rate of flowing water

**hotdogking** · March 29th 2010, 03:45 PM

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 !

**skeeter** · March 29th 2010, 04:37 PM

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?

**hotdogking** · March 30th 2010, 07:57 AM

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

**skeeter** · March 30th 2010, 02:05 PM

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 $s$ = side length of water triangle

$h$ = height of water triangle

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

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

Area of an equilateral triangle ...

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

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

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

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

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