What is the rate of change of the volume of a ball (V=(4/3)pi r^3) with respect to the radius when the radius is r=2?

LOL, it is the surface area 4 pi r^2

4 pi r^2 dr = dV
or
dV/dr = 4 pi r^2
or
dV/dt = 4 pi r^2 dr/dt

To find the rate of change of the volume of a ball with respect to the radius, we need to take the derivative of the volume function with respect to the radius and evaluate it at the given radius.

The volume of a ball is given by the formula V = (4/3)πr^3, where r represents the radius.

To find the derivative of V with respect to r, we can use the power rule for differentiation. According to the power rule, if we have a function f(x) = x^n, then the derivative of f(x) with respect to x is given by f'(x) = nx^(n-1).

Applying the power rule to the volume function V = (4/3)πr^3, we get:

dV/dr = (4/3)(3r^2)
dV/dr = 4πr^2

Now, we evaluate this derivative at the given radius, r = 2:

dV/dr = 4π(2^2)
dV/dr = 16π

Therefore, the rate of change of the volume of the ball with respect to the radius when the radius is r = 2 is 16π.