Question

if the radius of a sphere is increasing at a constant rate of 3 cm/sec, then the volume is increasing at a rate of cm³/sec when the radius is 3 cm. hint: $\frac{dv}{dt}=\frac{dv}{dr}cdot\frac{dr}{dt}$, and the volume of a sphere is $v = \frac{4}{3}pi r^{3}$.

Explanation:

Step1: Differentiate volume formula

The volume of a sphere is $V = \frac{4}{3}\pi r^{3}$. Differentiating with respect to $r$ using the power - rule, we get $\frac{dV}{dr}=4\pi r^{2}$.

Step2: Use the chain - rule

We know that $\frac{dV}{dt}=\frac{dV}{dr}\cdot\frac{dr}{dt}$. Given $\frac{dr}{dt} = 3$ cm/sec.

Step3: Substitute values

When $r = 3$ cm, $\frac{dV}{dr}=4\pi(3)^{2}=36\pi$ cm$^{2}$. Then $\frac{dV}{dt}=\frac{dV}{dr}\cdot\frac{dr}{dt}$. Substituting $\frac{dV}{dr}=36\pi$ and $\frac{dr}{dt}=3$ into the equation, we have $\frac{dV}{dt}=36\pi\times3 = 108\pi$ cm$^{3}$/sec.

Answer:

$108\pi$