The radius of a sphere is decreasing at a rate of 4 centimeters per second. At a certain instant, the radius is 10 centimeters. What is the rate of change of the surface area of the sphere at that instant (in square centimeters per second)? Choose 1 answer: (A) -64 pi (B) -320 pi (C) -400 pi (D) -160 pi The surface area of a sphere with radius r is 4pir^(2).

Question

The radius of a sphere is decreasing at a rate of 4 centimeters per second. At a certain instant, the radius is 10 centimeters. What is the rate of change of the surface area of the sphere at that instant (in square centimeters per second)? Choose 1 answer: (A)  -64 pi (B)  -320 pi (C)  -400 pi (D)  -160 pi The surface area of a sphere with radius  r is  4pir^(2).

Bytelearn · Accepted Answer

Formula Explanation: The formula for the surface area of a sphere is S = 4πr^2. We need to find the rate of change of the surface area, which is dS/dt.Rate of Change: Given that the radius is decreasing at a rate of dr/dt = -4 cm/s, we'll use the chain rule to find dS/dt: dS/dt = dS/dr * dr/dt.Chain Rule Application: First, find dS/dr by differentiating S = 4πr^2 with respect to r: dS/dr = 8πr.Differentiation with Respect to r: Now, plug in the values of r = 10 cm and dr/dt = -4 cm/s into the equation dS/dt = 8πr * dr/dt.Calculation of dS/dt: Calculate dS/dt: dS/dt = 8π(10) * (-4) = -320π square centimeters per second.

Full solution

More problems from Solve quadratic equations: word problems

Related topics