.4) 24 29 2.42 gal 12 4y2i 2 h(y2 4y2n 1 h. 12 54. (a) g(1) 2(2.4) 2 MRAMn 1 f (t) dt 2 1 (c) g( 1) 24.83 mi/gal (2)(1) 2704 ft. When her parachute opens, her 4296 ft. For t SA(t) 6144 6208 16 6208 388 sec. For t 45 4296 16(t so B lands at t 5224 16 53. (a) Area of the trapezoid 58) 16t, so A lands at 13 58 sec, B’s 5224 0 and the absolute minimum is 3 1 f (t) dt f (t) dt 3 1 (2)2 2 The range of g is [ 2 , 0]. 55. (a) NINT(e x 2/2 , x, 10, 10) , x, 2.506628275 20, 20) x 2/2 2.506628275 16t, (b) The area is 2. 326.5 sec. B lands first. 56. First estimate the surface area of the swamp. 1 (2h)(y1 2 (2h)y2 h(y1 h(y1 2(2hy2) (f) g (x) f (x), f (x) 0 at x 1 and f (x) is not defined at x 2. The inflection points are at x 1 and x 2. Note that g (x) f (x) is undefined at x 1 as well, but since g (x) f (x) is negative on both sides of x 1, x 1 is not an inflection point. NINT(e Area of the rectangle y3) (e) g ( 1) f ( 1) 2 The equation of the tangent line is y ( ) 2(x 1) or y 2x 2 1 position is given by SB(t) 1 (2)2 4 (d) g (x) f (x); Since f (x) 0 for 3 x 1 and f (x) 0 for 1 x 3, g(x) has a relative maximum at x 1. g( 3) 4 sec, A’s position is given by t f (t) dt 1 (g) Note that the absolute maximum is g(1) (c) Let t represent the number of seconds after A jumps. 4) 1 1 f (t) dt 1 (b) The distance B falls in 13 seconds is 16(t . 0 3 (b) g(3) 2704 Tn 3 f (t) dt 2.3] 6144 ft. altitude is 7000 … 1 … When her parachute opens, her altitude is 1 (32)(132) 2 y4) 1 12 52. (a) Using the freefall equation s gt from Section 3.4, 2 1 the distance A falls in 4 seconds is (32)(42) 256 ft. 2 256 4y3 y2n )] 1 1 6400 2 the ith of n rectangles plus the area of the ith of n trapezoids, S2n 2(2.3) 12 (b) (60 mi/h) h/gal 29 2y2n y2i ) is equal to twice the area of 1 3 1 51. (a) Each interval is 5 min … 2y4 Since each expression of the form 50. Graph (b). x y2) 2 4y3 y2n] 1 4y1 h(y2n Thus, it satisfies condition ii. y 2y2 4y2n 1 1 1 2 . 4y1 Thus, it satisfies condition i. 231 y3) 2hy2 4y2 y3) h(y1 y3) 20 [146 2 13] 2(122) 2(76) 2(54) 8030 ft2 (5 ft)(8030 ft2) 1 yd3 27 ft3 1500 yd3 2(40) 2(30) 2. 232 Section 6.1 57. (a) V 2 (V max)2 sin2 (120 t) 5. Again h 1 1 av(V ) 7 2 2 (Vmax) 2 2 sin (120 t) dt 240 21 (Vmax)2 2 2 2 1 2 Vmax ,3 1 2 ,5 1 2 , ..., 19 1 2 2 1 2 339.41 volts Chapter 6 Differential Equations and Mathematical Modeling 1 2 3 2 7 ,0 2 x 1, x 3 2 7 ,1 2 x 2, x 5 2 7 ,2 2 x 3, 1 2 y3 1 2 1 3 y2 s Section 6.1 Antiderivatives and Slope Fields (pp. 303–315) x 2 y1 Exploration 1 2 The 10 graphs are graphs of the functions. 2 2 19 . From left to right the 2 slopes of the line segments are (Vmax) 0 2 (b) Vmax 2 (Vmax) sin (120 t) dt (Vmax)2 Vrms 135 222 each line segment are , , , ..., 1 0 1 1. The y-coordinate of the midpoint of each line segment is . The x-coordinates of the midpoint of Using NINT: 2 k 5 2 1 Constructing a Slope Field 1. As i and j vary from 1 to 10, 100 ordered pairs are produced. Each ordered pair represents a distinct point in the viewing window. 2 y10 19 2 x 19 2 1 7 ,9 2 x 10. 2. The distance between the points with j fixed and i r and i r 1 is the distance between their x-coordinates. Xmin 2(r (Xmin 1) Xmin) 1 h 2 (2r Xmin 2 1 (2r 2r 1) 1) h 2 h 2 [0, 10] by [0, 10] h 3. The distance between the points with i fixed and j r and j r 1 is the distance between their y-coordinates. Ymin (2(r k 2 1) Ymin) (Ymin 1) (2r Ymin 2 1 (2r 2r 1) 1) k 2 k 2 6. For each line segment in part (5), make a column of parallel line segments as in part (4). 7. WL Quick Review 6.1 1. 100(1.06) k $106.00 2. 100 1 4. Here h k 0.06 4 4 3. 100 1 0.06 12 12 4. 100 1 0.06 365 365 1. Each line segment in the third column has 4 7 slope , because the x-coordinate of the midpoint of each line segment is 2.5. The y-coordinates are , , , …, 135 222 19 . 2 The 10 graphs are graphs of the functions y $106.14 $106.17 $106.18 n ,2 2 x 3, for n 1, 3, 5, …, 19. The length of the line segment can be increased or decreased by adjusting the restriction 2 [0, 10] by [0, 10] x dy dx d sin 3x dx (cos 3x)(3) 6. dy dx 5 d tan x 2 dx sec 2 x dy dx d Ce 2x dx (Ce 2x)(2) 8. 2.5) 5. 7. 4 (x 7 dy dx d ln (x dx 2) 5 2 3 cos 3x 5 2 5 5 sec 2 x 2 2 3. 2Ce 2x 1 x 2 Section 6.1 9. 6. 1 x dx 3 ln x 3 C Check: d [ln x dx [0.01, 5] by [ 3, 3] By setting the left endpoint at x 0.01 instead of x 7. (x 5 0, we avoid an error that occurs when our calculator attempts 1 to calculate NINT , x, 1, 0 . The graph appears to be the x same as the graph of y 3 ln x. 6x 8. ( x 3 9. x x 3 x6 6 3) dx 5 t2 e t/2 1 C] 3x 2 2 x 1) dx 2e t/2 10. [ 5, 0.01] by [ 3, 3] 11. By setting the right endpoint at x 0.01 instead of x 43 t dt 3 1 dx x3 x3 (x 3 1 , x, x to calculate NINT 1, 0 . The graph appears to be the same as the graph of y 1. (x 2x 13. x3 3 1) dx x 2 x C 3 dx dx 3 x2 x 4 2. ( 3x ) dx x2 C 3 x 2x 1 15. 17. d (x 3 dx C) 3x x) dx (x 2 4x x3 3 1/2 ) dx 8 3/2 x 3 Check: d x3 dx 3 4. (8 8 3/2 x 3 cos 1 x 1 C 1 x 1 4x e 4 2 x 2 8x 4x 1/2 x csc x 2 C dx cos 3x 3 C C C 2 ln x e cos 5x 5 tan 5r cot 7t 7 x C cos2 x dx 8 csc x cot x C 22. sin 2 x dx Check: d 1 4x e dx 4 x 2 2 sec t sin 5x 2 ln x 21. 5. e 4x dx C 3 cos x sin 20. csc 2 7t dt C C) ) dx C x dx 2 1/3 3 2/3 x 2 sin 3x) dx Check: csc x x 3 4/3 x 4 19. 5 sec 2 5r dr csc x cot x) dx d (8x dx (x 1/3 x 1/3 2 x C 1 ln x C 4 18. 4 2 C 2 1 2x 2 dx 1 2/3 x dx 3 C 2 x x 16. 2 sec t tan t dt C Check: 3. (x 3 14. (3 sin x Check: 2 1 x ln ( x). Section 6.1 Exercises 2 3 12. C x 3) dx x 4 x4 4 we avoid an error that occurs when our calculator attempts C C t 4/3 4 0, 1 5 t 4 1/3 t dt 3 x ) dt 5t t/2 C 2 5t 10. 2e x2 2 2 (e t/2 dt 3x 23. tan 2 d dx 2x e 2 C C C 1 cos 2x dx 2 1 cos 2x dx 2 2 sin 2x x C 4 2 1 1 2 e 4x 2x cos 2x dx 2 cos 2x dx 2 (sec 2 1) d x 2 tan sin 2x 4 C C 233 234 Section 6.1 24. cot 2 t dt (csc 2 t 1) dt cot t t C dy dx 29. dy dx dx 25. (a) Graph (b) (b) The slope is always positive, so (a) and (c) can be ruled out. 1 when x is positive since slope dy dx and (c) don’t show this slope pattern. y 2x 1 2 x. Graphs (a) tan 1 negative, zero slope when x is zero and negative slope dy dx dx 1 4 C C 4 C Solution: y (2x tan x 2 x2 1) dx x C , 0 by [ 8, 8] dy dx x 22 30. dy dx dx 2 y 1 x2 2/3 x 2/3 3x 1/3 dx C Initial condition: y ( 1) 5 3( 1)1/3 5 3C 2C y 3x 1/3 2 [ 4, 4] by [ 3, 3] dy dx C 1 Initial condition: y (2) 0 22 2 C 02C 2C Solution: y x 2 x 28. tan x Initial condition: y (b) The solution should have positive slope when x is dy dx sec 2 x dx y 26. (a) Graph (b) 27. sec 2 x 5 C x dy dx dx (x 2 x) dx y x 1 x2 2 y x2 2 1 x 1 1 1 2 Solution: y 2 2 3 2 [ 4, 4] by [ 8, 4] C Initial condition: y (2) 2 C 1 2 1 dy dx dy dx dx C C y C x2 2 31. 1 x [ 6, 6] by [ 4, 4] 1 2 9x 2 (9x 2 3x 3 4x 4x 2x 2 5 5) dx 5x C Initial condition: y ( 1) 0 0 3( 1)3 2( 1)2 5( 1) C 0 10 C 10 C Solution: y 3x 3 2x 2 5x 10 Section 6.1 dy dx 32. cos x dy dx dx (cos x y sin x cos x dy dt 2e dy dt dt C y t ln 2 0 2 2 0 1 y 0 C dy dx dx y t 1 1 x cos 1) d sin sin 0 C Initial condition: y(e 3) 0 0 ln (e 3) C 03C 3C Solution: y ln x 3 0 3 C2 C2 d 2y dx dx 2 2 dy dx dx 2 dy dx 1 dx x 1 C2 Solution: y 36. ln x dy d ( cos 3 dy dx C1 Initial condition: y (0) C 2e 0 C1 dy d d C 3 Solution: y C1 C1 C 1 34. cos 0 First derivative: Initial condition: y (ln 2) 2e 0 dt t d cos 1 2e 0 sin Initial condition: y (0) t 2e sin d 2y d d2 dy d sin x) dx Initial condition: y ( ) 1 1 sin cos C 11C 0C Solution: y sin x cos x 33. d 2y d2 35. sin x sin 2 3 6x (2 6x) dx 3x 2 2x C1 Initial condition: y (0) 4 4 3(0)2 2(0) 4 C1 First derivative: dy dx dx y x2 dy dx 4x 02 1 03 4) dx C2 Initial condition: y (0) 1 3x 2 2x 3x 2 (2x x3 C1 1 C2 4(0) C2 Solution: y x2 x3 4x x3 x2 4x 1 or y 1 4 235 ...
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