The surface area of a sphere is increasing at a rate of 14π square meters per hour.At a certain instant, the surface area is 36π 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) 36π(B) (14π)3(C) 127(D) 21πThe surface area of a sphere with radius r is 4πr2.The volume of a sphere with radius r is 34πr3.
Q. The surface area of a sphere is increasing at a rate of 14π square meters per hour.At a certain instant, the surface area is 36π 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) 36π(B) (14π)3(C) 127(D) 21πThe surface area of a sphere with radius r is 4πr2.The volume of a sphere with radius r is 34πr3.
Surface Area Calculation: Given the surface area S=36π square meters, we can find the radius r of the sphere using the formula for surface area S=4πr2.
Radius Calculation: Solve for r: 36π=4πr2.Divide both sides by 4π: r2=4π36π=9.Take the square root of both sides: r=3 meters.
Rate of Change Relation: The rate of change of the surface area is given as dtdS=14π square meters per hour.We can relate this to the rate of change of the volume dtdV using the chain rule in calculus: dtdV=drdV⋅dSdr⋅dtdS.
Volume Differentiation: First, find dV/dr using the volume formula V=34πr3. Differentiate V with respect to r: dV/dr=4πr2.
Surface Area Differentiation: Next, find dSdr using the surface area formula S=4πr2. Differentiate S with respect to r: drdS=8πr. Now, solve for dSdr: dSdr=drdS1=8πr1.
dr/dS Calculation: Substitute r=3 meters into dr/dS: dr/dS=1/(8π∗3)=1/(24π).
Substitute r into dSdr: Now, substitute drdV=4πr2 and dSdr=24π1 into the chain rule equation.dtdV=(4πr2)⋅(24π1)⋅(14π).
Substitute into Chain Rule: Simplify the equation: dV/dt=(4π⋅9)⋅(1/(24π))⋅(14π). dV/dt=(36π)⋅(1/(24π))⋅(14π).
Simplify Chain Rule: Cancel out the π terms and simplify: dtdV=2436×14.dtdV=23×14.dtdV=21.
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