The radius of a sphere is decreasing at a rate of 1 meter per hour. At a certain instant, the radius is 4 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) -(256 pi)/(3) (B) -64 pi (C) -(4pi)/(3)pi (D) -16 pi The volume of a sphere with radius r is (4)/(3)pir^(3).

Question

The radius of a sphere is decreasing at a rate of 1 meter per hour. At a certain instant, the radius is 4 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)  -(256 pi)/(3) (B)  -64 pi (C)  -(4pi)/(3)pi (D)  -16 pi The volume of a sphere with radius  r is  (4)/(3)pir^(3).

Bytelearn · Accepted Answer

Volume Derivative Formula: The formula for the volume of a sphere is V = (4/3)πr^3. We need to find the derivative of the volume with respect to time, dV/dt.Derivative with Respect to Radius: First, let's find the derivative of V with respect to r, which is dV/dr. Using the power rule, dV/dr = 4πr^2.Chain Rule Application: Now, we use the chain rule to find dV/dt. Since dr/dt = -1 m/hr (the radius is decreasing), dV/dt = dV/dr * dr/dt.Substitution and Calculation: Substitute dV/dr = 4πr^2 and dr/dt = -1 m/hr into the equation dV/dt = dV/dr * dr/dt to get dV/dt = 4πr^2 * (-1).Radius Substitution: Now plug in the value of the radius r = 4 meters into the equation dV/dt = 4π(4)^2 * (-1) to calculate the rate of change of volume.Final Volume Rate Calculation: The calculation is dV/dt = 4π(16) * (-1) = -64π cubic meters per hour.

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