Business Calc Homework w answers_Part_86

Business Calc Homework w answers_Part_86 - 426 70. (a)...

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Unformatted text preview: 426 70. (a) Section 10.6 Section 10.6 Exercises 1. [ 9, 9] by [ 6, 6] dy dx f ( ) sin f ( ) cos cos cos 2 sin cos2 sin cos cos sin2 1 1 f ( ) cos f ( ) sin ( 1 ( 1 cos sin sin )cos sin )sin The graphs are parabolas. (b) As k 0 , the limit of the graph is the negative x-axis. s Section 10.6 Calculus of Polar Curves (pp. 559568) Quick Review 10.6 dy dy/dt 5 cos t dx dx/dt 3 sin t 5 2. cot 2 0.763 3 dy dx dy dx dy dx 1, 0 dy dx 1 1 1 cos 2 cos . cos 2 sin 0 1 2. 5 cot t 3 f ( ) sin f ( ) cos f ( ) cos f ( ) sin 2 sin 2 sin 2 sin 2 cos 1. 0 1 dy , which is undefined; 0 dx 1 , which is undefined. 0 f ( ) cos f ( ) sin sin cos (2 (2 3 sin ) cos 3 sin ) sin cos sin2 ) 2 , 3 0 1 0; /2 3. Solve cot t 0: t 2 or 3 ; 2 2 and , 5 sin 2 dy dx f ( ) sin f ( ) cos 3 cos 3 cos 2 cos 2 sin the corresponding points are 3 cos 3 3 and 3 cos , 5 sin 2 2 (0, 5) 3. dy dx (0, 5) 0, , or 2 ; 4. 5 cot t is undefined when t 3 the corrresponding points are (3 cos 0, 5 sin 0) (3 cos , 5 sin ) (3 cos 2 , 5 sin 2 ) ( 3, 0). dy 2 dt dt 6 sin 3(cos2 2 3 (3, 0) and dy dx (2, 0) dy dx ( 1, /2) dy dx (2, ) dy dx (5, 3 /2) dy dx dy dx dy dx dy dx 0 5. Length 0 dx 2 dt dy dx 0, /2 9 sin2 t 0 25 cos2 t dt, 12.763. 2 , and 3 0 5 which using NINT evaluates to For questions 68, the graph is: 0. 3 /2 4. f ( ) sin f ( ) cos 3 sin2 3 sin cos f ( ) cos f ( ) sin 3 cos (1 cos ) 3 sin (1 cos ) [ 2, 4] by [ 2, 2] 3 cos 6 sin 1 2 3(cos2 sin2 ) cos 3 sin 1 2 3 2 1 2 3 2 3 2 6. The upper half of the outer loop 7. The inner loop 8. The lower half of the outer loop 9. y 0 for x 6 dy dx (1.5, /3) , which is undefined; 1 2 3 2 0 or 6. x 2) dx 3x 2 1 3 6 x 3 0 Area 0 (6x 36 dy dx (4.5, 2 /3) 0; 10. Use a graphing calculator's intersect function to find that the curves cross at x NINT to find 2.248 0.270 and x 2.248, then use dy dx (6, ) dy dx (3, 3 /2) 1 1 , which is undefined; and 0 0 0 0 ( 1) ( 1) Area 0.270 [2 sin x (x 2 1. 2x 1)] dx 2.403. Section 10.6 5. 7. 427 [ 3.8, 3.8] by [ 2.5, 2.5] [ 1.5, 1.5] by [ 1, 1] The graph passes through the pole when r which occurs when -interval 0 only consider 2 3 cos 3 . Since the 2 0, and when The polar solutions are 0, a given k, the line curve at 0, k 5 for k 0, 1, 2, 3, 4, and for 1 produces the entire graph, we need 2 k appears to be tangent to the 5 . At this point, there appears to be a 2 vertical tangent line with equation Confirm analytically: x y dy d (or x 0). (3 cos ) cos (3 cos ) sin ( 3 sin )sin dx d 3 cos2 (3 cos )cos 3(cos2 sin2 ) and 6 cos ( sin ). k . This can be confirmed analytically by 5 dy noting that the slope of the curve, , equals the slope of dx k the line, tan . So the tangent lines are 0 [y 0], 5 2 2 y tan x , y tan x , 5 5 5 5 3 3 4 4 y tan x , and y tan x . 5 5 5 5 8. dx , 0, and At 0, /2 2 d dy dx 3(02 12) 3. So at 0, , 0 /2 2 d d dy dy 0, so is undefined and the tangent line is and dx d [ 3, 3] by [ 2, 2] vertical. 6. The polar solutions are 0, a given k, the line curve at 0, k 2 for k 0, 1, 2, 3, 4, and for k appears to be tangent to the 2 [ 3, 3] by [ 2, 2] A trace of the graph suggests three tangent lines, one with positive slope for 6 k . This can be confirmed analytically by 2 dy noting that the slope of the curve, , equals the slope of dx k the line, tan . So the tangent lines are 0 [y 0] and 2 3 [x 0]. , and 2 are duplicate 2 2 , a vertical one for 5 . 6 2 , and one solutions. with negative slope for Confirm analytically: dy d dx d 6 sin 3 sin 6 sin 3 cos 6 2 cos 3 cos 2 cos 3 sin . 5 6 and 0, dy dx dy dx dy dx dy dx , 0, 2 , and 0, are all solutions. dy/d , and so dx/d 6(1)(1/2) 2(0)( 3/2) 6(1)( 3/2) 2(0)(1/2) /6 /2 1 3 ; 6( 1)(1) 6( 1)(0) 2(0)(0) , which is undefined; and 2(0)(1) 1 3 5 /6 6(1)(1/2) 2(0)( 3/2) 6(1)( 3/2) 2(0)(1/2) 6 1 3 . The tangent 2 lines have equations and 5 6 y 1 3 x , [x 0], y x . 428 dy d Section 10.6 dy d 9. cos sin ( 1 1) sin )cos 10. sin2 cos2 2 cos2 (1 cos cos cos cos )cos sin2 1 (1 cos )sin cos (2 sin sin 2 dx d cos ( 1 sin sin , sin )sin sin 2 cos2 cos 2 dx d sin sin (1 sin 2 dy d 2 cos ) sin 1 4 9 2 sin2 dy d 1 0) or when dx d 0 when cos , 3 1 or , i.e., when 0, , 2 1 2 0 when 5 (2 sin 6 6 3 (cos 2 2 or 5 dx . 3 d 0 when (then sin 0 when (then 1 7 11 , , or 6 6 3 ,r 2 1 2 5 ,r 6 1 2 0) or when 2 cos , 1 0). 2 4 , 3 3 3 sin 2 3 3 5 sin 2 3 3 3 and for 4 3 4 3 0). So there is a horizontal tangent line 3 the line y 2 sin 2 1 4 9 1 or 1, i.e., when 2 for 2 3 5 ,r 3 ,r . So there is a horizontal tangent line for 3 2 sin 2 1 sin 2 6 the line y the line y 2 , for 1 and for 4 6 ,r 3 the line y 2 . There is a vertical tangent line for [the line x line x 3 2 0, r 2], for 2 2 ,r 2 ,r 3 1 2 2 cos 0 2 [again, the again, the line y 1 5 sin 2 6 1 . 4 7 ,r 6 2 cos 2 2], for There is a vertical tangent line for the line x 11 ,r 6 3 7 cos 2 6 [the line x 3 3 and for 4 3 3 11 the line x cos 2 2 6 dx d 3 4 3 . For d dy d d d dx d d dy , 2 d 0, but sin cos cos 1 for 0 for 2 2 cos 2 4 sin and 2 1 2 1 4 1 cos ] and for ,r 2 3 4 3 2 1 2 1 again, the line x cos . 2 3 4 dy dx For , 0, but d d d dy 4 cos sin sin 0 for , and d d d dx 2 cos 2 cos 1 for , d d dy so by L'Hpital's rule 0 and the tangent line is horidx , so by zontal at ,r 0 [the line y 0]. dy is undefined and the tangent line is L'Hpital's rule dx This information can be summarized as follows. Horizontal at: 3 , 2 3 vertical at 2 ,r 0 [the line x 0]. y 3 4 3 , This information can be summarized as follows. Horizontal at: 1 , 2 6 1 5 , 2 6 (0, ) [y 3 5 , 2 3 0], y 2], x x 2] 1 , 4 1 , 4 3 3 4 y y y [x 2 0], x x 1 , 4 1 , 4 Vertical at: (2, 0) [x 1 , 2 1 , 2 2 3 4 3 2, Vertical at: 0, 2 3 2 3 7 , 2 6 3 11 , 2 6 3 3 , 4 3 3 4 (2, 2 ) [x Section 10.6 11. y dy d 429 2 sin2 4 sin 2 sin 2 cos 12. dy d 4 sin2 4(sin 2 (3 cos 2 4 cos )cos ) 3 cos 4 4 cos )sin 8 cos2 cos dx d 3 cos (3 3) 3 sin 3 x 2 sin sin 2 4 sin cos sin (8 cos 4 sin 2 0, 2 dy d dy d 2 cos 2 0 when , dx , and d 0 when 3 , . They are never both zero. 4 4 dy d 0 when cos 137 16 , i.e., when 0.405, 2.146, 4.137, or 5.878 (values solved for with a graphing calculator). (then sin 2 cos 1 For 0, 2 , the curve has horizontal asymptotes 0 sin 0 0], 2, 0]. For 2, 4 2 2 3 , the curve has 4 4 dx d 0 when cos 1 0, 3 8 or 2 1.186 or 3 0). So there 0.676 2.146, at (0, 0) [y (0, ) [ y y 2 sin 2 , and 0) or when 3 8 0 sin 5.097 (then 8 cos vertical asymptotes at 3 2, 4 [x 1 . 2 cos 4 1] and is a horizontal tangent line for [the line y r 0.676 sin 0.405 0.405, r 0.267], for x 3 2 cos 4 This information can be summarized as follows. Horizontal at: (0, 0) [y 2, 2 5.176 [the line y 4.137, r 5.176 5.176 sin 2.146 4.343], for 0], 2], 0] [x [x 1], 1] [y [the line y 5.878, r [the line y tangent for for r ,r 5.176 sin 4.137 0.676 0.676 sin 5.878 0, r 4.343], and for (0, ) [y Vertical at: 2, 4 3 2, 4 0.267]. There is a vertical 1 cos 0 7], for 1], for 1], 2 , 1 [the line x 7 cos 1 cos 2 7 [the line x 1 [again, the line x cos 2 1 3 9 the line x , and for 2 16 3 3 9 cos 1 ,r again, the line x . 8 2 16 3 ,r 8 This information can be summarized as follows. Horizontal at: ( 0.676, 0.405) [y (5.176, 2.146) [y (5.176, 4.137) [y ( 0.676, 5.878) [y Vertical at: ( 1, 0) [x (1.5, 1.186) (7, ) [x (1.5, 5.097) ( 1, 2 ) [x x 7], x 9 , 16 0.267], 4.343], 4.343], 0.267] 1], 9 , 16 1] 430 Section 10.6 2 (as can be verified 19. 13. The curve is complete for 0 by graphing). The area is 2 0 1 (4 2 2 2 cos )2 d (4 4 cos 4 sin 1 2 2 0 cos2 ) d 1 sin 2 4 2 [ 2.5, 5.2] by [ 2, 3.1] 2 4 18 0 The circles intersect at (x, y) coordinates (0, 0) and (1, 1). The area shared is twice the area inside the circle r 2 sin 2 0 /4 14. The curve is complete for 0 by graphing). The area is 2 0 2 (as can be verified between /4 0 and 4 . 1 2 a (1 cos )2 d 2 2 1 2 a (1 2 cos 2 0 1 2 1 a 2 sin 2 2 Shared area cos2 ) d 1 sin 2 4 2 0 1 (2 sin )2 d 2 4 sin2 d /4 0 0 3 2 a 2 4 4 20. 1 2 1 sin 2 4 1 4 15. Use r 4 2a2 cos 2 . One lobe is complete for 4 8 2 1. . The total area is 2 2 /4 2 1 ( /4 2 2a cos 2 ) d 2a 2 /4 cos 2 d /4 2a2 1 sin 2 2 /4 /4 2a2 (Integrating from 0 to 2 will not work, because r is not defined over the entire interval.) 16. One leaf covers /4 [ 3, 3] by [ 2, 2] The circles intersect at 1, 4 1 4 1 cos2 2 d /4 2 4 1 sin 4 16 6 and 1, /2 /6 5 . The shared area 6 . Its area is /4 /4 /6 is 2 8 . 4 2 0 1 (2 sin )2 d 2 2 /2 /6 1 2 (1) d 2 /6 0 sin2 d /6 0 d /2 /6 17. Use r Its area is 4 sin 2 . One loop is complete for 0 /2 0 . 1 ( 2 4 sin 2 )2 d 0 /2 4 2 3 1 2 2 sin 2 d /2 cos 2 0 2. 21. 3 1 sin 2 4 3 2 3 3 2 3 18. Use r 2 sin 3 . One leaf is complete for 0 . The total area is /3 6 0 1 ( 2 2 sin 3 )2 d /3 6 0 sin 3 d /3 [ 4.7, 4.7] by [ 3.1, 3.1] 2 cos 3 0 4. The shared area is half the circle plus two lobelike regions: 1 (2)2 2 /2 2 0 /2 1 [2(1 2 cos )]2 d 4 cos2 ) d 1 2 1 sin 2 4 /2 0 2 0 (4 4 8 8 cos 2 sin 2 5 ...
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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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