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

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

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=(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,

`(dV)/(dt)\propto S`

`(dV)/(dt)=\lambda S` ....(1)

`V=(4)/(3)\pi r^3`

`(dV)/(dt)=(4)/(3)\pi 3r^2(dr)/(dt)`

Substitute in (1),

`(4)/(3)\pi 3r^2(dr)/(dt)=\lambda 4\pi r^2`

`(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=-(2)/(5)`

The equation form is
`r=-(2)/(5)t+10`

The snowball be completely melted means radius became zero.

`0=-(2)/(5)t+10`

`(2)/(5)t=10`

`t=(10* 5)/(2)`

`t=25`

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