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.

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pieroraue pieroraue · Answer 1 · 2019-10-10T16:46:46+02:00

Answer:

According to the passage, we have the next equation:

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

where "K" is a proportional constant

Leaving at the end with the next equation:

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

Integrating the equation, we have:

[tex]r=K*t+C[/tex]

where "C" is a constant

Then, we have the 2 conditions for the problem:

1) t=0 → r=10

Replacing in the equation, we have C = 10

2) t=5 → r=8

Replacing in the equation, we have K = -0.4

Finally, the time which the snowball will be completely melted will be when r = 0. So replacing in the equation

[tex]0=-0.4*t+10[/tex]

t = 25 minutes