The rate of change of the volume of a snowball that is melting is proportional to the surface area of the snowball.

Burhan Hopper

Burhan Hopper

Answered question

2021-01-02

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 43πr3. 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 & Explanation

Daphne Broadhurst

Daphne Broadhurst

Skilled2021-01-03Added 109 answers

Step 1
Given:
The rate of change of the volume of a snowball that is melting is proportional to the surface area of the snowball is perfectly spherical
Then the volume (in centimetres cubed) of a ball of radius r centimetres is
v=43πr3
And the surface area is
s=4πr2
Set up the differential equation for how r is changing.
Then, suppose that at time t=0 minutes,the radius is 10 centimetres.After 5 minutes,the radius is 8 centimetres
Step 2
To find: At the what time t will be snowball be completely melted?
From the given conditions ,
dVdt(αS)
dVdt(λS)......(1)
V=43πr3
dVdt=43π3r2drdt
By putting this value in (1)
The equation must be,
43π3r2drdt=λ4πr2
drdt=λ
r=λt+c
Now, t=0 and r=10
so,
r=λt+c
10=λ(0)+c
c=10 and here
r=λt+10......(2)
After the 5 minutes, t=5 and r=8
8=λ(5)+10
5λ=2
λ=25
Now equation (2)becomes,
r=25t+10......(3)
This shows the differential equation for how r is changing.
As the snowball completely melted that means the radius of the snowball is zero.
From this by substituting r=0 in the equation (3) must be,
0=25t+10
25t=10
t=10×52 t=25 Hence the time required for melting the snowball is 25 minutes.

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