University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Practice Exercises - Page 205: 143


$2 m/sec$

Work Step by Step

Here, $(2s)\dfrac{ds}{dt}=[(2x) \dfrac{dx}{dt}+(2y) \dfrac{dy}{dt}]$ $s=\sqrt {(3)^2+(-4)^2}=5$ Plug $\dfrac{dx}{dt}=10m/sec; \dfrac{dy}{dt}=5m/sec$ Then , we have $\dfrac{ds}{dt}=(\dfrac{1}{s})(x \dfrac{dx}{dt}+y\dfrac{dy}{dt})$ This implies that $\dfrac{ds}{dt}=(\dfrac{1}{5})[(10)(3)+(-4)(5)]=2 m/sec$
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