# Thread: rate of flowing water

1. ## rate of flowing water

The ends of a trough are equilateral triangles with one vertex pointing down. The trough is 9 feet long. When the water in the trough is 2 feet deep, its depth is increasing at the rate of 1/2 foot per minute. At what rate is water flowing into the trough at that moment?

really could use some help getting this problem solved !

2. Originally Posted by hotdogking
The ends of a trough are equilateral triangles with one vertex pointing down. The trough is 9 feet long. When the water in the trough is 2 feet deep, its depth is increasing at the rate of 1/2 foot per minute. At what rate is water flowing into the trough at that moment?

really could use some help getting this problem solved !
are the side lengths of the equilateral triangle given?

3. No they are not, that is why it is I am kinda stumped on how to do this one.

4. Originally Posted by hotdogking
No they are not, that is why it is I am kinda stumped on how to do this one.
now that I think about it, doesn't matter anyway ... all equilateral triangles are similar.

let $\displaystyle s$ = side length of water triangle

$\displaystyle h$ = height of water triangle

for equilateral triangles, $\displaystyle \cos(30) = \frac{h}{s} = \frac{\sqrt{3}}{2}$

$\displaystyle s = \frac{2h}{\sqrt{3}}$

Area of an equilateral triangle ...

$\displaystyle A = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} \cdot \frac{4h^2}{3} = \frac{\sqrt{3}}{3} h^2$

$\displaystyle V = 9A = 3\sqrt{3} \cdot h^2$

$\displaystyle \frac{dV}{dt} = 6\sqrt{3} h \cdot \frac{dh}{dt}$

substitute and calculate the value of $\displaystyle \frac{dV}{dt}$