## Calculus 10th Edition

Published by Brooks Cole

# Chapter 2 - Differentiation - 2.6 Exercises: 14

#### Answer

a) 0.0707 cm/min b) 0.0177 cm/min

#### Work Step by Step

The volume of a sphere is $\frac{4}{3} \pi r^3$ The change in volume is given by $\frac {dV}{dt} = (4/3) (3) \pi r^2 = 4 \pi r^2 \frac{dr}{dt}$ We are told the volume is changing at a rate of + 800 $cm^3$ (positive since inflated) $800 = 4 \pi r^2 {dr}{dt}$ a) when the radius is 30cm, we solve for dr/dt $\frac {dr}{dt} = 0.0707 cm/min$ b) when the radius is 60cm, we solve for dr/dt $\frac {dr}{dt} = 0.0177 cm/min$

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