1. ## Related Rates problem

The ends of a water trough 8 feet long are equilateral triangle whose sides are 2 feet long. If water is being pumped into the trough at a rate of 5 ft^3/min, find the rate at which the water level is rising when the depth of the water is 8 inches.

2. Originally Posted by nystudent2729
The ends of a water trough 8 feet long are equilateral triangle whose sides are 2 feet long. If water is being pumped into the trough at a rate of 5 ft^3/min, find the rate at which the water level is rising when the depth of the water is 8 inches.

Recall that the area of an equilateral triangle is $A=\frac{\sqrt{3}}{4}s^2$, and its height is $h=\frac{\sqrt{3}}{2}s$, where s is the length of one of the sides. Thus, area can be written in terms of h as $A=\frac{\sqrt{3}}{3}h^2$

So the volume of this trough is $V=\frac{8\sqrt{3}}{3}h^2$

We know that $\frac{\,dV}{\,dt}=5~\frac{ft^3}{min}$. Thus, find $\frac{\,dh}{\,dt}$, when h is 8 inches [I would first convert this to feet]...

Can you take it from here?

--Chris

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# the end of a water trough 8 feets long are equilateral triangular

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