1. ## Calculus Applications

The cross-section of a water trough is an equilateral triangle with a horizontal top edge. If the trough is 5m long and 25 cm deep, and water is flowing in at a rate of 0.25m3/min, how fast is the water level rising when the water is 10 cm deep at the deepest point?

2. Originally Posted by mathamatics112
the cross-section of a water trough is an equilateral triangle with a horizontal top edge. If the trough is 5m long and 25 cm deep, and water is flowing in at a rate of 0.25m3/min, how fast is the water level rising when the water is 10 cm deep at the deepest point?

First we need to be careful with the units convert everything into meters

using the pythagorean theorem we can find the area of the front of the tank interms of only h.

$h^2 +\left( \frac{l}{2}\right)^2=l^2$

solving for l we get

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

Since we have two triangles to make the front of the trough we get

$A=2 \left( \frac{1}{2}\right)\left( \frac{l}{2}\right)h=\frac{h^2}{\sqrt{3}}$

So the volume is

$V=\frac{5m}{\sqrt{3}}h^2$

now taking the derivative with respect to time we get

$\frac{dV}{dt}=\frac{10 m}{\sqrt{3}}h\frac{dh}{dt}$

Now plugging in all of our data we get

$
\frac{0.25m^3}{min}=\frac{10m}{\sqrt{3}}(.1m)\frac {dh}{dt}
$

$
\frac{0.25m^3}{min}=\frac{m^2}{\sqrt{3}}\frac{dh}{ dt}
$

$\left( \frac{0.25m^3}{min}\right) \left( \frac{\sqrt{3}}{m^2}\right)=\frac{dh}{dt}$

$\frac{dh}{dt}=\frac{\sqrt{3}m}{4min}$

3. Thank you so much