Expert: Scotto - 4/11/2008

Question
QUESTION: The cross-section of a water trough is an equilateral triangle with a horizontal top edge. If the trough is 5m long and 25cm deep, and water is flowing in a rate of 0.25m^3/min, how fast is the water level rising when the water is 10cm deep at the deepest point?

ANSWER: The length is 500 cm and the width, from an equilateral triangle, is
10/root(3), so the surface area is 5,000/root(3).

To get the instantaneous rate of change, find d(.25m^3/min)/dm.
Convert m to cm by taking 1m^2 = 10,000cm^2.

Divide this by the surface area to get the rate of increase in depth at that point.

Have a great day and feel free to ask more questions.


---------- FOLLOW-UP ----------

QUESTION: I'm sorry, but I am still not understanding what exactly needs to happen. Correct me if I am wrong: length of trough = 500cm, width of trough = 10/root(3), surface area of trough = 5000/root(3). How'd you get those? Also, are you saying find the derivative of 0.25m^3/min? Thanks.

Answer
A meter is 100 cm for the length.

The trough is an equilateral triangle, so when it is cut in half the sides of that triangle are 1 unit and root(3) units with a hypoteneuse of 2 units.

The longer side is the depth, which is 25.  This would mean that the hypoteneuse (which is the side) has length 10/root(3) when the water is 10 cm deep.

The surface area is length * width, so divide the amount of water flowing in by this value to get how fast the depth is increasing per minute.  

No, I'm not sure why I put in the derivative.

Hope this helps you.