Q3.2-22E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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?

Step-by-Step Solution

Answer

The diameter of the snowball will be 2 in. 0.56 minutes and the snowball will not be disappeared.

Step 1: Analyzing the given statement

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.

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

Given, 

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

where, k is the constant of proportionality. Now as the snowball is a sphere and we know that The formula of the diameter of the sphere =$2r$             

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.

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

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.

 

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),

 

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

 

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

 

Consequently, the snowball will not be disappeared.