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I have a question similar to this:

The surface area of a snowball rolling down a hill is increasing at a rate of $12\pi\ cm^2/sec$. What is the radius of the snowball at the moment when the radius is increasing at $0.5\ cm/sec$?

But I have no idea how to solve it nor where to start. Thanks.

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1 Answer 1

up vote 5 down vote accepted

Differentiate $A=4\pi r^2$ to get $\frac{dA}{dt}=8\pi r\frac{dr}{dt}$. Substitute the values of $\frac{dA}{dt}$ and $\frac{dr}{dt}$ to solve for $r$.

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Thank you so much! –  Derek 朕會功夫 Nov 25 '12 at 21:09
    
@Will, someone has been collecting rep like a busy beaver in the last few months. Kudos! –  Matt N. Nov 25 '12 at 21:12
    
@Will you've been busy! –  amWhy Nov 26 '12 at 0:02

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