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 - Page 206: 79

Answer

The velocity of the particle after $t$ seconds is $$\frac{5}{2}\pi\cos(10\pi t) (cm/s)$$

Work Step by Step

$$s(t)=10+\frac{1}{4}\sin(10\pi t)$$ From physics, the velocity of the particle is the rate of change of its displacement. That means, to get the formula of the velocity, we need to get the derivative of the displacement formula. In other words, $$v(t)=s'(t)=[10+\frac{1}{4}\sin(10\pi t)]'$$ $$v(t)=\frac{1}{4}\frac{d(\sin(10\pi t))}{dt}$$ Apply the Chain Rule: $$v'(t)=\frac{1}{4}\frac{d(\sin(10\pi t))}{d(10\pi t)}\frac{10\pi dt}{dt}$$ $$v'(t)=\frac{1}{4}\cos(10\pi t)\times10\pi\times1$$ $$v'(t)=\frac{5}{2}\pi\cos(10\pi t)$$ In conclusion, the velocity of the particle after $t$ seconds is $$\frac{5}{2}\pi\cos(10\pi t) (cm/s)$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.