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HW_6_Key

# HW_6_Key - Miller Kierste – Homework 6 – Due Oct 2 2007...

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Unformatted text preview: Miller, Kierste – Homework 6 – Due: Oct 2 2007, 3:00 am – Inst: JEGilbert 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Find d 2 y dx 2 when x ( t ) = 6 + t 2 , y ( t ) = t ln( t 2 ) . 1. d 2 y dx 2 =- ln t t 3 2. d 2 y dx 2 = ln t 2 t 3 3. d 2 y dx 2 =- ln t 2 t 3 correct 4. d 2 y dx 2 =- ln t t 2 5. d 2 y dx 2 = ln t t 2 6. d 2 y dx 2 = ln t 2 t 2 Explanation: First notice that y ( t ) = 2 t ln t . After differentiation with respect to x , therefore, we see that x ( t ) = 2 t, y ( t ) = 2(1 + ln t ) . Thus dy dx = 1 + ln t t . On the other hand, by the Chain Rule, d dt ‡ dy dx · = dx dt n d dx ‡ dy dx ·o = dx dt d 2 y dx 2 , in which case d 2 y dx 2 = d dt ‡ dy dx ·. dx dt . But d dt ‡ dy dx · = t (1 /t )- (1 + ln t ) t 2 =- ln t t 2 . Consequently, d 2 y dx 2 =- ln t 2 t 3 . keywords: 002 (part 1 of 1) 10 points Find dy dx for the curve given parametically by x ( t ) = 3 + 2 t 2 , y ( t ) = t 2 + 2 t 3 . 1. dy dx = 1 + 3 2 t 2. dy dx = 1 2 + 3 2 t correct 3. dy dx = 4 2 + 3 t 4. dy dx = 1 + 3 4 t 5. dy dx = 2 1 + 3 t 6. dy dx = 1 1 + 3 t 7. dy dx = 2 2 + 3 t 8. dy dx = 1 2 + 3 t Explanation: Differentiating with respect to t we see that x ( t ) = 4 t, y ( t ) = 2 t + 6 t 2 . Consequently, dy dx = y ( t ) x ( t ) = 2 t + 6 t 2 4 t = 1 2 + 3 2 t . Miller, Kierste – Homework 6 – Due: Oct 2 2007, 3:00 am – Inst: JEGilbert 2 keywords: parametric curve, derivative, poly- nomial functions...
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HW_6_Key - Miller Kierste – Homework 6 – Due Oct 2 2007...

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