Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Practice Exercises - Page 441: 120

Answer

≈ $5$ $ft/sec$

Work Step by Step

$\frac{dy}{dx}$ = $9e^{\frac{-x}{3}}(\frac{-1}{3})$ = $-3e^{\frac{-x}{3}}$ $\frac{dy}{dt}$ = $\frac{dy}{dx}$*$\frac{dx}{dt}$ $\frac{-1}{4}(\sqrt{9-y})$ = $(-3e^{\frac{-x}{3}})$$\frac{dx}{dt}$ $\frac{-1}{4}(\sqrt{9-9e^{\frac{-x}{3}}})$ = $(-3e^{\frac{-x}{3}})$$\frac{dx}{dt}$ $\frac{dx}{dt}$ = $\frac{1}{4}(\frac{(\sqrt{1-e^{\frac{-x}{3}}}}{e^{\frac{-x}{3}}})$ $\frac{dx}{dt}$$|_{{\,x=9}}^{{\,}}$ = $\frac{1}{4}(\frac{(\sqrt{1-e^{-3}}}{e^{-3}})$ ≈ $5$ $ft/sec$
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