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?

Question

William Abdalla · Accepted Answer

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.

Emma Aldridge · Answer

undefined To find the rate at which the radius of the basketball is increasing, we can use the formula relating the volume $ V $ of a sphere to its radius $ r $:

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

Taking the derivative of both sides with respect to time $ t $, we get:

$$ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} $$

Given that the volume is increasing at a rate of $ \frac{dV}{dt} = 100 $ cm³/s when the radius $ r = 16 $ cm, we can plug in these values into the equation and solve for $ \frac{dr}{dt} $, the rate at which the radius is increasing:

$$ 100 = 4\pi (16)^2 \frac{dr}{dt} $$
$$ \frac{dr}{dt} = \frac{100}{4\pi (16)^2} $$
$$ \frac{dr}{dt} = \frac{100}{4\pi \times 256} $$
$$ \frac{dr}{dt} = \frac{100}{1024\pi} $$
$$ \frac{dr}{dt} ≈ \frac{100}{3207} $$
$$ \frac{dr}{dt} ≈ 0.0312 \, \text{cm/s} $$

Therefore, the radius of the basketball is increasing at a rate of approximately $ 0.0312 $ cm/s when the radius is 16 cm.