the radius ( r ) of a sphere is increasing at a rate of 4 inches per minute.\n(a) find the rate of change of…

the radius ( r ) of a sphere is increasing at a rate of 4 inches per minute.\n(a) find the rate of change of the volume (in ( mathrm{in}^{3} / mathrm{min} )) when ( r = 11 ) inches and when ( r = 34 ) inches.\n( r = 11 ) in ( mathrm{in}^{3} / mathrm{min} )\n( r = 34 ) in ( mathrm{in}^{3} / mathrm{min} )\n(b) explain why the rate of change of the volume of the sphere is not constant even though ( \frac{d r}{d t} ) is constant.\n( \bigcirc ) if ( \frac{d r}{d t} ) is constant, ( \frac{d v}{d r} ) is equal to ( t^{2} ).\n( \bigcirc ) if ( \frac{d r}{d t} ) is constant, ( \frac{d v}{d t} ) is equal to ( r^{2} ).\n( \bigcirc ) if ( \frac{d r}{d t} ) is constant, ( \frac{d v}{d t} ) is proportional to ( r^{2} ).\n( \bigcirc ) if ( \frac{d r}{d t} ) is constant, ( \frac{d v}{d r} ) is proportional to ( t^{2} ).

the radius ( r ) of a sphere is increasing at a rate of 4 inches per minute.\n(a) find the rate of change of the volume (in ( mathrm{in}^{3} / mathrm{min} )) when ( r = 11 ) inches and when ( r = 34 ) inches.\n( r = 11 ) in ( mathrm{in}^{3} / mathrm{min} )\n( r = 34 ) in ( mathrm{in}^{3} / mathrm{min} )\n(b) explain why the rate of change of the volume of the sphere is not constant even though ( \frac{d r}{d t} ) is constant.\n( \bigcirc ) if ( \frac{d r}{d t} ) is constant, ( \frac{d v}{d r} ) is equal to ( t^{2} ).\n( \bigcirc ) if ( \frac{d r}{d t} ) is constant, ( \frac{d v}{d t} ) is equal to ( r^{2} ).\n( \bigcirc ) if ( \frac{d r}{d t} ) is constant, ( \frac{d v}{d t} ) is proportional to ( r^{2} ).\n( \bigcirc ) if ( \frac{d r}{d t} ) is constant, ( \frac{d v}{d r} ) is proportional to ( t^{2} ).

Answer

Explanation:

Step1: Recall the volume formula of a sphere

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

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

Using the chain - rule (\frac{dV}{dt}=\frac{dV}{dr}\cdot\frac{dr}{dt}). Since (\frac{dV}{dr} = 4\pi r^{2}) and (\frac{dr}{dt}=4). So (\frac{dV}{dt}=4\pi r^{2}\times4 = 16\pi r^{2}).

Step3: Calculate (\frac{dV}{dt}) when (r = 11)

Substitute (r = 11) into (\frac{dV}{dt}=16\pi r^{2}). (\frac{dV}{dt}=16\pi\times(11)^{2}=16\pi\times121 = 1936\pi\approx1936\times3.14=6080.24).

Step4: Calculate (\frac{dV}{dt}) when (r = 34)

Substitute (r = 34) into (\frac{dV}{dt}=16\pi r^{2}). (\frac{dV}{dt}=16\pi\times(34)^{2}=16\pi\times1156 = 18496\pi\approx18496\times3.14 = 57077.44).

Step5: Analyze part (b)

From (\frac{dV}{dt}=16\pi r^{2}) and (\frac{dr}{dt}) is constant ((\frac{dr}{dt} = 4)). We can see that (\frac{dV}{dt}) is proportional to (r^{2}) (because (\frac{dV}{dt}=16\pi r^{2}), and (16\pi) is a constant).

Answer:

  • For (r = 11) in: (6080.24) (in^{3}/min)
  • For (r = 34) in: (57077.44) (in^{3}/min)
  • For part (b): If (\frac{dr}{dt}) is constant, (\frac{dV}{dt}) is proportional to (r^{2}).