Suppose you are blowing a spherical bubble, filling it in with air at a uniform rate of 400mm^3/s. How fast is the radius of the bubble increasing by the time it is already 20mm long?

Question

Samantha Adkison · Accepted Answer

#1/(4pi)# #"mm/s"#

#V=4/3pir^3# (volume of sphere)

if #r=20mm#, #V=4/3pi(20mm)^3=32000/3pi(mm)^3#

Differentiate the two sides in terms of time:

#(dV)/dt=d/dt(4/3pir^3)# #(dV)/dt=4pir^2*(dr)/dt#

volume of air increasing at uniform rate of 400mm^3/s = #(dV)/dt#

substitute: #400(mm)^3/s=4pir^2*(dr)/dt#

#(dr)/dt=400/(4pir^2)(mm)^3/s# #(dr)/dt=400/(4pi(20mm)^2)(mm)^3/s# (#r=20mm#) #(dr)/dt=1/(4pi)(mm)/s#

#(dr)/dt# is rate radius is increasing

Paisley Johnson · Answer

undefined To find the rate at which the radius of the bubble is increasing, we can use the formula for the volume of a sphere, which is $V = \frac{4}{3}\pi r^3$, where $V$ is the volume and $r$ is the radius.

We are given that the volume is increasing at a rate of $400 \, \text{mm}^3/\text{s}$. So, we can find the rate of change of the radius, $ \frac{dr}{dt} $, by differentiating the volume formula with respect to time $t$, assuming the radius is also a function of time.

$$
\frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)
$$

Using the chain rule, we get:

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

Given that $ \frac{dV}{dt} = 400 \, \text{mm}^3/\text{s} $ and $ r = 20 \, \text{mm} $ (when the bubble has a radius of $ 20 \, \text{mm} $), we can solve for $ \frac{dr}{dt} $:

$$
400 = 4\pi (20)^2 \frac{dr}{dt}
$$
$$
400 = 1600\pi \frac{dr}{dt}
$$
$$
\frac{dr}{dt} = \frac{400}{1600\pi}
$$
$$
\frac{dr}{dt} = \frac{1}{4\pi}
$$

So, the rate at which the radius of the bubble is increasing when it is 20 mm long is $ \frac{1}{4\pi} $ mm/s.