a spherical balloon is inflated with gas at a rate of 900 cubic centimeters per minute.\n(a) find the rate…

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

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

Answer

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}).