## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 3 - Section 3.9 - Related Rates - 3.9 Exercises: 6

#### Answer

The volume is increasing at $25600\pi \frac{mm^{3}}{s}$.

#### Work Step by Step

The radius of the sphere is $40 mm$. $V(r) = \frac{4}{3} \pi r^{3}$ $\frac{dV}{dt} = \frac{4}{3}\pi \times \frac{d(r^{3})}{dt}$ Use the chain rule $(\frac{dr}{dt})$: $\frac{dV}{dt} = \frac{4}{3}\pi \times \frac{d(r^{3})}{dt} \times \frac{dr}{dt}$ $\frac{dV}{dt} = \frac{4}{3}\pi \times 3r^{2} \times \frac{dr}{dt}$ $\frac{dV}{dt} = 4\pi r^{2} \times \frac{dr}{dt}$ $\frac{dr}{dt} = 4\frac{mm}{s}$ $\frac{dV}{dt} = 4\pi (40)^{2} \times (4)$ $\frac{dV}{dt} = 4\pi (1600) \times (4)$ $\frac{dV}{dt} = 4\pi (6400)$ $\frac{dV}{dt} = 25600\pi \frac{mm^{3}}{s}$

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