A snowball melts in such a way that the rate of change in its volume is proportional to its surface area. If the snowball was initially 4 in. in diameter and after 30 min its diameter is 3 in., when will its diameter be 2 in.? Mathematically speaking, when will the snowball disappear?

Given that the rate of change of the volume of a snowball is directly proportional to its surface area.

Let the volume of the snowball be V and its surface area be S.

Therefore,$\frac{\mathbf{dV}}{\mathbf{dt}}\mathbf{\propto }\mathit{S}$

Given that the initial diameter of the snowball is 4 in., which becomes 3 in. after 30 min. We have to find the time after which its diameter will be 2 in. and the time after which the snowball will disappear.

Given,

$$\begin{gathered} \frac{\mathbf{dV}}{\mathbf{dt}}\mathbf{\propto }\mathit{S} \\ \frac{\mathbf{dV}}{\mathbf{dt}}\mathbf{=}\mathit{k}\mathit{S}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{\left(}\mathbf{1}\mathbf{\right)} \end{gathered}$$

where, k is the constant of proportionality.

Now as the snowball is a sphere and we know that

The formula of the volume of the sphere =$\frac{4}{3}\pi r^3$

And Formula of the surface area of sphere =$4\pi {\mathrm{r}}^2$

Where, r is the radius of the sphere (here, snowball).

Thus, from equation (1),

$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{4}{3}\pi {\mathrm{r}}^3\right)=\mathrm{k}\left(4\pi {\mathrm{r}}^2\right) \\ \frac{4}{3}\pi \left(3{\mathrm{r}}^2\frac{\mathrm{dr}}{\mathrm{dt}}\right)=\mathrm{k}\left(4\pi {\mathrm{r}}^2\right) \\ \frac{\text{dr}}{\text{dt}}\text{= k}······\left(2\right) \end{gathered}$$

Now one will use this differential equation to solve the question.

Separating the variables in equation (2),

$dr=kdt$

Integrating both sides,

$r=kt+C······\left(3\right)$

Given that initially the diameter of the snowball = 4 in.

So, the radius of snowball, r = 2 in.

When t=0, r=2

Hence, from (3)

2=0+C

C =2

So, $r=kt+2······\left(4\right)$

Now, as given that diameter = 3 in. at t=30 min,

Accordingly, from (4)

$$\begin{gathered} 1.5=k\left(30\right)+2 \\ k=-0.0167 \end{gathered}$$

$r=(-0.0167)t+2······\left(5\right)$

When diameter = 2 in.

i.e., radius, r = 1 in.

$$\begin{gathered} 1=(-0.0167)t+2 \\ t=59.88\min \\ t\approx 1hour \end{gathered}$$

Consequently, the diameter of the snowball will be 2 in. 1 hour.

The snowball will disappear, when diameter = 0 in.

Therefore radius, r = 0 in.

From equation (5),

$$\begin{gathered} 0=(-0.0167)t+2 \\ t=119.76\min \\ t\approx 2hours \end{gathered}$$