Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.9 - Related Rates - 3.9 Exercises - Page 251: 42

Answer

Cart B is moving toward point Q at a speed of $~~0.867~ft/s$

Work Step by Step

Let $L$ be the length of rope between cart A and pulley P. Let $a$ be the distance between cart A and point Q. $L^2 = a^2+12^2$ $L = \sqrt{a^2+144}$ $(39-L)$ is the length of rope between cart B and pulley P. Let $b$ be the distance between cart B and point Q. $b^2+12^2 = (39-L)^2$ $b^2+144 = (39- \sqrt{a^2+144})^2$ $b^2 = (39- \sqrt{a^2+144})^2-144$ $b^2 = (39- \sqrt{5^2+144})^2-144$ $b^2 = (39- 13)^2-144$ $b^2 = 532$ $b = 23.065$ We can find $\frac{db}{dt}$: $b^2 = (39- \sqrt{a^2+144})^2-144$ $2b~\frac{db}{dt} = 2(39- \sqrt{a^2+144})(-\frac{2a}{2~\sqrt{a^2+144}})(\frac{da}{dt})$ $\frac{db}{dt} = (\frac{1}{b})(39- \sqrt{a^2+144})(-\frac{2a}{2~\sqrt{a^2+144}})(\frac{da}{dt})$ $\frac{db}{dt} = (\frac{1}{23.065})(39- \sqrt{5^2+144})(-\frac{(2)(5)}{2~\sqrt{5^2+144}})(2)$ $\frac{db}{dt} = -0.867~ft/s$ Cart B is moving toward point Q at a speed of $~~0.867~ft/s$
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