University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.10 - Related Rates - Exercises - Page 189: 33

Answer

$11\frac{ft}{sec}$

Work Step by Step

The given balloon is rising: $\frac{dy}{dt}=1\frac{ft}{sec}$ bicycle speed: $\frac{dx}{dt}=17\frac{ft}{sec}$ at $t=3sec$ balloon rising $y=68 ft $ high bicycle moved $x=51ft$ on applying the Pythagorean Theorem: $s^2=x^2+y^2$ $s=\sqrt{(68^2+51^2)}=85 ft$ on differentiating with respect to time: $2s\frac{ds}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}$ $\frac{ds}{dt}=\frac{1}{s}(x\frac{dx}{dt}+y\frac{dy}{dt})$ $\frac{ds}{dt}=\frac{1}{85}(51(17)+68(1))=11\frac{ft}{sec}$ Thus the final answer is: $11\frac{ft}{sec}$
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