Suppose that an air is being pumped into a spherical weather balloon. At any time $t$, the volume of the balloon is $V(t)$ and its radius is $R(t)$.
a.) What do the derivatives $\displaystyle \frac{dV}{dr}$ and $\displaystyle \frac{dV}{dt}$ represent?
$\displaystyle \frac{dV}{dr}$ represents how does the volume changes with respect to its radius. On the other hand, $\displaystyle \frac{dV}{dt}$ represents how quick the volume is changing with respect to time.
b.) Express $\displaystyle \frac{dV}{dt}$ in terms of $\displaystyle \frac{dr}{dt}$.
Recall that the volume of a sphere is $\displaystyle V(r) = \frac{4}{3} \pi r^3$ and by using Chain Rule we have,
$
\begin{equation}
\begin{aligned}
\frac{dV}{dt} =& \frac{dV}{dr} \left( \frac{dr}{dt} \right) = \frac{4 \pi}{3} \frac{d}{dt} (r^3) = \frac{4 \pi}{3} \frac{d}{dr} (r^3) \frac{dr}{dt} = \frac{4 \pi}{\cancel{3}} (\cancel{3}r^2) \frac{dr}{dt}
\\
\\
\frac{dV}{dt} =& 4 \pi r^2 \frac{dr}{dt}
\end{aligned}
\end{equation}
$
Sunday, April 23, 2017
Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 80
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