The Derivatives as a Rate of Change

Question

A trough is 20ft long and has a cross-section in the shape of an equilateral triangle 2ft on each side. If the water is being pumped in at the rate of 10 cubic feet per minute, how fast is the water level rising when the water is  

sqrt of(3) feet deep?

Answer 1

The 2-ft measurement is irrelevant to the question.

An equilateral triangle with height = h has area = h²/√3 (since the base = 2h/√3).

This makes V(h) = 20h²/√3. Differentiating with respect to t gives ...

dV/dt = 40h/√3·dh/dt. The rate in is dV/dt, so ...

10 = 40√3/√3·dh/dt

dh/dt = 1/4 ft/min