
Quick problem
There is a trough that is 7 feet long, and its vertical cross sections are inverted isosceles triangles with a height 5 feet and base 2 feet. There is water in it and and the water is being siphoned out at the rate of 3 cubic feet per minute. At any time t, let d be the depth and V be the volume of the water in the trough.
1. Find the volume of the trough when its full.
2. What is the rate of change in the area of the surface of the water at the instant when the trough is .25 full by volume?

The area of the triangle formed by a cross section of the trough is
$\displaystyle \frac{1}{2}bh$
By similar triangles $\displaystyle \frac{b}{h}=\frac{2}{5}$
$\displaystyle b=\frac{2h}{5}$
The volume at some height h is then
$\displaystyle V=7\cdot\frac{1}{2}(\frac{2h}{5})h=\frac{7h^{2}}{5 }$
The full volume of the trough is $\displaystyle V=\frac{1}{2}(2)(5)(7)=35$
Therefore, when it is 1/4 full by volume:
$\displaystyle \frac{35}{4}=\frac{7h^{2}}{5}$ and $\displaystyle h=\frac{5}{2}$ and $\displaystyle b=1$
Now, can you take it from there by differentiating and finding dh/dt.
The surface area of the water at some height h would be given by the length of the trough times b (the width of the water surface at some height).
$\displaystyle S=7b=\frac{14h}{5}$