A spherical snowball is melting so that its diameter is decreasing at rate of 0.2 cm/min. At what is the rate is the volume of the snowball changing when the diameter is 18 cm?
The volume is changing at a rate of
To find the rate at which the volume of the snowball is changing, we can use the formula for the volume of a sphere:
V = (4/3) * π * r^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 find the rate at which the radius is changing:
rate of change of radius = rate of change of diameter / 2
rate of change of radius = 0.2 cm/min / 2 = 0.1 cm/min
Now, we need to find the rate at which the volume is changing when the diameter is 18 cm. Since the radius is half the diameter, the radius is 18 cm / 2 = 9 cm.
Substituting the value of the radius into the volume formula, we have:
V = (4/3) * π * (9 cm)^3
V = (4/3) * π * 729 cm^3
Differentiating the volume formula with respect to time (t), we get:
dV/dt = (4/3) * π * 3 * (9 cm)^2 * (dr/dt)
Now, we can substitute the value of the rate of change of the radius (dr/dt = 0.1 cm/min) and the radius (9 cm) into the formula:
dV/dt = (4/3) * π * 3 * (9 cm)^2 * (0.1 cm/min)
dV/dt = (4/3) * π * 3 * 81 cm^2 * 0.1 cm/min
Simplifying this expression, we find the rate at which the volume is changing:
dV/dt = (4/3) * π * 243 cm^3/min
Therefore, the volume is changing at a rate of 324π cm^3/min.