Question

what is the rate of change of the volume of a ball (v = \frac{4}{3}\pi r^{3}) with respect to the radius when the radius is r = 3? the volume changes at a rate of . (type an exact answer, using \pi as needed.)

Explanation:

Step1: Differentiate volume formula

The volume formula of a ball is $V = \frac{4}{3}\pi r^{3}$. Using the power - rule for differentiation $\frac{d}{dr}(x^{n})=nx^{n - 1}$, we get $\frac{dV}{dr}=\frac{4}{3}\pi\times3r^{2}=4\pi r^{2}$.

Step2: Substitute radius value

We are given $r = 3$. Substitute $r = 3$ into $\frac{dV}{dr}$. So $\frac{dV}{dr}\big|_{r = 3}=4\pi\times3^{2}$.

Step3: Calculate the result

$4\pi\times3^{2}=4\pi\times9 = 36\pi$.

Answer:

$36\pi$