Question 667241
A trough of length 6 m has a uniform cross section which is an equilateral triangle with sides 1 m. Water leaks from the bottom of the trough at a constant rate of 0.1 m^3 per minute. Find the rate at which the water level is falling at the instant when it is 20 cm deep?
<pre>
This is a cross section of the trough.  The blue line is the water
level.  The red line h is the height of the water level.  We are
looking for the rate at which h is shrinking when h=20cm.   


{{{drawing(600,345,-2,2,-2,.3, triangle(0,-sqrt(3),-1,0,1,0),
blue(line(-.65,-.6062,.65,-.6062)), red(line(0,-.6062,0,-sqrt(3))),
locate(.05,-1,h), locate(.25,-.47,x), locate(-.34,-.47,x),
locate(-.59,-.61,"60°")
 )}}}

The volume of the water is the area of the equilateral triangle
whose base is the blue line times the trough length of 6m, The
area of the triangle is {{{1/2}}}(2x)(h) or xh and multiplying
this by the trough length of 6 m, we have

                      V = 6xh

We also know that {{{h/x}}} = tan(60°) = <font face="symbol">Ö</font><span style="text-decoration: overline">3</span>.

So {{{h/x=sqrt(3)/1}}}, therefore x = {{{h/sqrt(3)}}} = {{{h*sqrt(3)/3}}}, so V = 6xh becomes 

                     V = 6{{{h*sqrt(3)/3}}}h
or
                     V = 2·<font face="symbol">Ö</font><span style="text-decoration: overline">3</span>·h²

Differentiating with respect to time t

                    {{{(dV)/(dt)}}} = 4·<font face="symbol">Ö</font><span style="text-decoration: overline">3</span>·h·{{{(dh)/(dt)}}} 

We are given that {{{(dV)/(dt)}}} = -0.1 m³ taken negative because the
volume of water is decreasing.  And we want the particular value of
{{{(dh)/(dt)}}} when h = 20 cm = 0.2 m.

                    -0.1 = 4·<font face="symbol">Ö</font><span style="text-decoration: overline">3</span>·0.2·{{{(dh)/(dt)}}}

Solve that for {{{(dh)/(dt)}}} and we get

                   {{{-0.1/(4sqrt(3)*.2)}}} = {{{(dh)/(dt)}}}

Multiply top and bottom by 10

                   {{{-1/(4sqrt(3)*2)}}} = {{{(dh)/(dt)}}}                   

                   {{{-1/(8sqrt(3))}}} = {{{(dh)/(dt)}}}

Rationalize the denominator:

                   {{{-1sqrt(3)/(8*3)}}} = {{{(dh)/(dt)}}}

                   {{{-sqrt(3)/24}}} = {{{(dh)/(dt)}}}

That's in meters/minute, so to change it to centimeters/minute,
multiply by 100

                   {{{-100sqrt(3)/24}}} = {{{(dh)/(dt)}}}

                   {{{-25sqrt(3)/6}}} = {{{(dh)/(dt)}}}

which is about -7.2 centimeters/minute, which means that the water 
level is falling at {{{25sqrt(3)/6}}} centimeters/minute or 
about 7.2 centimeters/minute.

Edwin</pre>