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Expert-verified Found in: Page 964 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # Use Theorem 12.32 to find the indicated derivatives in Exercises 21–26. Express your answers as functions of a single variable.$\frac{dy}{dt}wheny=r\mathrm{sin}\theta ,r={t}^{3},and\theta =\sqrt{t}$

The value is $\frac{dy}{dt}={t}^{2}\left(\sqrt{t}·\mathrm{cos}\sqrt{t}+2\mathrm{sin}\sqrt{t}\right)$

See the step by step solution

## Step 1. Given Information:

Given:

$y=r\mathrm{sin}\theta ,\phantom{\rule{0ex}{0ex}}r={t}^{3},and\phantom{\rule{0ex}{0ex}}\theta =\sqrt{t}$

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

## Step 2. Solution:

$Usingr={t}^{3}and\theta =\sqrt{t}iny=r\mathrm{sin}\theta weget\phantom{\rule{0ex}{0ex}}y={t}^{3}\mathrm{sin}\sqrt{t}\phantom{\rule{0ex}{0ex}}Diff.w.r.t.tweget\phantom{\rule{0ex}{0ex}}\frac{dy}{dt}={t}^{3}\frac{d}{dt}\mathrm{sin}\sqrt{t}+\mathrm{sin}\sqrt{t}\frac{d}{dt}{t}^{3}\phantom{\rule{0ex}{0ex}}\frac{dy}{dt}={t}^{3}·\mathrm{cos}\sqrt{t}·\frac{1}{2\sqrt{t}}+\mathrm{sin}\sqrt{t}·2{t}^{2}\phantom{\rule{0ex}{0ex}}\frac{dy}{dt}={t}^{2}\left(\sqrt{t}·\mathrm{cos}\sqrt{t}+2\mathrm{sin}\sqrt{t}\right)$ ### Want to see more solutions like these? 