The radius of a sphere is decreasing at a rate of 4mm/s. How fast is the volume decreasing when...

Question:

The radius of a sphere is decreasing at a rate of 4mm/s. How fast is the volume decreasing when the diameter is 90 mm?

Volume of a sphere, derivative of a function, chain rule

The rate of change of a function {eq}f {/eq} at time {eq}t {/eq} is the derivative of this function evaluated at {eq}t {/eq}, also referred to as {eq}f^\prime(t) {/eq}. It is the slope of the tangent line to the curve at time {eq}t {/eq}.

Let us denote {eq}V(t) {/eq} the volume of the sphere at time {eq}t {/eq} in cube millimiters

and {eq}r(t) {/eq} the radius of the sphere at time {eq}t {/eq} in millimiters. Then using the formula for the volume of a sphere, the following holds :

$$V(t)=\frac{4}{3}\pi r(t)^3$$

Since we know that the radius of the sphere is decreasing at a rate of 4mm/s, we have that {eq}r^\prime(t)=-4 {/eq}. Using the chain rule, the rate of change of the volume at time t is given by

{eq}V^\prime(t)=\frac{d}{dt}\left[\frac{4}{3}\pi r(t)^3\right] = 4\pi r^\prime(t) r(t)^2 {/eq}

Since the radius is equal to one half of the diameter, at the moment {eq}t_s {/eq} when the diameter is 90mm, {eq}r(t_s)=45 {/eq}.

Plugging {eq}r^\prime(t_s)=-4 {/eq} yields

{eq}V^\prime(t_s)=4\pi\times(-4)\times 45^2=-8100\pi {/eq}. in mm^3. An approximate value for the above is {eq}-25446.90\,\text{mm}^3 {/eq}.