Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Section 3.9 - Related Rates - 3.9 Exercises - Page 249: 5


$\frac{dh}{dt} = \frac{3}{25\pi} \frac{m}{min}$

Work Step by Step

Substitute the $r$ value and differentiate the equation to the volume of a cylinder. $V = \pi r^{2}h$ $V = \pi 5^{2}h$ $V = 25\pi h$ $\frac{dV}{dt} = 25\pi \frac{dh}{dt}$ Now substitute $\frac{dV}{dt}$ and solve it for $\frac{dh}{dt}$. $ 3 = 25\pi \frac{dh}{dt}$ Now solve for $\frac{dh}{dt}$: $\frac{dh}{dt} = \frac{3}{25\pi} \frac{m}{min}$
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