Shania Houston

2023-03-12

A spherical snowball melts so that its radius decreases at a rate of 4 in/sec. At what rate is the volume of the snowball changing when the radius is 8 in?

### Answer & Explanation

Jazlene Martin

The formula for volume of a sphere is $V=\frac{4}{3}{r}^{3}\pi$
Differentiating with respect to $t$, time.
$\frac{dV}{dt}=4{r}^{2}\left(\frac{dr}{dt}\right)$
The rate of change of the snowball is given by $\frac{dV}{dt}$. We know $\frac{dr}{dt}=-4$. We want to find the rate of change when $r=8$. Thus,
$\frac{dV}{dt}=4{\left(8\right)}^{2}\left(-4\right)$
$\frac{dV}{dt}=4\left(64\right)\left(-4\right)$
$\frac{dV}{dt}=-1024$
Consequently, the snowball's volume is shrinking at a rate of $-1024\frac{{\phantom{\rule{1ex}{0ex}}\text{in}}^{3}}{\text{sec}}$

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