Calculating the Rate of Water Level Increase in a Trapezoid-Shaped Trough

[cscott]

A water trough is 6 m long and has a cross-section in the shape of an isosceles trapezoid (dimension shown in the diagram.) Water is being pumped into the trough at a rate of 0.5 m^3/min. How fast is the water level rising when it is 0.5 m deep.

AFAIK the volume of a trapazoid like this should be

V = \frac{1}{2}h(a + b)l

where a, b, and l are all given constants, therefore

\frac{dV}{dt} = \frac{1}{2}\frac{dh}{dt}(a + b)l

But this can't be right. I don't think I'm approaching this correctly...

 

[vsage]

What do you personally think you're doing wrong? you should be very careful to note that V = 1/2h(a+b)l makes no distinction as to what is actually a function of t, although you try to differentiate with respect to t the height later. Yes that's the volume of a filled container, but differentiating h with respect to t would yield 0 because it is a constant. Perhaps construct the volume of a trapezoid whose height is dependent on time t.

 

[cscott]

Would I have to express all the other variables in terms of h?

 

[vsage]

No, no. Just think about the shape of a volume that was being filled up by water and what dimensions are variable with time. Certainly the height filled with water will be variable with time, but what about the width of the surface? Obvioously the base of the trough will not change though.

 

[cscott]

I'm assuming I can relate the width to height using (1/4)w/h = 0.6/0.9 (i.e. similar triangles)?

 

[vsage]

Test it out :). I suggest using extreme cases such as where h = 0 what is w? (better equal the width of the base)

 

[cscott]

Well mine isn't going to work because at h = 0, w = 0.

Trying w = 1.2/0.9h + 0.6 ...

 

[cscott]

Woot, I got it. Thanks for your help/patience!