Business Calc Homework w answers_Part_92

Business Calc Homework w answers_Part_92 - 456 Appendix...

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Unformatted text preview: 456 Appendix A5.1 L and 0 M) M g(x) M by the M, for . lim f (x) xc 18. If L, M, and c are real numbers and lim f (x) xc 23. By the Quotient Rule, lim (L L g(x) lim g(x) xc M, show that for any x g(x) M c (L g(x) xc 0, there is a g(x) f (x) L f (x) xc g(x) lim g(x) xc f (c) . g(c) such that 0 Proof: f (x) f (x) L f (x) M) f (x) 24. From the continuity of g, for any such that 0 f (x) f (c) 0 there is a g( f (c)) f (c), so 0 . g( f (x)) The second inequality also holds when f (x) f (x) f (c) g( f (x)) g( f (c)) 0 such that f (c) triangle inequality. Since lim f (x) any 0 0 there exist x x c c 1 2 1 L, lim g(x) xc . But from the , 2 such that L M 2 2 continuity of f, there is a and . 0 x c f (x) 0 such that c g( f (x)) f (x) g(x) 2 . So for any 0 there is a 0 x Choose 0 x c min { 1, }. Then L g(x) g(x) (L M M) f2(x)] 2 2 g( f (c)) , which means f (x) f (x) that g( f (x)) g f is continuous at c. . s Appendix A5.1 19. lim [ f1(x) xc f2(x) f3(x)] xc lim [ f1(x) L3 (pp. 593606) 1. (x (x x2 h)2 (y 0)2 (y (y 2)2 y 5 L1 L2 L3, by two applications of the Sum Rule. To generalize: Step 1 (n 1): lim f1(x) xc k)2 2)2 4 a2 22 L1 as given. f2(x) ... fk(x)] Step 2: Suppose lim [ f1(x) L1 xc (0, 4) L2 ... xc Lk. Then ... f2(x) ... fk 1(x)] ... fk(x)] Lk 1 lim [ f1(x) f2(x) (0, 2) lim[ f1(x) L1 20. Step 1 (n L2 3 (0, 0) 3 x Lk 1, by the Sum Rule. L1, as given. 2. (x [x (x h)2 (y k)2 a 2 ( 1)]2 (y 5)2 ( 1)2 (y 5)2 10 y 10 (0, 8) 1 1): lim f1(x) xc xc 10)2 Step 2: Suppose lim ( f1(x) f2(x) ... fk(x)) L1 L2 ... Lk. Then xc lim ( f1(x) f2(x) ... fk 1(x)) xc lim (f1(x)) f2(x) ... fk(x)) Lk L1 L2 ... L k 1 (1, 5) , by the Product Rule. c c ... c 1 21. lim x n xc lim (x x ... x) xc cn 5 (0, 2) 5 x 22. lim f (x) xc xc xc n n factors lim (anxn an 1xn xc n factors ... ax 1 xc a0) xc lim anx xc n lim an 1x aa 1 1 xc n 1 1 ... ... a0 lim a1x a1 lim x xc lim a0 xc an lim x n an c an lim x n lim a0 1 cn ...a c 1 f (c), where in addition to the items given in the problem, the Constant Multiple Rule was used (to move the coefficients out of the scope of the limit signs). Appendix A5.1 3. Complete the squares. x 2 y 2 4x 4y 4 0 x 2 4x 4 y 2 4y 4 4 (x 2)2 (y 2)2 22 Center ( 2, 2); radius 2 y 457 13. x2 2 y2 1c 2, 0) 2 1 1 foci are ( 1, 0); vertices are ( 14. y2 4 x2 1c 4 1 5 foci are (0, 2x 5); vertices are (0, 4 2); asymptotes are y 12 p 15. y 2 2 12x 4p 3 y 4 3; focus is (3, 0), directrix is x 2 x 4 2 y 2 = 12x F 4. Complete the squares. x 2 y 2 4x 4y 0 x 2 4x 4 y 2 4y 4 8 (x 2)2 (y 2)2 (2 2)2 Center (2, 2); radius 2 2 y 5 x = 3 4 2 2 2 2 4 4 x 16. y 5 x 4x 2 x 2 1 y 4p 4 1 p 4 1 16 1 ; 16 focus is 0, 1 , directrix is y 16 y 5. The circle with center at (1, 0) and radius 2 plus its interior. 6. The region exterior to the unit circle and interior to the circle with center at (0, 0) and radius 2. 7. y 2 8x 4p 8p 2 4p 1; focus is ( 1, 0), 17. 16x 2 6p 3 ; focus is 0, 2 3 , 2 1 4 1 4 1 8 y = 4x2 F 1 4 2; focus is (2, 0), y= 1 16 1 8 1 4 x directrix is x 8. y 2 4x 4p 1 directrix is x 9. x 2 25y 2 a2 400 b2 x2 25 y2 16 1 3 6y 4p 3 2 c 25 16 foci are ( 3, 0) y directrix is y 10. x 2 2y 4p 2p 1 2 1 1 ; focus is 0, , 2 2 2 F1 F2 2 2 4 directrix is y x2 11. 4 y2 9 1c 4 9 13 4 2 x foci are ( asymptotes are y 12. x2 4 y2 9 13, 0); vertices are ( 2, 0); 3 x 2 1c 3) 9 4 5 foci are (0, 5); vertices are (0, 458 Appendix A5.1 2y 2 a2 6 b2 1) y 2 F2 2 1 F1 5 2 V F 5 1 2 x x = 1 x 18. 3x 2 c x2 2 y2 3 1 2 1 24. Vertices: ( 3, 0), Asymptotes: y b and a 4 b 3 4 (3) 3 x 4 9 2 3 4 xa 3 y2 1 16 3 foci are (0, 25. (a) y 2 8x 4p 8 p 2 directrix is x 2, focus is (2, 0), and vertex is (0, 0); therefore the new directrix is x 1, the new focus is (3, 2), and the new vertex is (1, 2). y 5 (y + 2)2 = 8(x 1) 19. Foci: ( b2 20. Foci: (0, b2 a2 2, 0), Vertices: ( 2, 0) a c2 4 ( 2)2 x2 2 4 2, c y2 2 2 1 26. (a) x2 16 y2 9 1 center is (0, 0), vertices are ( 4, 0) a2 b2 7 foci are ( 7, 0) 4), Vertices: (0, 25 16 9 a2 x, x 9 2 5) a y 25 2 5, c 1 1 1 4 and (4, 0), c and ( 2; (b) 8 7, 0); therefore the new center is (4, 3), the 21. x 2 y 2 1 c asymptotes are y foci are ( 2, 0) y 4 2 F2 2 2 4 b2 new vertices are (0, 3) and (8, 3), and the new foci are (4 7, 3). y y= y= F1 4 2 x x 4 x V1 F1 (4, 3) F2 V2 x 8 22. 8y 2 c 2x 2 a2 16 b2 y2 2 x2 8 1 27. (a) 8 10; 10) 2 x2 16 y2 9 1 center is (0, 0), vertices are ( 4, 0) 3x , 4 asymptotes are y y 5 y= x , foci are (0, 2 and (4, 0), and the asymptotes are y c a2 b2 25 5 foci are ( 5, 0) and (5, 0); therefore the new center is (2, 0), the new vertices are ( 2, 0) and (6, 0), the new foci are ( 3, 0) y= 2 x F2 x 2 and (7, 0), and the new asymptotes are y x 3(x 4 2) . 5 F1 (b) 10 y y = 3(x 2) 4 (x 2)2 y2 =1 16 9 F1 C V1 V2 F2 10 23. Foci: (0, and a b a 2), Asymptotes: y 1a b c2 1 y2 a2 x2 1 xc b2 2 2a 2 10 x 2a 2 2 1b 10 y = 3(x 2) 4 Appendix A5.1 28. Original parabola: y 2 4x; vertex is (0, 0); y 2 4x 4p 4 p 1, so focus is (1, 0) and directrix is x 1. New parabola: (y 3)2 4(x 2); vertex is ( 2, 3), focus is ( 1, 3), directrix is x 3. 29. Original ellipse: (0, 3); c 2 9 x 6 2 459 37. Volume of the Parabolic Solid: b/2 V1 0 2 xh x2 2 x4 b2 b/2 0 4h 2 x dx b2 hb 2 ; 8 1 3 b/2 2 h 0 x 4x 3 dx b2 y 9 2 2 h 1; vertices are (0, 3) and 3 foci are (0, 3); 6c Volume of the Cone: V2 therefore V1 3 V 2 2 b 2 h 2 1 3 b2 h 4 hb 2 ; 12 center is (0, 0). New ellipse: (x 6 2)2 (y 9 1)2 1; vertices are 1 3); center is 38. (a) y 2 kx x y2 ; the volume of the solid formed by k ( 2, 2) and ( 2, ( 2, 1). 4); foci are ( 2, revolving A about the y-axis is kx x2 30. Original hyperbola: 4 y2 5 1; vertices are (2, 0) and 3 foci are (3, 0) and 5 2 V1 0 y2 2 dy k kx k2 0 y 4 dy x 2 kx ; the 5 ( 2, 0); c 2 4 5c volume of the right circular cylinder formed by revolving the rectangle about the y-axis is V2 x2 kx the volume of the solid formed by V2 V1 ( 3, 0); center is (0, 0); asymptotes are y New hyperbola: (x 4 5 2 2)2 (y 5 2)2 x. 1; vertices are (4, 2) and (0, 2); foci are (5, 2) and ( 1, 2); center is (2, 2); asymptotes are y (x 2) 2. revolving B about the y-axis is V3 4 x 2 kx . Therefore we can see the ratio of V3 to V1 is 5 31. Original hyperbola: y 2 x 2 1; vertices are (0, 1) and (0, 1); c 2 1 1 c 2 foci are (0, 2); center is (0, 0); asymptotes are y x. New hyperbola: (y 1)2 (x 1)2 1; vertices are ( 1, 2) and ( 1, 0); foci are ( 1, 1 2); center is ( 1, 1); asymptotes are y (x 1) 1. 32. x 2 4x y 2 12 x 2 4x 4 y 2 12 4 (x 2)2 y 2 16; this is a circle: center at C( 2, 0), a 4 33. 2x 2 2y 2 28x 12y 114 0 x 2 14x 49 y 2 6y 9 57 49 9 (x 7)2 (y 3)2 1; this is a circle: center at C(7, 3), a 1 4y 34. x 2 2x 4y 3 0 x 2 2x 1 (x 1)2 4(y 1); this is a parabola: V( 1, 1), F( 1, 0) 35. x 2 (x 5y 2 2)2 4x 1 x2 5y 2 5 (x 5 4:1. (b) The volume of the solid formed by revolving B about x the x-axis is V1 0 ( kt)2 dt x k 0 t dt kx 2 . 2 The volume of the right circular cylinder formed by revolving the rectangle about the x-axis is V2 ( kx)2x kx 2 the volume of the solid 3 1 formed by revolving A about the x-axis is V3 V2 V1 kx 2 kx 2 2 kx 2 . Therefore the 2 4x 4 2)2 5y 2 y2 4 1 ratio of V3 to V1 is 1:1. 1; this is an ellipse: the center is ( 2, 0), the vertices are ( 2 5, 0); c a2 b2 5 1 2 the foci are ( 4, 0) and (0, 0) 36. x 2 y 2 2x 4y 4 x 2 2x 1 (y 2 4y 4) 1 (x 1)2 (y 2)2 1; this is a hyperbola: the center is (1, 2), the vertices are (2, 2) and (0, 2); c a2 b2 1 1 2 the foci are (1 2, 2); the asymptotes are y 2 (x 1) 460 Appendix A5.2 p, and let P(x, y) be a 4px. Now (b) Around the y-axis: 9x 2 x V 2 3 39. Let P1( p, y1) be any point on x point where a tangent intersects y 2 y2 4px 2y dy dx 4y 2 36 x 2 4 4 2 y 9 4 2 4y 0 4p y x dy dx y1 ( p) 2p ; then the slope of a y dy dx 2p y tangent line from P1 is y2 y2 1 2 4 2 y and we use the positive root 9 3 4 2 2 4 2 4 y dy 2 4 y dy 0 9 9 4 3 3 y 16 27 0 9x 2 4 4 2 yy1 yy1 2p yy1 2y1 2px y2 4p 2p 2. Since x 2p 2 y 2 0 16p 2 y2 , we have 4p 42. 9x 2 4y 2 36 y 2 x 4) dx 16 24) 1x 24 8 3 36 y 3 2 4 2 3 2 x2 2 4 on yy1 1 2 y 2 2p 2 the interval 2 9 4 4 2 4V 9 x3 4 3 x2 4 dx y2 y 2p 2 4y12 2 (x 2 4x y1 y12 4p 2. Therefore 9 4 64 3 8 9 56 4 3 8 the slopes of the two tangents from P1 are m1 2p y1 y12 y1 2 4p 2 4p (y12 2 3 (56 4 4p 2 and m2 4p 2) 2p y1 y12 43. x 2 V y2 3 1 y 2)2 dy 2 C y y 2 on the interval 3 3 y 3 m1m2 1 the lines are 3 3 ( 1 y 2) dy w x2 H 2 w(0)2 2H 2 perpendicular. 40. Let y 1 x2 on the interval 0 4 2 44. y x 0 (1 y 3 3 0 wx 2 2H 0 3 ( 1 24 y 2)2 dy x 2. The area of w x dx H C; y 0 when wx 2 is the 2H the inscribed rectangle is given by A(x) 2x 2 1 x2 4 00 CC 0; therefore y 4x 1 and the height is 2y) A (x) x2 (since the length is 2x 4 x2 x2 4 1 x2 . 4 1 4 equation of the cable's curve. 45. drA dt drB dt d (r dt A rB) 0 rA rB a constant Thus A (x) 4 1 x 4 2 04 1 x2 x2 4 x 1 2 x2 4 0 2 (only the 46. PF will always equal PB because the string has constant length AB FP PA AP PB. 0 x2 2x s Appendix A5.2 (pp. 606611) 1. 16x 2 25y 2 a2 c a positive square root lies in the interval). Since A(0) A(2) 0 we have that A( 2) 4 is the maximum 2. x 41. (a) Around the x-axis: 9x 2 y V 2 2 400 b2 x2 25 y2 16 1 3 c e 25 16 area, when the length is 2 2 and the height is 4y 2 36 y 2 0 y2 3 ; F( 3, 0); directrices are 5 5 a 25 3 = e 3 5 9 9 2 x 4 2. 2x 2 c e 2 x2 a2 b2 1 2 a e 2 1 2 9 2 9x 0 9 2 x and we use the positive root 4 2 9 2 2 9 2 9 x dx 2 9 x dx 0 4 4 y2 2 1 1 1 2 c a ; F(0, 1); directrices are 3 3 2 x 4 0 24 y 0 2 ...
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