## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 3 - Practice Exercises - Page 205: 147

#### Answer

$0.6$ km/sec and $\dfrac{18}{\pi}rev/min$

#### Work Step by Step

Since, velocity $v=\dfrac{dx}{dt}$ Then $v=\sec^2 \theta\dfrac{d\theta}{dt}=(-0.6)\sec^2 \theta$ Plug $\theta=0$ Then , we have $v=(-0.6)\sec^2 (0)=-0.6$ km/sec Thus, the speed is $0.6$ km/sec Since, 1 rev=2 $\pi$ radians Thus, $0.6 rad/sec=(\dfrac{0.6 rad}{s}) (\dfrac{1 rev}{2 \pi rad}) (\dfrac{60 sec}{1 min}) =\dfrac{18}{\pi}rev/min$ Hence, $0.6$ km/sec and $\dfrac{18}{\pi}rev/min$

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