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The surface area of a sphere is increasing at a rate of 
14 pi square meters per hour.
At a certain instant, the surface area is 
36 pi square meters.
What is the rate of change of the volume of the sphere at that instant (in cubic meters per hour)?
Choose 1 answer:
(A) 
36 pi
(B) 
(sqrt(14 pi))^(3)
(C) 
(7)/(12)
(D) 
21 pi
The surface area of a sphere with radius 
r is 
4pir^(2).
The volume of a sphere with radius 
r is 
(4)/(3)pir^(3).

The surface area of a sphere is increasing at a rate of 14π 14 \pi square meters per hour.\newlineAt a certain instant, the surface area is 36π 36 \pi square meters.\newlineWhat is the rate of change of the volume of the sphere at that instant (in cubic meters per hour)?\newlineChoose 11 answer:\newline(A) 36π 36 \pi \newline(B) (14π)3 (\sqrt{14 \pi})^{3} \newline(C) 712 \frac{7}{12} \newline(D) 21π 21 \pi \newlineThe surface area of a sphere with radius r r is 4πr2 4 \pi r^{2} .\newlineThe volume of a sphere with radius r r is 43πr3 \frac{4}{3} \pi r^{3} .

Full solution

Q. The surface area of a sphere is increasing at a rate of 14π 14 \pi square meters per hour.\newlineAt a certain instant, the surface area is 36π 36 \pi square meters.\newlineWhat is the rate of change of the volume of the sphere at that instant (in cubic meters per hour)?\newlineChoose 11 answer:\newline(A) 36π 36 \pi \newline(B) (14π)3 (\sqrt{14 \pi})^{3} \newline(C) 712 \frac{7}{12} \newline(D) 21π 21 \pi \newlineThe surface area of a sphere with radius r r is 4πr2 4 \pi r^{2} .\newlineThe volume of a sphere with radius r r is 43πr3 \frac{4}{3} \pi r^{3} .
  1. Surface Area Calculation: Given the surface area S=36πS = 36 \pi square meters, we can find the radius rr of the sphere using the formula for surface area S=4πr2S = 4 \pi r^2.
  2. Radius Calculation: Solve for rr: 36π=4πr236 \pi = 4 \pi r^2.\newlineDivide both sides by 4π4 \pi: r2=36π4π=9r^2 = \frac{36 \pi}{4 \pi} = 9.\newlineTake the square root of both sides: r=3r = 3 meters.
  3. Rate of Change Relation: The rate of change of the surface area is given as dSdt=14π\frac{dS}{dt} = 14 \pi square meters per hour.\newlineWe can relate this to the rate of change of the volume dVdt\frac{dV}{dt} using the chain rule in calculus: dVdt=dVdrdrdSdSdt\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dS} \cdot \frac{dS}{dt}.
  4. Volume Differentiation: First, find dV/drdV/dr using the volume formula V=43πr3V = \frac{4}{3} \pi r^3. Differentiate VV with respect to rr: dV/dr=4πr2dV/dr = 4 \pi r^2.
  5. Surface Area Differentiation: Next, find drdS\frac{dr}{dS} using the surface area formula S=4πr2S = 4 \pi r^2. Differentiate SS with respect to rr: dSdr=8πr\frac{dS}{dr} = 8 \pi r. Now, solve for drdS\frac{dr}{dS}: drdS=1dSdr=18πr\frac{dr}{dS} = \frac{1}{\frac{dS}{dr}} = \frac{1}{8 \pi r}.
  6. dr/dS Calculation: Substitute r=3r = 3 meters into dr/dSdr/dS: dr/dS=1/(8π3)=1/(24π)dr/dS = 1 / (8 \pi * 3) = 1 / (24 \pi).
  7. Substitute rr into drdS:\frac{dr}{dS}: Now, substitute dVdr=4πr2\frac{dV}{dr} = 4 \pi r^2 and drdS=124π\frac{dr}{dS} = \frac{1}{24 \pi} into the chain rule equation.\newlinedVdt=(4πr2)(124π)(14π).\frac{dV}{dt} = (4 \pi r^2) \cdot \left(\frac{1}{24 \pi}\right) \cdot (14 \pi).
  8. Substitute into Chain Rule: Simplify the equation: dV/dt=(4π9)(1/(24π))(14π)dV/dt = (4 \pi \cdot 9) \cdot (1 / (24 \pi)) \cdot (14 \pi). dV/dt=(36π)(1/(24π))(14π)dV/dt = (36 \pi) \cdot (1 / (24 \pi)) \cdot (14 \pi).
  9. Simplify Chain Rule: Cancel out the π\pi terms and simplify: dVdt=3624×14\frac{dV}{dt} = \frac{36}{24} \times 14.dVdt=32×14\frac{dV}{dt} = \frac{3}{2} \times 14.dVdt=21\frac{dV}{dt} = 21.

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