Question

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?

Answer 1

Answer:

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

Step-by-step explanation:

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 $V=\frac{4}{3}\pi r^3$

The surface area is $S=4\pi r^2$

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,

$\frac{dV}{dt}\propto S$

$\frac{dV}{dt}=\lambda S$ ....(1)

$V=\frac{4}{3}\pi r^3$

$\frac{dV}{dt}=\frac{4}{3}\pi 3r^2\frac{dr}{dt}$

Substitute in (1),

$\frac{4}{3}\pi 3r^2\frac{dr}{dt}=\lambda 4\pi r^2$

$\frac{dr}{dt}=\lambda$

$r=\lambda t+c$

Now, t=0 , r=10

So, $10=\lambda(0)+c$

$c=10$

i.e. $r=\lambda t+10$

After 5 minutes, t=5 , r=8

$8=\lambda (5)+10$

$5\lambda=-2$

$\lambda=-\frac{2}{5}$

The equation form is $r=-\frac{2}{5}t+10$

The snowball be completely melted means radius became zero.

$0=-\frac{2}{5}t+10$

$\frac{2}{5}t=10$

$t=\frac{10\times 5}{2}$

$t=25$

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