Question

The radius of a sphere is increasing at a rate of 4 mm/s. How fast is the volume increasing when the diameter is 40 mm?

Answer 1

Using `r` to represent the radius and `t` for time, you can write the first rate as:

`(dr)/(dt) = 4 "mm"/"s"`

or

`r = r(t) = 4t`

The formula for a solid sphere's volume is:

`V = V(r) = 4/3pir^3`

When you take the derivative of both sides with respect to time...

`(dV)/(dt) = 4/3pi(3r^2)((dr)/(dt))`

...remember the Chain Rule for implicit differentiation. The general format for this is:

`(dV(r))/(dt) = (dV(r))/(dr(t))*(dr(t))/(dt)`

with `V = V(r)` and `r = r(t)`.

So, when you take the derivative of the volume, it is with respect to its variable `r` `((dV(r))/(dr(t)))`, but we want to do it with respect to `t` `((dV(r))/(dt))`. Since `r = r(t)` and `r(t)` is implicitly a function of `t`, to make the equality work, you have to multiply by the derivative of the function `r(t)` with respect to `t` `((dr(t))/(dt))`as well. That way, you're taking a derivative along a chain of functions, so to speak (`V -> r -> t`).

Now what you can do is simply plug in what `r` is (note you were given diameter) and what `(dr)/(dt)` is, because `(dV)/(dt)` describes the rate of change of the volume over time, of a sphere.

`(dV)/(dt) = 4/3pi(3(20 "mm")^2)(4 "mm"/"s")`

`= 6400pi "mm"^3/"s"`

Since time just increases, and the radius increases as a function of time, and the volume increases as a function of a constant times the radius cubed, the volume increases faster than the radius increases, so we can't just say the two rates are the same.