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Pre-Calc Homework Solutions 166

# Pre-Calc Homework Solutions 166 - 166 Section 4.4 0 0 a 3 6...

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53. (a) According to the graph, y 9 (0) 5 0. (b) According to the graph, y 9 ( 2 L ) 5 0. (c) y (0) 5 0, so d 5 0. Now y 9 ( x ) 5 3 ax 2 1 2 bx 1 c , so y 9 (0) implies that c 5 0. Therefore, y ( x ) 5 ax 3 1 bx 2 and y 9 ( x ) 5 3 ax 2 1 2 bx . Then y ( 2 L ) 5 2 aL 3 1 bL 2 5 H and y 9 ( 2 L ) 5 3 aL 2 2 2 bL 5 0, so we have two linear equations in the two unknowns a and b . The second equation gives b 5 } 3 a 2 L } . Substituting into the first equation, we have 2 aL 3 1 } 3 a 2 L 3 } 5 H , or } a 2 L 3 } 5 H , so a 5 2 } L H 3 } . Therefore, b 5 3 } L H 2 } and the equation for y is y ( x ) 5 2 } L H 3 } x 3 1 3 } L H 2 } x 2 , or y ( x ) 5 H 3 2 1 } L x } 2 3 1 3 1 } L x } 2 2 4 . 54. (a) The base radius of the cone is r 5 } 2 p 2 a p 2 x } and so the height is h 5 ˇ a w 2 w 2 w r w 2 w 5 ! a § 2 § 2 § 1 § } 2 § p § 2 a p § 2 § x } § 2 2 § . Therefore, V ( x ) 5 } p 3 } r 2 h 5 } p 3 } 1 } 2 p 2 a p 2 x } 2 2 ! a § 2 § 2 § 1 § } 2 § p § 2 a p § 2 § x } § 2 2 § . (b) To simplify the calculations, we shall consider the volume as a function of r: volume 5 f ( r ) 5 } p 3 } r 2 ˇ a w 2 w 2 w r w 2 w , where 0 , r , a .
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