Math CentralQuandaries & Queries


my name is Julia, and I'm having a lot of trouble with this question:

A circular blob of molasses of uniform thickness has a volume of 1 m3.
The thickness of the molasses is decreasing at a rate of 0.1 cm/hour.
At what rate is the radius of the molasses increasing when the radius is 8 m?

I know that:
volume = pi r2 times thickness = 1
d(thickness) / d(time) = 0.001 m/h
Area = pi r2
d(area) / d(time) = 2 pi r (d(r)/d(time))
and that we're looking for d(r) / d(time) when r =8

unfortunately I don't know how to put all of this information together.
Any help would be greatly appreciated!


Hi Julia,

Actually d(thickness) / d(time) is negative, d(thickness) / d(time) = -0.001 m/h.

The piece of information you don't have listed is that the volume is not changing (V(t) = 1 m3) so
dV/dt = 0 where t is time in hours. Let T(t) be the thickness of the blob in metres at time t. The radius r(t) and the thickness T(t) are functions of time so differentiating V with respect to time gives

dV/dt = 2 pi r T dr/dt + pi r2 dT/dt = 0

Can you see how to complete the problem now?

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