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: 13

Answer

(a) The altitude $y = 1~mile$ The velocity is $\frac{dx}{dt} = 500~mi/h$ The distance $D$ between the plane and the station is two miles. (b) The unknown is $\frac{dD}{dt}$ (c) We can see a sketch below. (d) $x^2+y^2 = D^2$ (e) The rate at which the distance from the plane to the station is increasing when it is 2 miles away from the station is $~~433~mi/h$
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Work Step by Step

(a) The altitude $y = 1~mile$ The velocity is $\frac{dx}{dt} = 500~mi/h$ The distance $D$ between the plane and the station is two miles. (b) The unknown is $\frac{dD}{dt}$ (c) We can see a sketch below. (d) $x^2+y^2 = D^2$ (e) We can find $x$ when $D = 2~miles$: $x^2+y^2 = D^2$ $x^2 = D^2-y^2$ $x = \sqrt{D^2-y^2}$ $x = \sqrt{(2)^2-(1)^2}$ $x = \sqrt{3}~miles$ $2x~\frac{dx}{dt}+2y~\frac{dy}{dt} = 2D~\frac{dD}{dT}$ $\frac{dD}{dt} = \frac{x~\frac{dx}{dt}+y~\frac{dy}{dt}}{D}$ $\frac{dD}{dt} = \frac{(\sqrt{3})(500)+(1)(0)}{2}$ $\frac{dD}{dt} = 433~mi/h$ The rate at which the distance from the plane to the station is increasing when it is 2 miles away from the station is $~~433~mi/h$
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