Q3.2-21E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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?

Step-by-Step Solution

Answer

The diameter of the snowball will be 2 in. after 1 hour and the snowball will disappear after 2 hours.

Step 1: Analyzing the given statement

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.

 

Step 2: Determining the differential equation using the given proportionality relation

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.

Step 3: Finding the time after which the diameter of the snowball will be 2 in.

Separating the variables in equation (2), 

$dr=k dt$ 

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)

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

Now equation (4) becomes,

$r=(-0.0167)t+2······\left(5\right)$                  

When diameter = 2 in.

 i.e.,     radius, r = 1 in.

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

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

Step 4: Finding the time after which the snowball will disappear

The snowball will disappear, when diameter = 0 in.

 

Therefore radius, r = 0 in.

 

From equation (5),

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

Hence, the snowball will disappear after 2 hours.