# A snowball, in the shape of a sphere, is melting so that the radius is decreasing at a uniform rate of 1cm/s. How fast is the volume decreasing when the radius equals 6cm?

Aug 25, 2015

We can write the first relationship (in $\text{cm/s}$) as:

$\frac{\mathrm{dr}}{\mathrm{dt}} = 1$

We know that the formula for the volume of a sphere is:

$V \left(r\right) = \frac{4}{3} \pi {\left(r \left(t\right)\right)}^{3}$

Also, the derivative of $V$ with respect to $t$, $\frac{\mathrm{dV}}{\mathrm{dt}}$, implies that we also multiply by $\frac{\mathrm{dr}}{\mathrm{dt}}$, since $r = r \left(t\right)$, whose slope is $1$, and $\frac{\mathrm{dV} \left(r\right)}{\mathrm{dt}} = \frac{\mathrm{dV} \left(r\right)}{\mathrm{dr} \left(t\right)} \cdot \frac{\mathrm{dr} \left(t\right)}{\mathrm{dt}} = \frac{\mathrm{dV}}{\mathrm{dr}} \cdot \frac{\mathrm{dr}}{\mathrm{dt}}$.

Thus, if we differentiate this with respect to $t$, we can find a use for $\frac{\mathrm{dr}}{\mathrm{dt}}$ through the Chain Rule.

$\frac{\mathrm{dV}}{\mathrm{dt}} = \frac{4}{3} \pi \frac{d}{\mathrm{dt}} \left[{\left(r \left(t\right)\right)}^{3}\right]$

$= \frac{4}{3} \pi \frac{d}{\mathrm{dr}} \left[{r}^{3}\right] \left(\frac{\mathrm{dr}}{\mathrm{dt}}\right)$

$= \frac{4}{3} \pi \cdot 3 {r}^{2} \left(\frac{\mathrm{dr}}{\mathrm{dt}}\right)$

Finally, we can plug it in:

$= \frac{4}{3} \pi \left(3 {r}^{2}\right) \left(1\right) = \textcolor{g r e e n}{4 \pi {r}^{2}}$

Therefore, when $r = 6$, we get the rate of decrease in the volume while the radius is currently $\text{6 cm}$:

$\frac{\mathrm{dV} \left(6\right)}{\mathrm{dt}} = V ' \left(6\right) = 4 \pi \left(36\right)$

$= \textcolor{b l u e}{144 \pi \text{cm"^3"/s}}$

Aug 25, 2015

The volume is decreasing at a rate of $144 \pi c {m}^{3.} {s}^{-} 1$

#### Explanation:

We are told that $\frac{\mathrm{dr}}{\mathrm{dt}} = - 1 \text{cm/s}$

We need to find $\frac{\mathrm{dV}}{\mathrm{dt}}$ when $r = 6 \text{cm}$

We know:

$V = \frac{4}{3} . \pi {r}^{3}$

So:

$\frac{\mathrm{dV}}{\mathrm{dr}} = \frac{12}{3} . \pi {r}^{2} = 4 \pi {r}^{2}$

$\frac{\mathrm{dV}}{\mathrm{dt}} = \frac{\mathrm{dV}}{\mathrm{dr}} . \frac{\mathrm{dr}}{\mathrm{dt}}$

$\frac{\mathrm{dV}}{\mathrm{dt}} = 4 \pi {r}^{2} \times \left(- 1\right) = - 4 \pi {r}^{2}$

So when the radius $r = 6 \text{cm}$ we can substitute this in :

$\frac{\mathrm{dV}}{\mathrm{dt}} = - 4 \pi .6 \times 6 = - 144 \pi c {m}^{3.} {s}^{-} 1$