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 249: 19

Answer

The people are moving apart at a rate of $8.987~ft/s$

Work Step by Step

Let $x$ be the east-west distance between the two people. Then $x = 500~ft$ Let $y$ be the north-south distance between the two people. We can find $y$ fifteen minutes after the woman starts walking: $y = (4~ft/s)(20 \cdot 60~s)+ (5~ft/s)(15\cdot 60~s)$ $y = 9300~ft$ Let $z$ be the distance between the two people. We can find $z$ fifteen minutes after the woman starts walking: $z^2 = x^2+y^2$ $z = \sqrt{x^2+y^2}$ $z = \sqrt{(500~ft)^2+(9300~ft)^2}$ $z = 9313.43~ft$ We can differentiate both sides of the equation with respect to $t$: $z^2 = x^2+y^2$ $2z~\frac{dz}{dt} = 2x~\frac{dx}{dt} + 2y~\frac{dy}{dt}$ $\frac{dz}{dt} = \frac{1}{z}~(x~\frac{dx}{dt} + y~\frac{dy}{dt})$ $\frac{dz}{dt} = \frac{1}{9313.43}~[(500)(0) + (9300)(4+5)]$ $\frac{dz}{dt} = 8.987~ft/s$ The people are moving apart at a rate of $8.987~ft/s$.
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