(1 pt) A spherical snowball is melting in such a way that its diameter is decreasing at rate of 0.3 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).

V = (4/3) pi r^3

so
dV/dt = 4 pi r^2 dr/dt

To find the rate at which the volume of the snowball is decreasing when its diameter is 8 cm, we can use the derivative to relate the change in volume to the change in diameter.

The formula for the volume of a sphere is V = (4/3)πr³, where V is the volume and r is the radius.

Given that the diameter is decreasing at a rate of 0.3 cm/min, we can determine that the rate of change of the radius (dr/dt) is half of that, or -0.15 cm/min (because diameter is twice the radius).

We are looking for the rate at which the volume is decreasing, which is dV/dt. We can calculate this with the chain rule:

dV/dt = (dV/dr) * (dr/dt)

To find dV/dr, we differentiate the volume formula with respect to the radius:

dV/dr = 4πr²

Plugging in the radius at the given diameter (8 cm), we get:

dV/dr = 4π(8)^2 = 256π cm²

Now we can calculate dV/dt by multiplying dV/dr and dr/dt:

dV/dt = (256π cm²) * (-0.15 cm/min)

Simplifying this expression, we obtain:

dV/dt = -38.4π cm³/min

Since we are looking for the magnitude of the rate (a positive number), we take the absolute value of -38.4π:

dV/dt ≈ 120.678 cm³/min

Therefore, when the diameter of the snowball is 8 cm, the volume is decreasing at a rate of approximately 120.678 cm³/min.