Suppose the snowball in Problem 21 melts so that the rate of change in its diameter is proportional to its surface area. Using the same given data, determine when its diameter will be 2 in. Mathematically speaking, when will the snowball disappear?

Given that the rate of change of the diameter of a snowball is directly proportional to its surface area. Let the diameter of the snowball be D and its surface area be S.

Therefore, $\frac{\mathbf{d}\mathbf{D}}{\mathbf{d}\mathbf{t}}\mathbf{\propto }\mathit{S}$. 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,

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

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

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

Thus, from equation (1),

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

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

Separating the variables in equation (2),

$\frac{1}{{\mathrm{r}}^2}\mathrm{dr}=2\mathrm{k}\pi \mathrm{dt}$

Integrating both sides,

$-\frac{1}{\mathrm{r}}=2\mathrm{k}\pi \mathrm{t}+\mathrm{C}······\left(3\right)$

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

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

When t=0, r=2

Therefore, from (3)

$$\begin{gathered} C=-\frac{1}{2} \\ -\frac{1}{\mathrm{r}}=2\mathrm{k}\pi \mathrm{t}-\frac{1}{2} \end{gathered}$$

Hence,

$\mathrm{r}=\frac{2}{1-2\mathrm{k}\pi \mathrm{t}}······\left(4\right)$

Now as given that diameter = 3 in. at t=30 min and taking $\pi =3.14$,

Hence, from (4)

$$\begin{gathered} \text{r =}\frac{\text{2}}{\text{1 - 2k}\pi \text{t}} \\ \text{k = - 1.769} \end{gathered}$$

Now equation (4) becomes,

$r=\frac{2}{1+\left(1.769\right)t}······\left(5\right)$

When diameter = 2 in.

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

So, from (5)

$$\begin{gathered} r=\frac{2}{1+\left(1.769\right)t} \\ t=0.56\min \end{gathered}$$

Accordingly, the diameter of the snowball will be 2 in. 0.56 minutes.

The snowball will disappear, when diameter = 0 in.

Therefore radius, r = 0 in.

From equation (5),

$0=\frac{2}{1+\left(1.769\right)t}$

So, we see that we can’t find the value there.

Consequently, the snowball will not be disappeared.