# If a snowball melts so that its surface area decreases at a rate of 3 cm2/min, how do you find the rate at which the diameter decreases when the diameter is 12 cm?

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Steve M
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Dec 10, 2016

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diameter is decreasing at rate of

#### Explanation:

Assuming the snowball is a perfect sphere, then if

# A = 4pir^2 = 4pi(D/2)^2 = piD^2 #

Differentiating wrt

# (dA)/(dD) = 2piD #

We are told that

By the chain rule we have:

# (dA)/(dD) = (dA)/(dt)*(dt)/(dD) = ((dA)/(dt)) / ((dD)/dt)#

# :. 2piD = -3/ ((dD)/dt) #

# :. (dD)/dt = -3/ (2piD) #

When

The sign confirms that

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