# Calculus

A piece of ice stays in the shape of a sphere as it melts. The volume of the ice is decreasing at a constant rate of 3π/2 cubic feet per hour. What is the rate of change of the surface area of the piece of ice when the radius is 10 feet?

I found the rate of change of the radius to be -3/800 feet per hour and the rate of change of the surface area to be -3π/10 square feet per hour.

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1. A = 4πr^2
dA/dt = 8πr dr/dt

v = 4/3 πr^3
dv/dt = 4πr^2 dr/dt

when r=10, we have

-3π/2 = 400π dr/dt
dr/dt = -3/800
dA/dt = 80π * -3/800 = -3π/10

You are correct

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