calculus
posted by Kelsey .
A trough is 12 feet long and 3 feet across at the top. It ends are isosceles triangles with a height of 3 feet. If water is being pumped into the trough at 2.5 cubic feet per minute, how fast is the water level rising when the water is 1 foot deep?

At a given time of t minutes,
let the width of the water level be x ft
let the height of the water level be h ft
but by ratios, x/h = 3/3
x = h
V = area of triangle x 12
= (1/2)xh(12
= 6xh
= 6h^2
given : dV/dt = 2.5 ft^3/min
find dh/dt when h = 1
dV/dt = 12h dh/dt
2.5 = 12(1) dh/dt
dh/dt = 2.5/12 = 5/24 ft/min or appr. .208 ft/min 
Dezz nuts

*deez