SOLUTION: A triangular trough is 10ft. Long, 6 ft. across the top, and 3 ft deep. If water flows in at the rate of 12 cu. ft. per min, find how fast the surface is rising when the water is

Algebra ->  Finance -> SOLUTION: A triangular trough is 10ft. Long, 6 ft. across the top, and 3 ft deep. If water flows in at the rate of 12 cu. ft. per min, find how fast the surface is rising when the water is      Log On


   

Question 1194690: A triangular trough is 10ft. Long, 6 ft. across the top, and 3 ft deep. If
water flows in at the rate of 12 cu. ft. per min, find how fast the surface is
rising when the water is 6 in. deep.
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The depth of the trough is 3 feet, and it is 6 feet wide at the top. The sides of the trough are straight, so when the depth of the water in the tank is h, the width of the surface of the water is 2h.

So the cross sectional area of the trough when the depth of the water is h is then (one-half base times height) = %281%2F2%29%282h%29%28h%29+=+h%5E2.

The trough is 10 feet long, so the volume of water in the tank when the depth is h feet is 10h^2.

V=10h%5E2
dV%2Fdt=20h%28dh%2Fdt%29
12=20h%28dh%2Fdt%29
dh%2Fdt=12%2F%2820h%29=3%2F%285h%29

Note that the form of that expression for the rate of change in the depth makes sense; as the depth of water h gets greater, the rate at which the depth of the water rises decreases.

The problem asks for how fast the surface of the water is changing when the depth of the water is 6 inches = 0.5 feet:

3%2F%285%280.5%29%29=3%2F2.5=6%2F5

ANSWER: the surface of the water is rising at a rate of 1.2 feet per minute when the depth of the water is 6 inches.