Business Calc Homework w answers_Part_94

Business Calc Homework w answers_Part_94 - 466 Appendix...

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Unformatted text preview: 466 Appendix A5.3 A B C 4 4 1 3 cos 2 4 3 (if we 5 7 2 9 28. cot 2 34. tan 2 12 0.38, cos 45 0.92; 10.45, D 22.5 choose 2 in Quadrant II); thus sin 1 cos 2 2 1 5 1 cos 2 2 1 (3/5) 2 1 1 5 2 5 1 2 2 (3/5) 2 2 5 sin and cos then A E 0.55, B 0.00, C 86 2 18.48, 7.65, and F 2 or sin 29. tan 2 sin C 1 1 3 and cos 0.55x an ellipse 13.28 0.88, B 3 3 0, an (b) A D 0.00, 35. (a) A C x 1 2 2 10.45y 18.48x 7.65y 86 0, 26.57 0.97; then A cos 45 sin 45 sin 45 cos 45 1 2 y 2 2 2 2 1 ,F 2 2 2 1 ,B 2 0, 0.23, cos 3.12, D 2 1 2 0.74, E 3.12y 2 1.20, and F 0.74x 1.20y 1x y 2 0.88x ellipse 30. tan 2 sin B F 1 ,C 2 E 2 B y 2 2 1 ( 3) 1 2 5 1 (see part (a) above), 2 1 0, F a x 2 2 1 2 y 2 a 11.31 5.65 2.05, 0.30, and x 2a 0 we have 0.10, cos 0.995; then A 2.99, E 0.00, C 7 2 3.05, D 36. Yes, the graph is a hyperbola: with AC 4AC 0 and B 2 4AC 0. 2.05x 3.05y 2 2.99x 0.30y 7 0, a hyperbola 31. tan 2 sin C 4 1 4 4 2 3 53.13 0.89; then A 0, and F 5 26.57 0.00, B 0.00, 0.45, cos 5.00, D 2 0, E 5 0 or y 5.00y 32. tan 2 sin C 1.00, parallel lines 36.87 18.43 0.00, B 49 0.00, 2 12 18 3 2 4 0.32, cos 20.00, D 2 0.95; then A 0, and F 37. The one curve that meets all three of the stated criteria is the ellipse x 2 4xy 5y 2 1 0. The reasoning: The symmetry about the origin means that ( x, y) lies on the graph whenever (x, y) does. Adding Ax 2 Bxy Cy 2 Dx Ey F 0 and A( x)2 B( x)( y) C( y)2 D( x) E( y) F 0 and dividing by the result by 2produces the equivalent equation Ax 2 Bxy Cy 2 F 0. Substituting x 1, y 0 (because the point (1, 0) lies on the curve) shows further that A F. Then Fx 2 Bxy Cy 2 F 0. By implicit differentiation, 2Fx By Bxy 2Cyy 0, so substituting x 2, y 1, and y 0 (from Property 3) gives 4F B 0 B 4F the conic is Fx 2 4Fxy Cy 2 F 0. Now substituting x 2 and y 1 again gives 4F 8F C F 0 C 5F the equation is now Fx 2 4Fxy 5Fy2 F 0. Finally, dividing through by F gives the equation x 2 4xy 5y 2 1 0. 38. If A C, then B B cos 2 (C A) sin 2 0, E 49 20.00y 33. tan 2 sin C 0, parallel lines B cos 2 . 52 78.69 0.77; then A 5.07, E 2 5 3 2 39.35 5.05, B 0.00, 1 0, 39. Then 4 2 2 B B cos 2 0 so the 0.63, cos 0.05, D 2 xy-term is eliminated. 90 x x cos 90 y sin 90 and y x sin 90 y cos 90 x x2 b2 y2 (b) 2 a 6.19, and F 6.19y 1 y 5.05x 0.05y 5.07x a hyperbola (a) y2 a2 x2 b2 1 1 a2 m( y ) y m( y ) 1 x m (c) x (d) y (e) y 2 y 2 mx x mx bx by 1 x m b m Appendix A6 40. 180 x and y (a) (b) x a2 2 467 x cos 180 y sin 180 y x x sin 180 y b2 2 y cos 180 43. (a) B 2 4AC 42 4(1)(4) 0, so the discriminant indicates that this conic is a parabola. (b) The left-hand side of x 2 4xy 4y 2 6x 12y 9 0 factors as a perfect square: (x 2y 3)2 0 x 2y 3 0 2y x 3; thus the curve is a degenerate parabola (i.e., a straight line). 44. (a) B 2 4AC 62 4(9)(1) 0, so the discriminant indicates that this conic is a parabola. 1 1 a2 y b 1 y m( x ) y y m( x ) mx by mx b x2 a2 2 y2 b2 (c) x (d) y (e) y 41. (a) B y 2 mx mx 2 4AC 2x 2x x 1 1 4(0)(0) 0 y(x 1 hyperbola 1) 2x y and we want 2x x 1 (b) The left-hand side of 9x 2 6xy y 2 12x 4y 4 0 factors as a perfect square: (3x y 2)2 0 3x y 2 y 3x 2; thus the curve is a degenerate parabola (i.e., a straight line). 45. Assume the ellipse has been rotated to eliminate the xy-term the new equation is A x the semi-axes are 1 and A 2 AC 4A C 2 0 (b) xy (c) y dy dx 2 (x 1) 2 Cy 2 1 1 the area is C dy dx 2, the slope of y 2 (x 2 1)2 2x 1)2 4 3 (3, 3) is a 1 A 1 C . 4A C 2 4AC B2 x (x Since B2 (because B 4AC B 2 4A C 0) we find that the area is as 3 or x 1; x 3y claimed. point on the hyperbola where the line with slope m y x 3 2 is normal the line is 2(x 1y 3) or y 1 ( 1, 2x 3; 46. (a) A C (A cos2 B cos sin C sin2 ) 2 2 (A sin B cos sin C cos ) A(cos2 sin2 ) C(sin2 cos2 ) A C E sin )2 ( D sin (b) D 2 E 2 (D cos 2 cos ) D 2 cos2 2DE cos sin E2 sin2 D 2 sin2 2DE sin cos E 2 cos2 2 2 2 2 2 D (cos sin ) E (sin cos2 ) 2 2 D E E 1) is a point on the 2 is normal hyperbola where the line with slope m the line is y 1 2(x 1) or y s Appendix A6 2x 3 (pp. 618627) 1. sinh x 1 3 cosh x 1 sinh2 x 4 9 25 5 sinh x , tanh x 16 4 16 cosh x 3 , coth x 5 1 sin x 1 tanh x 4 3 1 3 2 4 [ 9.4, 9.4] by [ 6.1, 6.1] 42. (a) False: let A parabola C 1, B 2 B2 4AC 0 3 4 5 4 5 , sech x 3 1 cosh x (b) False: see part (a) above (c) True: AC 0 hyperbola 4AC 0 B2 4AC 0 4 , and csch x 5 468 Appendix A6 ex 2 e x x 2 2. sinh x 1 4 cosh x 3 16 9 sinh x cosh x 1 tanh x 25 9 4 3 5 3 1 5 , 3 4 , 5 sinh2 x 1 4 2 3 12. cosh2 x 1 x (e 4 sinh2 x e x ex 2 e x x 2 ) (e x e ) (e x 1 (4) 4 x 3 1 3 e 1 ) (e x e x ) 1 (2e x )(2e x ) 4 1 (4e 0) 4 tanh x 13. y 1 cosh x 3 , 5 6 sinh dy x dx 3 6 cosh dy dx 2 cosh x 3 coth x 5 , sech x 4 3 4 14. y and csch x 1 sinh x 1 sinh (2x 2 1) 1 [cosh (2x 2 1)](2) cosh (2x cosh2 x 64 225 8 , 15 1 tanh x 1 sinh x 15 8 1) t 2t 1/2 tanh t 1/2 1 1/2 t (2t 1/2) 2 t 3. cosh x 17 ,x 15 0 sinh x 289 225 8 15 17 15 1 15. y 2 dy dt t tanh 17 2 15 1 1 [sech2(t 1/2)] 2 (tanh t 1/2)(t 1/2 ) sech 8 , coth x 17 15 , and csch x 17 t 1 t 1 tanh t tanh x sinh x cosh x 16. y t 2 tanh [sech2 (t sech2 1 t t 2 tanh t )]( t 2 1 dy dt 1 17 , sech x 8 1 cosh x )(t 2) 1 t (2t)(tanh t ) 2t tanh dy dz dy dz 4. cosh x 169 25 13 ,x 5 0 sinh x 144 25 12 5 13 5 cosh2 x 1 17. y 18. y ln (sinh z) ln (cosh z) (sech )(1 1 sinh x cosh x 12 , 5 12 , coth x 13 5 , and csch x 13 1 tanh x cosh z sinh z sinh z cosh z coth z tanh z tanh x 19. y 1 sinh x 5 12 ln sech ) (sech ) ln sech ) tanh )(1 ln sech )] ln sech ) 13 , sech x 12 1 cosh x dy d sech tanh sech In Exercises 510, graphical support may consist of showing that the graph of the original expression minus the simplified one is the line y 0. 5. 2 cosh (ln x) 2 e e ln x 2 e 2 x4 1 2x 2 2 ln x ( sech sech (sech (sech tanh tanh )(1 (sech (1 e ln x e ln x e ln x 2 1 e ln x e 2 lnx 2 x 1 x tanh )[1 tanh )(ln sech ) ln csch ) csch coth csch 2 ln x 6. sinh (2 ln x) x2 2 1 x2 20. y (csch )(1 dy d (csch ) (1 csch 7. cosh 5x 8. cosh 3x 9. (sinh x 10. ln (cosh x sinh 5x sinh 3x cosh x)4 sinh x) e 5x 2 e3x 2 ex e 5x e 5x 2 e 3x 2 ex 2 e 5x e 5x e 3x ln csch )( csch coth (1 1 coth ) coth ) e 3x e 3x ln csch )(csch ln csch ) (csch e 4x 21. y (csch ln cosh x dy dx coth )(1 e 2 x e x 4 (e x)4 coth )(ln csch ) ln (cosh x ln 1 x) 0 sinh x) ln (cosh2 x sinh2 x) 1 tanh2 x 2 1 sinh x (2 tanh x)(sech2 x) 2 cosh x 11. (a) sinh 2x sinh (x 2 sinh x cosh x sinh x cosh x cosh x cosh x cosh x sinh x sinh x sinh x tanh x (tanh x)(sech2 x) tanh3 x (tanh x)(1 sech2 x) (tanh x)(tanh2 x) (b) cosh 2x cosh (x x) cosh2 x sinh2 x Appendix A6 1 coth2 x 2 cosh x 1 (2 coth x)( csch2 x) sinh x 2 469 22. y ln sinh x dy dv 31. y csch x) 2 cos 1x dy dx 1 1 x2 x sech 1 1 x 1 x 1 x2 1 coth x (coth x)(csch x) 2 2 (coth x)(1 1 x2 1 1 x (1) sech x sech 1 x (coth x)(coth x) 23. y (x (x 2 2 coth x (x (x 2 2 3 x2 sech 1 1 1 x 1) sech (ln x) 2 1) x x 1 e ln x 2x 1) 2 1 x 1) 2 e ln x 32. y dy 2x dx 2 ln x ln x (1 1 x 1 x 1 x 2 sech x x 1 1 x2 1 x 2)1/2 sech (1 1 (1 2 x 1 x2 2 dy dx 1 x x 2)1/2 x 24. y (4x 2 (4x 2 1) csch (ln 2x) 1) 4 1 (4x 2 (4x 2 1) 1) x ) 2 1/2 ( 2x) sech x 1 x x x2 e ln 2x e 1 ln 2x 2x 2 (2x) 4x 4x 2 sech 1 sech 1 x 1 4x 33. y dy dx csch 1 1 2 ln 1 2 1 2 1 2 2 25. y sinh dy dx x sinh 1 (x 1/2 ) 1 x 1 1 x 1/2 2 2 x(1 x) dy d 1 2 x 1 1 2 1 1 1 (x 1/2 )2 ln (1) ln (2) 1 2 2 1 ln 2 1 dy d dy dx 1 2 2 26. y cosh (2 x 1) 1) 1)1/2]2 1/2 cosh (2(x 1 1)1/2 ) 34. y 4x 3 1 (2) dy dx 1 4x 2 1 (x 2 csch 2 (ln 2)2 2 1 (2 ) sec2 x 2 ln 2 1 22 [2(x 1 x 1 35. y 7x 3 1 sinh sec2 x 1 (tan x) sec2 x sec x 27. y (1 ) tanh (1 ) tanh 2 sec2 x 1 2 1 (tan x)2 sec x sec x sec x sec x (sec x)(tan x) sec2 x 1 dy d 1 1 1 1 ( 1) tanh 1 36. y cosh 1 (sec x) dy dx (sec x)(tan x) tan2 x 1 (sec x)(tan x) tan x 1 sec x, 0 dy dx x 2 28. y ( dy d 2 2 ) tanh ( 2 ( 1 ( 1) 1)2 1 37. (a) If y (2 ( 1) 2) tanh 1 tan (sinh x) cosh x cosh2 x C, then 2 ) (2 1 1 ( 1) 1 cosh x sinh2 x sech x, which verifies the 2 2 2 2) tanh ( t 1) (1 1/2 formula. (b) If y sin 1 1 (2 29. y (1 dy dt 1 2 t 2) tanh t) coth 1 t) coth 1 (tanh x) 2 C, then sech2 x 1 (t 1/2 ) tanh x sech2 x sech x dy dx sech x, which verifies the formula. (1 t) (1/2)t 1 (t 1/2 )2 1 ( 1) coth 1 (t 1/2 ) 38. If y dy dx x2 sech 1 x 2 coth t 1 x sech x sech 1 1 x x 1 2 x 1 x2 2 1 2 1 x2 C, then 2x 4 1 x2 x, which verifies the formula. 1 30. y (1 dy dt t 2) coth (1 2t coth t 2) 1 t 1 t2 1 ( 2t) coth 1 t 39. If y dy dx x2 2 coth 1 1 x x 2 1 t x coth x x 2 1 2 1 C, then 1 x2 1 2 x coth 1 x, which verifies the formula. 470 Appendix A6 1 ln (1 2 40. If y x tanh 1 x 1 x2) 1 x2 C, 1 2x 2 1 x2 49. Let u tanh 1 t t tanh t t 1/2 and du t 2 dt t . dy then dx tanh x x 1 x, sech dt C 2 sech u tanh u du 2 sech dt . t which verifies the formula. 41. Let u 2x and du 2 dx. cosh u 2 2( sech u) 50. Let u C cosh 2x 2 t C ln t and du sinh 2x dx 42. Let u sinh 43. Let u 1 sinh u du 2 1 dx. 5 C csch (ln t) coth (ln t) dt t csch u coth u du C x and du 5 x dx 5 csch u 5 cosh u 1 dx. 2 C csch (ln t) 5 sinh u du ln 3 and du ln 3 dx ln 3 C 5 cosh x 5 C 51. Let u sinh x, du e ln 2 2 e ln 4 2 ln 4 15/8 ln 2 3/4 cosh x dx, the lower limit is e ln 2 x 2 x 6 cosh 2 x 12 sinh 2 2 2 ln 4 sinh (ln 2) 12 sinh u C 1 2 12 cosh u du C 3 and the upper 4 4 2 1 4 limit is sinh (ln 4) ln 2 and du ln 2) dx ln 2) C 52. Let u cosh 0 7 ln cosh u C1 cosh (2 ln 2) 4 2 1 4 e 15 . 8 44. Let u 3x 3 dx. 4 cosh u du 3 4 sinh u 3 ln 4 4 cosh (3x 4 sinh (3x 3 C ln 2 coth x dx ln u 15/8 3/4 cosh x dx sinh x 15 3 ln ln 8 4 1 du u 15 4 ln 8 3 ln 5 2 0.916 45. Let u 1 x and du dx. 7 7 x sinh u tanh dx 7 du 7 cosh u x 7 ln cosh C1 7 x/7 x/7 e e 7 ln C1 2 cosh 2x, du 2 sinh (2x) dx, the lower limit is 1 and the upper limit is cosh (ln 4) 17 . 8 1 sinh 2x dx 0 2 cosh 2x 1 17 ln ln 1 2 8 ln 2 ln 2 17/8 1 e ln 4 2 e ln 4 7 ln e x/7 7 ln e x/7 e e x/7 x/7 7 ln 2 C d C1 ln 2 0 tanh 2x dx 1 17/8 ln u 2 1 1 du u 1 17 ln 2 8 0.377 46. Let u 3 and du d 3 . 3 ln sinh u e / 3 ln 2 coth 3 cosh u 3 du sinh u 3 53. C1 C1 ln 4 2e cosh ln 2 d 1) d ln 4 2e e 2 e ln 2 ln 4 d ln 4 (e 2 e 2 e 2 ln 4 2 3 ln sinh 3 ln e 3 ln e 47. Let u sech 2 C1 e e / 3 3 ln e 2 / 3 e 1 8 3 32 ln 2 2 ln 2 2 ln 2 ln 2 ln 2 1 32 2 ln 4 3 32 / 3 3 ln 2 C dx. 2 C1 ln 4 ln 2 2 ln 2 / 3 / 3 0.787 ln2 x x 1 and du 2 1 dx 2 1 C 2 54. tanh u C 0 4e ln 2 sinh (1 e 1 8 d e 2 1 2 2 0 4e 2 0 e 2 e 2 e 0 2 e ln 2 0 d sech u du 2 0 )d tanh x 48. Let u (5 2 2 ln 2 2 ln 2 dx. csch2 u du ( coth u) C 2 ln 2 x) and du x) dx C x) C csch2 (5 coth u coth (5 2 ln 2 1 4 1 ln 4 3 4 0.636 ...
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