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.(adsbygoogle = window.adsbygoogle || []).push({});

AFAIK the volume of a trapazoid like this should be

[tex]V = \frac{1}{2}h(a + b)l[/tex]

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

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

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

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# Homework Help: Rate problem

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