A spherical snowball is melting in such a way that its diameter is decreasing at rate of 0.2 cm/min. At what rate is the volume of the snowball decreasing when the diameter is 8 cm? (Note the answer is a positive number).

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To find the rate at which the volume of the snowball is decreasing, we can use the formula for the volume of a sphere: V = (4/3)πr^3, where V is the volume and r is the radius.

We are given that the diameter is decreasing at a rate of 0.2 cm/min. Since the diameter is twice the radius, we can say that the radius is decreasing at a rate of 0.1 cm/min.

To find the rate at which the volume is decreasing, we need to find dV/dt, which represents the derivative of the volume with respect to time.

We can find dV/dt using the chain rule. Since V = (4/3)πr^3, we can differentiate both sides with respect to time (t):

dV/dt = d/dt [(4/3)πr^3]

Using the chain rule, we have:

dV/dt = (4/3)π * d/dt (r^3)

To find d/dt (r^3), we differentiate r^3 with respect to time (t):

d/dt (r^3) = 3r^2 * dr/dt

Substituting this back into the equation, we have:

dV/dt = (4/3)π * 3r^2 * dr/dt

Since we know that r is decreasing at a rate of 0.1 cm/min, we can substitute this value:

dV/dt = (4/3)π * 3(8^2) * (-0.1)

Simplifying, we have:

dV/dt = (4/3)π * 3 * 64 * (-0.1)

dV/dt = -256π * 0.3

Finally, calculating the value:

dV/dt ≈ -241.92 cm^3/min

Therefore, the rate at which the volume of the snowball is decreasing when the diameter is 8 cm is approximately 241.92 cm^3/min.