A snowball melts at a rate of 2 cubic inch an hour. When the volume is 36π in^3, how fast is the radius shrinking? V=(4/3)(pi)(r^3)
well, you have the formula
v = 4/3 pi r^3
dv/dt = 4pi r^2 dr/dt
Now just find r from v, and you know that dv/dt = -2, so solve for dr/dt
To find the rate at which the radius is changing, we can start by differentiating the volume formula with respect to time.
Given: V = (4/3)πr^3 and dV/dt = -2 in^3/hr (since the volume is melting at a rate of 2 in^3/hr)
Differentiating both sides of the equation with respect to time (t) using implicit differentiation, we get:
dV/ dt = (dV/ dr) * (dr/ dt)
Now, let's solve for (dV/ dr). Taking the derivative of the volume formula with respect to r, we have:
dV/ dr = 4πr^2
Substituting this value along with the given value of dV/ dt into the equation, we get:
-2 = (4πr^2) * (dr/ dt)
Now, we can solve for the rate at which the radius is shrinking, dr/dt:
dr/dt = -2 / (4πr^2)
Simplifying this expression further, we get:
dr/dt = -1 / (2πr^2)
So, the rate at which the radius is shrinking is -1 / (2πr^2) units per hour.