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



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