Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 2 - The Derivative - 2.8 Related Rates - Exercises Set 2.8 - Page 174: 25

Answer

$\frac{9}{20 \pi} $ $ft/min$

Work Step by Step

Ref to the fig $\frac{BD}{AB}=\frac{24}{10}=\frac{CD}{r}$ Let $CD=h$ $\Longrightarrow\frac{h}{r}=\frac{12}{5}$ $ 12r=5h$ $r=\frac{5}{12} h$ ................... eq (1) Let $h$ be the height of the water tank and r be the radius of the same. Let $V$ be the volume of the conical water tank. $V=\frac{1}{3} \pi r^2 h$ .................. eq (2) Putting equation (1) in equation (2) $V=\frac{1}{3} \pi (\frac{5h}{12})^2 \times h=\frac{25 \pi h^3}{432}$ Taking derivative with respect to $t$ $\frac{dV}{dt}=\frac{d ( \frac{25 \pi h^3}{432} ) }{dt}= \frac{25 \pi}{432}\frac{dh^3}{dt} = \frac{25 \pi}{432}\frac{dh^3}{dh} \frac{dh}{dt}=\frac{25 \pi \times 3 h^2}{432} \frac{dh}{dt} $ $\frac{dV}{dt}=\frac{25 \pi \times 3 h^2}{432} \frac{dh}{dt} $ ..................... eq (3) Putting $ \frac{dV}{dt}=20 $ $ ft^3/min$, $h=16$ ft in equation (3) $ 20 =\frac{25 \pi \times 3 \times 16^2}{432} \frac{dh}{dt} = \frac{ 19200\pi}{432} \frac{dh}{dt}= \frac{400\pi}{9}\frac{dh}{dt}$ $\frac{dh}{dt}= \frac{9}{400\pi}\times 20=\frac{9}{20 \pi}$ $ ft/min$
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