A spherical snowball is melting so that its diameter is decreasing at rate of 0.1 cm/min. At what is the rate is the volume of the snowball changing when the diameter is 12 cm?
if diameter decreases at .01 cm/min
then dr/dt = -.005 cm/min
Vol = (4/3)π r^3
d(Vol)/dt = 4π r^2 dr/dt
d(Vol)/dt = 4π(36)(-.005) = .....
To find the rate at which the volume of the snowball is changing, you need to use related rates and the formula for the volume of a sphere.
The formula for the volume of a sphere is given by V = (4/3)πr^3, where V is the volume and r is the radius.
We are given that the diameter of the snowball is decreasing at a rate of 0.1 cm/min. The rate of change of the diameter is the same as the rate of change of the radius, because the diameter is twice the radius.
Let's denote the radius of the snowball as r(t), where t is the time in minutes. We are given that dr/dt = -0.1 cm/min.
First, let's find the relation between the rate of change of the radius and the rate of change of the volume. We can do this by differentiating the volume formula with respect to time:
dV/dt = dV/dr * dr/dt
Taking the derivative of the volume formula with respect to r, we get:
dV/dr = 4πr^2
Now, substitute the given value for dr/dt and the value for the current radius (12 cm) into the equation:
dV/dt = (4π(12 cm)^2)(-0.1 cm/min)
Simplifying, we get:
dV/dt = -576π cm^3/min
Therefore, the rate at which the volume of the snowball is changing when the diameter is 12 cm is approximately -576π cm^3/min.