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?
Answers
In mathematicsland, basketballs are perfect spheres with 0 thickness (inside radius = 0utside radius).
Air is being pumped in at a rate of 100cm3/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)÷Dr.
Now, cm3 is a measure of volume, so we see that we have:
dV÷dt=100cm3/sec
We are asked to find how fast radius, r is.
changing, that is, we are asked to find dr÷dt when r=16cm.
We need an equation with V and r. For a sphere, V=4/3πr3
We understand that both V and r are functions of the (unmentioned) variable t.We will differentiate with respect to t :
1. d÷dt(V)=d÷dt(43πr3)
2. dV÷dt=43π⋅3r^2dr÷dt
(In implicit differentiation, we use the chain rule.)
dV÷dt=4πr2dr÷dt
So , dr÷dt=1004π16^2cm. 3/sec ÷ cm2
dr÷dt=25π16^2cm/sec
Do whatever arithmetic you like or are required to do to get the answer in an acceptable form.