Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.6 Exercises: 13

Answer

a) 3053.63 $in^{3}$, 48858.049 $in^{3}$ b) $\frac{d}{dr}$$\frac{4}{3}$ $\pi$$r^{3}$ = 4$\pi$$r^{2}$ $\frac{dr}{dt}$

Work Step by Step

a) r = 9 in $\frac{dr}{dt}$ = 3 in/min V = $\frac{4}{3}$ $\pi$$r^{3}$ $\frac{d}{dr}$ V = $\frac{d}{dr}$$\frac{4}{3}$ $\pi$$r^{3}$ = 4$\pi$$r^{2}$ $\frac{dr}{dt}$ =4$\pi$ $(9^{2})$ (3) = 3053.63 $in^{3}$ r = 36 in $\frac{dr}{dt}$ = 3 in/min = 4$\pi$$r^{2}$ $\frac{dr}{dt}$ =4$\pi$ $(36^{2})$ (3) =48858.049 $in^{3}$ b) $\frac{dr}{dt}$ is constant but $\frac{dV}{dt}$ is not because when you find the derivative of the volume, the volume is not constant. r increases when $\frac{dV}{dt}$ increases $\frac{d}{dr}$$\frac{4}{3}$ $\pi$$r^{3}$ = 4$\pi$$r^{2}$ $\frac{dr}{dt}$
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.