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: 43

Answer

(a) The distance from the television camera to the rocket is increasing at a rate of $~~360~ft/s$ (b) The angle of elevation is increasing at a rate of $~~0.096~rad/s$

Work Step by Step

(a) Let $y$ be the rocket's height. Let $x$ be the distance from the camera to the launching pad. Let $D$ be the distance between the camera and the rocket. We can find $D$: $D^2 = x^2+y^2$ $D = \sqrt{x^2+y^2}$ $D = \sqrt{4000^2+3000^2}$ $D = 5000~ft$ We can find $\frac{dD}{dt}$: $D^2 = x^2+y^2$ $2D~\frac{dD}{dt} = 2x~\frac{dx}{dt}+ 2y~\frac{dy}{dt}$ $\frac{dD}{dt} = \frac{x~\frac{dx}{dt}+ y~\frac{dy}{dt}}{D}$ $\frac{dD}{dt} = \frac{(4000)(0)+ (3000)(600)}{5000}$ $\frac{dD}{dt} = 360~ft/s$ The distance from the television camera to the rocket is increasing at a rate of $~~360~ft/s$ (b) Let $\theta$ be the camera's angle of elevation. We can find $\frac{d\theta}{dt}$: $\frac{y}{D} = sin~\theta$ $\frac{1}{D}~\frac{dy}{dt}-\frac{y}{D^2}~\frac{dD}{dt} = cos~\theta~\frac{d\theta}{dt}$ $\frac{d\theta}{dt}=(\frac{1}{cos~\theta})(\frac{1}{D}~\frac{dy}{dt}-\frac{y}{D^2}~\frac{dD}{dt})$ $\frac{d\theta}{dt}=(\frac{1}{4/5})[\frac{1}{5000}~(600)-\frac{3000}{5000^2}~(360)]$ $\frac{d\theta}{dt}= 0.096~rad/s$ The angle of elevation is increasing at a rate of $~~0.096~rad/s$
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