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Q. 23

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Calculus

Found in: Page 964

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

Use Theorem 12.32 to find the indicated derivatives in Exercises 21–26. Express your answers as functions of a single variable. $\frac{d x}{d t} w h e n x = r \cos θ, r = t^{2} - 5, a n d θ = t^{3} + 1$

The value is $\frac{d x}{d t} = 2 t \cdot \cos (t^{3} + 1) - 3 t^{2} (t^{2} - 5) \sin (t^{3} + 1)$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information:

Given:

$x = r \cos θ, r = t^{2} - 5, a n d θ = t^{3} + 1$

We have to find the indicated derivatives and express your answers as functions of a single variable.

Step 2. Solution:

$U \sin g r = t^{2} - 5 a n d θ = t^{3} + 1 i n x = r \cos θ w e g e t x = (t^{2} - 5) \cdot \cos (t^{3} + 1) D i f f . w . r . t . t w e g e t \frac{d x}{d t} = (t^{2} - 5) \frac{d}{d t} \cos (t^{3} + 1) + \cos (t^{3} + 1) \frac{d}{d t} (t^{2} - 5) \frac{d x}{d t} = (t^{2} - 5) (- \sin (t^{3} + 1) \cdot 3 t^{2}) + \cos (t^{3} + 1) \cdot 2 t \frac{d x}{d t} = 2 t \cdot \cos (t^{3} + 1) - 3 t^{2} (t^{2} - 5) \sin (t^{3} + 1)$

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