Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises: 88

Answer

(a) $\frac{dV}{dr}$ represents the rate of change of the volume of the balloon according to its radius. $\frac{dV}{dt}$ represents the rate of change of the volume of the balloon according to time. (b) $$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$$

Work Step by Step

(a) $\frac{dV}{dr}$ represents the rate of change of the volume of the balloon according to its radius. That means, as the balloon reaches radius $r_0$, the volume of the balloon is expanding at a rate of $\frac{dV}{dr_0}(volume/radius)$. $\frac{dV}{dt}$ represents the rate of change of the volume of the balloon according to time. That means, at time $t_0$, the volume of the balloon is expanding at a rate of $\frac{dV}{dt_0}(volume/time)$. (b) The weather balloon is spherical. The volume of the balloon with radius $r$ would be $$V=\frac{4}{3}\pi r^3$$ The derivative of $V(t)$ then would be $$\frac{dV}{dt}=\frac{d(\frac{4}{3}\pi r^3)}{dt}$$ $$\frac{dV}{dt}=\frac{4}{3}\pi\frac{d(r^3)}{dt}$$ Here we apply the Chain Rule, $$\frac{dV}{dt}=\frac{4\pi}{3}\frac{d(r^3)}{dr}\frac{dr}{dt}$$ $$\frac{dV}{dt}=\frac{4\pi}{3}\times3r^2\frac{dr}{dt}$$ $$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$$
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