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).
To find the rate at which the volume of the snowball is decreasing, we need to relate the rate at which the diameter is decreasing to the rate at which the volume changes.
The volume of a sphere is given by the formula V = (4/3)πr^3, where r is the radius of the sphere.
In this problem, the diameter is decreasing at a rate of 0.2 cm/min. Since the radius is half of the diameter, the radius is also decreasing at a rate of 0.2 cm/min.
We are given that the diameter is 8 cm, so the radius is half of that, which is 4 cm.
To find the rate at which the volume is decreasing, we need to differentiate the volume formula with respect to time (t) and then substitute the values.
Differentiating the volume formula with respect to t, we get:
dV/dt = d/dt [(4/3)πr^3]
Using the chain rule, we can write this as:
dV/dt = (4/3)π(3r^2)(dr/dt)
Substituting the given values, where r = 4 cm and dr/dt = -0.2 cm/min (since the diameter is decreasing), we get:
dV/dt = (4/3)π(3(4 cm)^2)(-0.2 cm/min)
Simplifying this expression, we find:
dV/dt = (4/3)π(48 cm^2)(-0.2 cm/min)
dV/dt = -32π cm^3/min
So, the volume of the snowball is decreasing at a rate of 32π cm^3/min when the diameter is 8 cm.