Question

How fast is the radius of the basketball increasing when the radius is 16 cm if air is being pumped into a basketball at a rate of 100 cm3/sec?

Answer 1

In mathematicsland, basketballs are perfect spheres with `0` thickness (inside radius = 0utside radius).

Air is being pumped in at a rate of `100 " cm"^3"/sec"`.

This rate of change is a rate of change of something with respect to some (other) thing.
We expect related rates (of change) problems to involve derivatives with respect to time, `t`.

So, we expect to see a `(d("something"))/(dt)`.

Now, `" cm"^3"` is a measure of volume, so we see that we have:

`(dV)/dt = 100 " cm"^3"/sec"`

We are asked to find how fast radius, `r` is changing, that is, we are asked to find `(dr)/dt` when `r = 16" cm"`.

We need an equation with `V` and `r`. For a sphere,

`V=4/3 pir^3`

We understand that both `V` and `r` are functions of the (unmentioned) variable `t`. We will differentiate with respect to `t`:

`d/dt(V)= d/dt(4/3 pir^3)`

`(dV)/dt= 4/3 pi *3r^2 (dr)/dt`
(In implicit differentiation, we use the chain rule.)

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

Now, plug in what we know and solve for what we don't know:

`100 " cm"^3"/sec" = 4 pi (16" cm")^2 (dr)/dt`

So `(dr)/dt = 100/(4 pi 16^2) (" cm"^3"/sec")/" cm"^2`

`(dr)/dt = 25/( pi 16^2) " cm/sec"`

Do whatever arithmetic you like or are required to do to get the answer in an acceptable form.