University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.5 - Derivatives of Trigonometric Functions - Exercises - Page 152: 61


a) $x(0)=10(cm)$, $x(\pi/3)=5(cm)$ and $x(3\pi/4)=-5\sqrt2(cm)$ b) $v(0)=0(cm/s)$, $v(\pi/3)=-5\sqrt3(cm/s)$ and $v(3\pi/4)=-5\sqrt2(cm/s)$.

Work Step by Step

$$x=10\cos t$$ a) For $t=0$: $$x=10\cos0=10\times1=10(cm)$$ - For $t=\pi/3$: $$x=10\cos\frac{\pi}{3}=10\times\frac{1}{2}=5(cm)$$ - For $t=3\pi/4$: $$x=10\cos\frac{3\pi}{4}=10\times\Big(-\frac{\sqrt2}{2}\Big)=-5\sqrt2(cm)$$ b) First, we need to deduce the formula for velocity here: $$v=\frac{dx}{dt}=(10\cos t)'=-10\sin t$$ - For $t=0$: $$v=-10\sin0=-10\times0=0(cm/s)$$ - For $t=\pi/3$: $$v=-10\sin\frac{\pi}{3}=-10\times\frac{\sqrt3}{2}=-5\sqrt3(cm/s)$$ - For $t=3\pi/4$: $$v=-10\sin\frac{3\pi}{4}=-10\times\frac{\sqrt2}{2}=-5\sqrt2(cm/s)$$
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