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 175: 43

Answer

$4.5$ cm/s; away

Work Step by Step

$\frac{1}{s} +\frac{1}{S} =\frac{1}{f} $................ eq (1) $f=6$cm, $ \frac{ds}{dt}=-2$ cm/s, $ \frac{dS}{dt}=? $, $s=10$cm Equation (1) may be written as $s^ {-1} +S^ {-1} =f^ {-1} $ Taking derivative with respect to t $\frac{d(s^{-1})}{dt} + \frac{d(S^{-1})}{dt} = \frac{d(f^{-1})}{dt} $ $\frac{d(s^{-1})}{ds}\frac{ds}{dt} + \frac{d(S^{-1})}{dS} \frac{dS}{dt} = \frac{d(f^{-1})}{df}\frac{df}{dt} $ As focal length f is constant $-s^{-2}\frac{ds}{dt} - S^{-2} \frac{dS}{dt} = 0 $ $-s^{-2}\frac{ds}{dt} = S^{-2} \frac{dS}{dt} $ Or $S^{-2} \frac{dS}{dt} =-s^{-2}\frac{ds}{dt} $ ................ eq (2) Putting $s=10$cm, $f=6$cm in equation (1) $\frac{1}{10}+\frac{1}{S}=\frac{1}{6}$ $\frac{1}{S}=\frac{1}{6}-\frac{1}{10}=\frac{1}{15}$ $\frac{1}{S}=\frac{1}{15}$ $S=15$ cm Putting $ \frac{ds}{dt}=-2$ cm/s, $s=10$cm, $S=15$ cm in equation (2) $15^{-2} \frac{dS}{dt} =- 10^{-2} \times (-2) $ $ \frac{1}{15^2} \frac{dS}{dt} =- \frac{1}{10^2} \times (-2) =\frac{1}{50}$ $ \frac{1}{225} \frac{dS}{dt} =\frac{1}{50}$ $ \frac{dS}{dt} =225 \times\frac{1}{50}$ $ \frac{dS}{dt} =4.5$ cm/s Since image distance from the lens is positive, image is moving away from the lens.
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