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A spherical snowball is melting at the rate of 2 cubic inches per minute (and it's staying spherical). How fast is the radius of the snowball decreasing when the radius is one-half inch?

User Koshinae
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1 Answer

5 votes
Volume of Sphere V = (4/3)πr³

Given dV/dt = 2 in³/minute, r = 1/2 = 0.5 inch

dV/dt = dV/dr * dr/dt

V = (4/3)πr³

dV/dr = 3*(4/3)πr³ ⁻ ¹ = 4πr² = 4π*0.5² = 4π*0.25 = π in²

dV/dt = dV/dr * dr/dt

2 in³/minute = π * (dr/dt)

π in² * (dr/dt) = 2 in³/minute

(dr/dt) = (2 in³/minute) / (π in²)

(dr/dt) = (2/π) in/minute.

The radius of the snowball is reducing at (2/π) inches/minute.
User Rolf Bartstra
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