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A spherical snowball is rolled in fresh snow, causing it to grow so that its volume increases at a rate of 2???? cm^3/sec. How fast is the diameter of the snowball increasing when the radius is 2 cm?

User Kametrixom
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Let V be the volume of the snowball, r be the radius of the snowball, and d be the diameter of the snowball. We know that the volume of a sphere is given by V = (4/3)πr^3, and the radius and diameter are related by d = 2r.

We are given that dV/dt = 2 cm^3/sec, and we want to find dD/dt when r = 2 cm.

Using the chain rule, we can write:

dV/dt = dV/dr * dr/dt

To find dV/dr, we differentiate the volume formula with respect to r:

dV/dr = 4πr^2

Substituting the given values, we get:

2 = (4/3)π(2)^2 * dD/dt

Simplifying the equation, we get:

dD/dt = 3 / (4π) cm/sec

Therefore, when the radius is 2 cm, the diameter of the snowball is increasing at a rate of 3 / (4π) cm/sec.
User GeorgeP
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