Business Calc Homework w answers_Part_88

Business Calc Homework w answers_Part_88 - 436 Chapter 10...

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23. } d d u y } 5 } d d u } 31 1 2 cos 1 } u 2 } 22 sin u 4 5 } 1 2 } sin 1 } u 2 } 2 sin u 1 cos u 2 cos 1 } u 2 } 2 cos u } d d u x } 5 } d d u } 31 1 2 cos 1 } u 2 } 22 cos u 4 5 } 1 2 } sin 1 } u 2 } 2 cos u 2 sin u 1 cos 1 } u 2 } 2 sin u Solve } d d u y } 5 0 for u with a graphing calculator: the solutions are 0, < 2.243, < 4.892, < 7.675, < 10.323, and 4 p . Using the middle four solutions to y 5 r sin u reveals horizontal tangent lines at y < 6 0.443 and y < 6 1.739. Solve } d d u x } 5 0 for u with a graphing calculator: the solutions are 0, < 1.070, < 3.531, 2 p , < 9.035, < 11.497, and 4 p . Using the middle five solutions to find x 5 r cos u reveals vertical tangent lines at x 5 2, x < 0.067, and x < 2 1.104. Where } d d y t } and } d d x t } both equal zero ( u 5 0, 4 p ), close inspection of the plot shows that the tangent lines are horizontal, with equation y 5 0. (This can be confirmed using L’Hôpital’s rule.) 24. } d d u y } 5 } d d u } [2(1 2 sin u )sin u ] 5 2 4 sin u cos u 1 2 cos u } d d u x } 5 } d d u } [2(1 2 sin u )cos u ] 5 2 2 cos 2 u 2 2 sin u 1 2 sin 2 u 5 4 sin 2 u 2 2 sin u 2 2 Solve } d d u y } 5 0 for u : the solutions are } p 6 } , } p 2 } , } 5 6 p } , and } 3 2 p } . Using the first, third, and fourth solutions to find y 5 r sin u reveals horizontal tangent lines at y 5 } 1 2 } and y 5 2 4. Solve } d d u x } 5 0 for u (by first using the quadratic formula to find sin u ): the solutions are } p 2 } , } 7 6 p } , and } 11 6 p } . Using the last two solutions to find x 5 r cos u reveals vertical tangent lines at x 5 6 } 3 ˇ 2 3 w } < 6 2.598. Where } d d y t } and } d d x t } both equal zero 1 u 5 } p 2 } 2 , inspection of the plot shows that the tangent line is vertical, with equation x 5 0. (This can be confirmed using L’Hôpital’s rule.) 25. [ 2 1.5, 1.5] by [ 2 1, 1] The tips have Cartesian coordinates 1 } ˇ 1 2 w } , } ˇ 1 2 w } 2 , 1 2 } ˇ 1 2 w } , } ˇ 1 2 w } 2 , 1 2 } ˇ 1 2 w } , 2 } ˇ 1 2 w } 2 , and 1 } ˇ 1 2 w } , 2 } ˇ 1 2 w } 2 . From the curve’s symmetries, it is evident that the tangent lines at those points have slopes of 2 1, 1, 2 1, and 1, respectively. So the equations of the tangent lines are y 2 } ˇ 1 2 w } 5 2 1 x 2 } ˇ 1 2 w } 2 or y 5 2 x 1 ˇ 2 w , y 2 } ˇ 1 2 w } 5 x 1 } ˇ 1 2 w } or y 5 x 1 ˇ 2 w , y 1 } ˇ 1 2 w } 5 2 1 x 1 } ˇ 1 2 w } 2 or y 5 2 x 2 ˇ 2 w , and y 1 } ˇ 1 2 w } 5 x 2 } ˇ 1 2 w } or y 5 x 2 ˇ 2 w 26. [ 2 3, 3] by [ 2 2, 2] As the plot shows, the curve crosses the x -axis at ( x , y )-coordinates ( 2 1, 0) and (1, 0), with slope 2 1 and 1, respectively. (This can be confirmed analytically.) So the equations of the tangent lines are y 2 0 5 2 ( x 1 1) y 5 2 x 2 1 and y 2 0 5 x 2 1 y 5 x 2 1. 27. r cos u 5 r sin u x 5 y , a line 28. r 5 3 cos u r 2 5 3 r cos u x 2 1 y 2 5 3 x x 2 2 3 x 1 } 9 4 } 1 y 2 5 } 9 4 } 1 x 2 } 3 2 } 2 2 1 y 2 5 1 } 3 2 } 2 2 a circle 1 center 5 1 } 3 2 } , 0 2 , radius 5 } 3 2 } 2 436 Chapter 10 Review
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29. r 5 4 tan u sec u r cos u 5 4 } r r c s o in s u u } x 5 4 } y x } or x 2 5 4 y , a parabola 30. r cos 1 u 1 } p 3 } 2 5 2 ˇ 3 w r cos u cos 1 } p 3 } 2 2 r sin u sin 1 } p 3 } 2 5 2 ˇ 3 w } 1 2 } r cos u 2 } ˇ 2 3 w } r sin u 5 2 ˇ 3 w } 1 2 } x 2 } ˇ 2 3 w } y 5 2 ˇ 3 w x 2 ˇ 3 w y 5 4 ˇ 3 w or y 5 } ˇ x 3 w } 2 4, a line 31. x 2 1 y 2 1 5 y 5 0 r 2 1 5 r sin u 5 0 r 5 2 5 sin u 32. x 2 1 y 2 2 2 y 5 0 r 2 2 2 r sin u 5 0 r 5 2 sin u 33. x 2 1 4 y 2 5 16 ( r cos u ) 2 1 4( r sin u ) 2 5 16 r 2 cos 2 u 1 4 r 2 sin 2 u 5 16, or r 2 5 } cos 2 u 1 16
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