Solutions3 - MATH 2007B Test 3 Solutions November 7, 2011...

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MATH 2007B Test 3 Solutions November 7, 2011 [Marks] 1. Consider the parametric curve x = t 2 1 ,y =2 t 3 + t 2 . [6] (a) Find dy dx and d 2 y dx 2 . Solution: dy dx = dy/dt dx/dt = 6 t 2 +2 t 2 t =3 t +1. d 2 y dx 2 = 1 dx/dt d dt ± dy dx ² = 3 2 t . (b) Determine the point(s) where the tangent is horizontal/vertical. Solution: The tangent is horizontal when dy dx =3 t + 1 = 0, i.e., t = 1 3 . The tangent is never vertical since dy dx =3 t +1 6 = ±∞ for any t . (c) Determine the intervals of t where the graph is concave up/down. Solution: d 2 y dx 2 = 3 2 t > 0for t> 0and < 0for t< 0. Hence, the graph is concave up for t> 0and concave down for t< 0. 2. Find the length of the parametric curve x =3 t 2 ,y =2 t 3 , 3 t 8. [6] Solution: L = Z 8 3 q ( dx/dt ) 2 +( dy/dt ) 2 dt = Z 8 3 p (6 t ) 2 +(6 t 2 ) 2 dt =3 Z 8 3 2 t 1+ t 2 dt =3 Z 9 4 udu =2 u 3 / 2 ³ ³ ³ ³ 9 4 =2(27 8) = 38, where u =1+ t 2 . 3.
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This note was uploaded on 01/18/2012 for the course MATH 2007 taught by Professor S.melkonian during the Fall '11 term at Carleton CA.

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Solutions3 - MATH 2007B Test 3 Solutions November 7, 2011...

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