   SEARCH HOME Math Central Quandaries & Queries  Question from David, a student: A rectangular trough is 2 meter long, 0.5 meter across the top and 1 meter deep. At what rate must water be poured into the trough such that the depth of the water is increasing at 1m/min. when the depth of the water is 0.7m. I know this involves implicit differentiation somehow, but the 3 variables, since V=l*w*h for a rectangle is confusing me. I'm not sure whether one of the variables should be fixed or not, since I'm not getting anywhere with this right now. Any help would be great. We have two responses for you

Hi David,

Suppose that the height of the water at some time t minutes after you started your clock is h(t) minutes. The water in the trough at that time thus has volume

V(t) = length × width × height = 2 × 0.5 × h(t) = h(t) cubic meters

Thus dV/dt = dh/dt cubic meters per minute.

I hope this helps,
Penny

Hi David.

If it is rectangular (and oriented conventionally, with a flat side on the bottom) then the depth of the water doesn't matter: the water level will rise at a constant rate for a constant flow rate.

So if you start from empty and fill the trough, the water level will rise in 1 minute. What's the volume of the trough? That divided by 1 minute is the flow rate you need.

This question doesn't actually require any differentiation at all!

Cheers,
Stephen La Rocque.     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.