The trough shown in the figure is 5 feet long, and its vertical cross sections are inverted isosceles triangles with base 2 feet and height 3 feet. Water is being siphoned out of the trough at the rate of 2 cubic feet per minute. At any time t, let h be the depth and V be the volume of water in the trough. Read the given situation off slide and point out measurements on diagram of trough.

a) Find the volume of water in the trough when it is full. b) What is the rate of change in h at the instant when the trough is ¼ full by volume? Tell what we have to solve. c) What is the rate of change in the area of the surface of the water (shaded in the figure) at the instant when the trough is ¼ full by volume?

a) Find the volume of water in the trough when it is full. Explain that the volume of the water when the trough is full is simply the volume of the entire container. Explain how we solved for the volume. We couldn’t remember the formula for the volume of a prism,but we realized that it was simply the area of the triangle multiplied by the number of triangles there (the depth, or length, of the prism). Explain which letters represented which numbers in our formula.

b) What is the rate of change in h at the instant when the trough is ¼ full by volume? First, we realized that we would need a derivative and that this would be a related rate equation. Like we did in class to make the derivative easier to find, we substituted one variable for another (in this case, b for 2/3h). Next, we rewrote the volume formula with our substitution. Note that length, l, is constant, so we just replaced it with 5 (the length), once again to make the derivative easier to obtain.

b) What is the rate of change in h at the instant when the trough is ¼ full by volume? From there, all we had to do was use implicit differentiation to differentiate the volume equation with respect to t. Then we replaced dV/dt with -2, which was given in the set up of the problem. Then we went back to the original Volume equation, plugged in 15/4, for ¼ the total volume, and solved for h. Then we plugged h into the derivative and solved for dh/dt

[ -2 , 0 ] [ 0 , 3 ] Solving for dh/dt graphically... To solve graphically, we simply put the equation for the derivative in the y= screen and then traced along the resulting graph to about 1.5 to find the derivative at that height.

Solving for dh/dt numerically... To solve numerically, all we had to do was look on the table provided by the calculator.

c) What is the rate of change in the area of the surface of the water (shaded in the figure) at the instant when the trough is ¼ full by volume? First we had to write an equation for the surface area, which was simply the rectangle shaded more lightly in the diagram. After we had the equation, we again substituted b for 2/3 h and then simplified the equation. Next, we differentiated the equation with respect to t. Then we substituted -3/5 (found in the previous step of the problem) for dh/dt and solved the equation.

Any Questions?

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