Posted by **Kelsey** on Saturday, October 15, 2011 at 8:35pm.

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?

- calculus -
**Reiny**, Saturday, October 15, 2011 at 9:53pm
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

- calculus -
**rami**, Thursday, December 3, 2015 at 3:20pm
Dezz nuts

- calculus -
**not rami**, Monday, November 14, 2016 at 10:15am
*deez

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