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: 16

Answer

(a) Initially the east-west distance $x = 150~miles$ Initially the north-south distance $y = 0$ Ship A's speed to the east is $35~km/h$ so $\frac{dx}{dt} = -35~km/h$ Ship B's speed to the north is $25~km/h$ so $\frac{dy}{dt} = 25~km/h$ (b) The unknown is $\frac{dD}{dt}$ where $D$ is the distance between the ships. (c) We can see a sketch below. (d) $x^2+y^2 = D^2$ (e) At 4:00 pm, the distance between the two ships is increasing at a rate of $~~21.4~km/h$
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Work Step by Step

(a) Initially the east-west distance $x = 150~miles$ Initially the north-south distance $y = 0$ Ship A's speed to the east is $35~km/h$ so $\frac{dx}{dt} = -35~km/h$ Ship B's speed to the north is $25~km/h$ so $\frac{dy}{dt} = 25~km/h$ (b) The unknown is $\frac{dD}{dt}$ where $D$ is the distance between the ships. (c) We can see a sketch below. (d) $x^2+y^2 = D^2$ (e) We can find $x$ after $4~hours$: $x = 150-(4)(35) = 10$ We can find $y$ after $4~hours$: $y = (4)(25) = 100$ We can find $D$ after $4~hours$: $D^2 = x^2+y^2$ $D = \sqrt{x^2+y^2}$ $D = \sqrt{10^2+100^2}$ $D = 100.5~miles$ We can find $\frac{dD}{dt}$: $2x~\frac{dx}{dt}+2y~\frac{dy}{dt} = 2D~\frac{dD}{dT}$ $\frac{dD}{dt} = \frac{x~\frac{dx}{dt}+y~\frac{dy}{dt}}{D}$ $\frac{dD}{dt} = \frac{(10)(-35)+(100)(25)}{100.5}$ $\frac{dD}{dt} = 21.4~km/h$ At 4:00 pm, the distance between the two ships is increasing at a rate of $~~21.4~km/h$
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