91影视

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, $\frac{\mathbf{d}\mathbf{D}}{\mathbf{d}\mathbf{t}}\mathbf{鈭�}\mathit{S}$. 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,

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

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

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

Thus, from equation (1),

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

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

Separating the variables in equation (2),

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

Integrating both sides,

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

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)

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

Hence,

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

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

Hence, from (4)

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

Now equation (4) becomes,

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

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鈥檛 find the value there.

Consequently, the snowball will not be disappeared.