University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.10 - Related Rates - Exercises - Page 188: 20

Answer

$\frac{dA}{dt}=\pi\frac{cm^2}{min}$

Work Step by Step

Given $\frac{dr}{dt}=0.01\frac{cm}{min}$ and radius (r)=50 cm area (A)= $\pi r^2$ on differentiating both sides, we get: $\frac{dA}{dt}=2\pi r \frac{dr}{dt}$ r=50cm, so: $\frac{dA}{dt}=2\pi (50)\times 0.01$ $\frac{dA}{dt}=\pi\frac{cm^2}{min}$ Thus the final answer is:$\frac{dA}{dt}=\pi\frac{cm^2}{min}$
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.