Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises: 80

Answer

(a) The velocity of the particle at time $t$ is $$-A\omega\sin(\omega t+\delta)$$ (b) The velocity is 0 when $$t=\frac{n\pi-\delta}{\omega}$$ $(n\in Z)$

Work Step by Step

$$s(t)=A\cos(\omega t+\delta)$$ (a) The formula for the velocity of the particle is the rate of change of the formula of its motion. That means the formula for the velocity of the particle is the derivative of the formula of its motion. In other words, $$v(t)=s'(t)=[A\cos(\omega t+\delta)]'$$ $$v(t)=A\frac{d\cos(\omega t+\delta)}{dt}$$ Apply the Chain Rule: $$v(t)=A\frac{d\cos(\omega t+\delta)}{d(\omega t+\delta)}\frac{d(\omega t+\delta)}{dt}$$ $$v(t)=-A\sin(\omega t+\delta)[\frac{\omega dt}{dt}+\frac{d\delta}{dt}]$$ $$v(t)=-A\sin(\omega t+\delta)(\omega\times1+0)$$ $$v(t)=-A\omega\sin(\omega t+\delta)$$ Therefore, the velocity of the particle at time $t$ is $$-A\omega\sin(\omega t+\delta)$$ (b) The velocity is 0 when $$-A\omega\sin(\omega t+\delta)=0$$ $$\sin(\omega t+\delta)=0$$ (here we only consider $t$ a variable, so $A$ and $\omega$ are unchanged values) $$\omega t+\delta=n\pi$$ $(n\in Z)$ $$t=\frac{n\pi-\delta}{\omega}$$ Therefore, the velocity is 0 when $$t=\frac{n\pi-\delta}{\omega}$$ $(n\in Z)$
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