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Unformatted text preview: Problem Set 6 Calculus 1 Abdullah Khalid, Hassan Bukhari December 31, 2010 1. We have v ( t ) = 0 . 001302 t . 09029 t 2 + 23 . 61 t 3 3 . 083. We find the acceleration by differentiating this w.r.t. t . a ( t ) = v ( t ) = 0 . 001302 2 × . 09029 t + 3 × 23 . 61 t 2 = 0 . 001302 . 18058 t + 70 . 83 t 2 To find maximum velocity, we find all critical points, by differentiating the above expres sion again and putting it equal to zero a ( t ) = 0 . 18058 + 141 . 66 t = 0 Which gives us t = 0 . 001275 Hence we have a (0 . 001275) = 0 . 001187 Now we evaluate the a(t) at the endpoints, t = 0 and t = 126. a (0) = 0 . 001302 a (126) = 1124474 So, the maximum acceleration is just before the fuel runs out at t = 126 and is equal to 1124474. 2. For f ( x ) = 1 x 2 / 3 , we compute the derivative, which comes out to be f ( x ) = 1 2 3 x 1 / 3 . This is undefined for x = 0. Rolle’s theorem requires that the function have a derivative at every point in the given domain. Since, this condition is not true, Rolle’s theorem does not apply to this situation....
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 Spring '10
 RazaSuleman

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