Question

A spherical snowball is melting in such a way that its diameter is decreasing at rate of 0.3 cm/min. At what rate is the volume of the snowball decreasing when the diameter is 9 cm. (Note the answer is a positive number).

Answer 1

Answer:

Approximately $38.2\; {\rm cm^{3}}$ per minute.

Step-by-step explanation:

This question is an example of related rates.

Let $V(t)$ and $d(t)$ denote the volume and diameter of this spherical snowball at time $t$, respectively. The goal is to find the rate of change in the volume of the snowball at the current moment $t_{0}$, $V^{\prime}(t_{0})$ despite the question did not provide an expression for $V(t)\!$ in terms of time $t$. Nevertheless, since this snowball is a sphere it is possible to find an expression for $V(t)$ in terms of $d(t)$:

$\displaystyle V(t) = \frac{1}{6}\, \pi\, (d(t))^{3}$.

It is also given that at the current moment, $d(t_{0}) = 9\; {\rm cm}$ and $d^{\prime}(t_{0}) = 0.3\; {\rm cm\cdot \text{min}^{-1}}$.

To find $V^{\prime}(t_{0})$, make use of the fact that $V(t)$ can be expressed in terms of $d(t)$. Specifically, apply the chain rule of differentiation to implicitly differentiate $V(t)\!$ and express $V^{\prime}(t)$ in terms of $d(t)$ and $d^{\prime}(t)$:

$\begin{aligned}V^{\prime}(t) &= \frac{d}{d t}\left[\frac{1}{6}\, \pi\, (d(t))^{3}\right] \\ &= \frac{1}{6}\, \pi\, \left(\frac{d}{d t}\left[(d(t))^{3}\right]\right) \\ &= \frac{1}{6}\, \pi\, \left(3\, (d(t))^{2}\, \frac{d}{d t}\left[d(t)\right]\right) && (\text{chain rule}) \\ &= \frac{1}{2}\, \pi \, (d(t))^{2}\, d^{\prime}(t)\end{aligned}$.

Substitute in the value of $d(t_{0})$ and $d^{\prime}(t_{0})$ to find $V^{\prime}(t_{0})$:

$\begin{aligned}V^{\prime}(t) &= \frac{1}{2}\, \pi \, (d(t))^{2}\, d^{\prime}(t) \\ &= \frac{1}{2}\, \pi\, \left(9)^{2}\, (0.3) \\ &\approx 38.2\end{aligned}$.

In other words, the rate of change in th evolume of this snowball would be approximately $38.2\; {\rm cm^{3}}$ at the moment.