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) -160 pi (C) -320 pi (D) -400 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)  -160 pi (C)  -320 pi (D)  -400 pi The surface area of a sphere with radius  r is  4pir^(2).

Bytelearn · Accepted Answer

Surface Area Formula: 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.Chain Rule Application: To find dS/dt, we use the chain rule from calculus: dS/dt = dS/dr * dr/dt. We know dr/dt = -4 cm/s (since the radius is decreasing).Calculate dS/dr: First, we find dS/dr by differentiating S = 4πr^2 with respect to r. dS/dr = 8πr.Substitute r=10 cm: Now we plug in the value of r = 10 cm into dS/dr to get dS/dr = 8π(10) = 80π.Find dS/dt: Finally, we multiply dS/dr by dr/dt to find dS/dt: dS/dt = 80π * (-4) = -320π square centimeters per second.

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