Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 3 - Section 3.9 - Related Rates - 3.9 Exercises - Page 251: 15

Answer

(a) The light is at a height of 15 feet. The man is 6 feet tall. The man is walking away from the pole at a rate of $5~ft/s$ (b) The unknown is $\frac{dz}{dt}$ where $z$ is the distance between the tip of the shadow and the pole. (c) We can see a sketch below. (d) $\frac{x}{6} = \frac{x+y}{15}$ $z = x+y$ (e) The tip of the shadow is moving at a speed of $~~\frac{25}{3}~ft/s$

Work Step by Step

(a) The light is at a height of 15 feet. The man is 6 feet tall. The man is walking away from the pole at a rate of $5~ft/s$ (b) The unknown is $\frac{dz}{dt}$ where $z$ is the distance between the tip of the shadow and the pole. (c) We can see a sketch below. (d) Using similar triangles: $\frac{x}{6} = \frac{x+y}{15}$ $z = x+y$ (e) We can find $\frac{dx}{dt}$: $\frac{x}{6} = \frac{x+y}{15}$ $15x = 6x+6y$ $9x = 6y$ $x = \frac{2}{3}~y$ $\frac{dx}{dt} = \frac{2}{3}~\frac{dy}{dt}$ $\frac{dx}{dt} = \frac{2}{3}~(5~ft/s)$ $\frac{dx}{dt} = \frac{10}{3}~ft/s$ We can find $\frac{dz}{dt}$: $z = x+y$ $\frac{dz}{dt} = \frac{dx}{dt}+\frac{dy}{dt}$ $\frac{dz}{dt} = (\frac{10}{3}~ft/s)+(5~ft/s)$ $\frac{dz}{dt} = \frac{25}{3}~ft/s$ The tip of the shadow is moving at a speed of $~~\frac{25}{3}~ft/s$
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