The rate of change of the volume of a snowball that is melting is proportional to the surface area of the snowball. Suppose the snowball is perfectly spherical. Then the volume (in centimeters cubed) of a ball of radius r centimeters is 4/3πr3. The surface area is 4πr2. Set up the diﬀerential equation for how r is changing. Then, suppose that at time t = 0 minutes, the radius is 10 centimeters. After 5 minutes, the radius is 8 centimeters. At what time t will the snowball be completely melted?

Given : The rate of change of the volume of a snowball that is melting is proportional to the surface area of the snowball. Suppose the snowball is perfectly spherical.

Then the volume (in centimeters cubed) of a ball of radius r centimeters is [tex]V=\frac{4}{3}\pi r^3[/tex]

The surface area is [tex]S=4\pi r^2[/tex]

Set up the diﬀerential equation for how r is changing. Then, suppose that at time t = 0 minutes, the radius is 10 centimeters. After 5 minutes, the radius is 8 centimeters.

To find : At what time t will the snowball be completely melted?

Solution :

Using given condition,

[tex]\frac{dV}{dt}\propto S[/tex]

[tex]\frac{dV}{dt}=\lambda S[/tex] ....(1)

[tex]V=\frac{4}{3}\pi r^3[/tex]

[tex]\frac{dV}{dt}=\frac{4}{3}\pi 3r^2\frac{dr}{dt}[/tex]

Substitute in (1),

[tex]\frac{4}{3}\pi 3r^2\frac{dr}{dt}=\lambda 4\pi r^2[/tex]

[tex]\frac{dr}{dt}=\lambda[/tex]

[tex]r=\lambda t+c[/tex]

Now, t=0 , r=10

So, [tex]10=\lambda(0)+c[/tex]

[tex]c=10[/tex]

i.e. [tex]r=\lambda t+10[/tex]

After 5 minutes, t=5 , r=8

[tex]8=\lambda (5)+10[/tex]

[tex]5\lambda=-2[/tex]

[tex]\lambda=-\frac{2}{5}[/tex]

The equation form is [tex]r=-\frac{2}{5}t+10[/tex]

The snowball be completely melted means radius became zero.

[tex]0=-\frac{2}{5}t+10[/tex]

[tex]\frac{2}{5}t=10[/tex]

[tex]t=\frac{10\times 5}{2}[/tex]

[tex]t=25[/tex]

The time will be 25 minutes in which snowball be completely melted.