a151q4wi - M ATH 151 Answers to Assignment 4 Autumn 2009 1....

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Unformatted text preview: M ATH 151 Answers to Assignment 4 Autumn 2009 1. Let u and v be differentiable functions of x such that u(0) = 9, u (0) = 2, v (0) = 4 and v (0) = −3. The function w is defined by w = u5 /v 3 . Find w (0). w = u5/2 v −3/2 . w = 5 u3/2 u v −3/2 + u5/2 −3 v −5/2 v ; 2 2 w (0) = 5 {u(0)}3/2 u (0){v (0)}−3/2 + {u(0)}5/2 −3 {v (0)}−5/2 v (0) = 2 2 5 {9}3/2 2{4}−3/2 + {9}5/2 −3 {4}−5/2 (−3). Answer: 3267/64 2 2 d 1 + cos2 t. dt d 2 1/2 dt {1 + cos t} 2. Find Answer: 1 2 {1 + cos2 t}−1/2 2 cos t (− sin t) 3. Find f (−1) where f is the function defined by f (u) = (1 − u)8 (1 + u2 )4 (1 + u4 )2 . f (u) = 8(1 − u)7 (−1)(1 + u2 )4 (1 + u4 )2 8 23 f (−1) = 8(2)7 (−1)(2)4 (2)2 + (1 − u) 4(1 + u ) 2u(1 + u ) 42 + (2)8 4(2)3 (−2)(2)2 + (2)8 (2)4 2(2)(−4) + (1 − u)8 (1 + u2 )4 2(1 + u4 )(4u3 ) = 3 × (−8 · 213 ) Answer: −3 × 216 or −196608 4. Find the equation of the tangent to the curve x3 + y 3 = 3xy at the point (2/3, 4/3). Differentiate both sides with respect to x: 3x2 + 3y 2 y = 3y + 3xy . Substituting x = 2/3 and y = 4/3 gives 3(2/3)2 + 3(4/3)2 y = 3(4/3) + 3(2/3)y . So at (2/3, 4/3), 4/3 + 16y /3 = 4 + 2y ; (16/3 − 2)y = 4 − 4/3 ; (10/3)y = 8/3; y = 8/10. Answer: y − 4/3 = 4 (x − 2/3) or 4x − 5y + 4 = 0 5 5. Find g (x) where g (x) = sec3 (2x). g (x) = 3 sec2 (2x) sec(2x) tan(2x) 2 = 6 tan(2x) sec3 (2x) g (x) = {6 sec2 (2x) 2} × sec3 (2x) + 6 tan(2x) × {3 sec2 (2x) tan(2x) sec(2x) 2} Answer: 12 sec5 (2x) + 36 tan2 (2x) sec3 (2x) end Typeset using AMS-TEX. ...
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This note was uploaded on 03/13/2010 for the course MATH 151 taught by Professor Any during the Winter '08 term at Ohio State.

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