Assume that the radius r of a sphere is expanding at a rate of 40 cm/min. The volume of a sphere is

V = 4/3πr^3 and its surface area is 4πr^2. Determine the rate of change in surface area when r = 10 cm.

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Mathematics

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31 December, 05:03

Assume that the radius r of a sphere is expanding at a rate of 40 cm/min. The volume of a sphere is

V = 4/3πr^3 and its surface area is 4πr^2. Determine the rate of change in surface area when r = 10 cm.

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Answers (1)

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Abram Coffey · Answer 1

S = 4pir^2

dS / dt = [dS / dr] * [dr/dt]

dS / dr = 8pir

r = 10 cm

dr / dt = 40 cm/min

=> dS/dt = [8pir]*40cm/min

=> dS = [8pi*10cm]*40cm/min = 10,053 cm^2 / min

Answer: 10,053 cm^2 / min.

Assume that the radius r of a sphere is expanding at a rate of 40 cm/min. The volume of a sphere isV = 4/3πr^3 and its surface area is 4πr^2. Determine the rate of change in surface area when r = 10 cm.

Assume that the radius r of a sphere is expanding at a rate of 40 cm/min. The volume of a sphere is

V = 4/3πr^3 and its surface area is 4πr^2. Determine the rate of change in surface area when r = 10 cm.