the volume of spherical tumor is observed to be decreasing at a rate of 2 cubic centimeters per month when the radius is 5 centimeters. at what rate is the radius of the tumor decreasing?

Question

the volume of a sphere is this formula(V=4/3pir^3)

Answer 1

We have our volume equation: V = 4/3 * pi * r³

We also know that Volume is a function of time which means the radius is a function of time. We can use chain rule to obtain:

dV/dt = 4 * pi * r(t)² * (dr/dt)

From the problem we have a value for r(t) and dV/dt:

r(t) = 5, dV/dt = -2

Substituting:

-2 = 4 * pi * (5)² * (dr/dt)

-2 = 4 * 25 * pi * (dr/dt)

-2/(100*pi) = dr/dt

This means that when the volume is decreasing by 2 cm³/month, the radius of the tumor is 5 cm and at that time, the radius is decreasing by 2/(100*pi) cm/month.