If a snowball melts so that its surface area decreases at a rate of 1cm^2/min, find the rate at which the diameter decreases when the diameter is 10cm.

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A snowball melts so that the surface area decreases at a rate equal to 1 cm^2/min. The surface area of a ball with diameter D is `S = 4*pi*(d/2)^2 = pi*D^2`

`(dS)/(dt) = 1`

`(d (pi*D^2))/dt = 1`

=> `pi*2*D*(dD)/dt = 1`

=> `(dD)/dt = 1/(2*pi*D)`

**When D = 10 cm, the rate of change of diameter is `1/(2*pi*10) ~~ 0.0159` cm/minute.**

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