Question

a spherical balloon is inflated with gas at a rate of 900 cubic centimeters per minute.
(a) find the rate of change of the radius (in cm/min) when ( r = 20 ) centimeters and ( r = 45 ) centimeters.
( r = 20 ) cm/min
( r = 45 ) cm/min
(b) explain why the rate of change of the radius of the sphere is not constant even though ( \frac{dv}{dt} ) is constant.
the volume only appears constant; it is actually a rational relationship.
( \frac{dr}{dt} ) as a function runs parallel to the volume function, which is not linear.
the rate of change of the radius is a linear relationship whose slope is ( \frac{dv}{dt} ).
if ( \frac{dr}{dt} ) is constant, ( \frac{dv}{dt} ) depends on ( r^{2} ).
the rate of change of the radius is a cubic relationship.

Explanation:

Step1: Recall the volume formula for a sphere

The volume \(V\) of a sphere is given by \(V=\frac{4}{3}\pi r^{3}\).

Step2: Differentiate the volume formula with respect to time \(t\)

Using the chain - rule, \(\frac{dV}{dt} = 4\pi r^{2}\frac{dr}{dt}\).

Step3: Solve for \(\frac{dr}{dt}\)

We get \(\frac{dr}{dt}=\frac{1}{4\pi r^{2}}\frac{dV}{dt}\). Given \(\frac{dV}{dt} = 900\) \(cm^{3}/min\).

When \(r = 20\)

Substitute \(r = 20\) and \(\frac{dV}{dt}=900\) into \(\frac{dr}{dt}=\frac{1}{4\pi r^{2}}\frac{dV}{dt}\).
\(\frac{dr}{dt}=\frac{900}{4\pi\times(20)^{2}}=\frac{900}{1600\pi}=\frac{9}{16\pi}\approx\frac{9}{16\times3.14}\approx\frac{9}{50.24}\approx0.18\) \(cm/min\).

When \(r = 45\)

Substitute \(r = 45\) and \(\frac{dV}{dt}=900\) into \(\frac{dr}{dt}=\frac{1}{4\pi r^{2}}\frac{dV}{dt}\).
\(\frac{dr}{dt}=\frac{900}{4\pi\times(45)^{2}}=\frac{900}{8100\pi}=\frac{1}{9\pi}\approx\frac{1}{9\times3.14}\approx\frac{1}{28.26}\approx0.04\) \(cm/min\).

Step4: Analyze part (b)

From \(\frac{dr}{dt}=\frac{1}{4\pi r^{2}}\frac{dV}{dt}\), if \(\frac{dV}{dt}\) is constant, \(\frac{dr}{dt}\) depends on \(r^{2}\).

Answer:

a. When \(r = 20\), \(\frac{dr}{dt}\approx0.18\) \(cm/min\); when \(r = 45\), \(\frac{dr}{dt}\approx0.04\) \(cm/min\).
b. The correct option is: If \(\frac{dr}{dt}\) is constant, \(\frac{dV}{dt}\) depends on \(r^{2}\).