# A horizontal trough 12 feet long has a vertical cross section in the form of a trapezoid. The bottom is 3 feet wide, and the sides are inclined.....to the vertical at an angle with sine 4/5. Given...

A horizontal trough 12 feet long has a vertical cross section in the form of a trapezoid. The bottom is 3 feet wide, and the sides are inclined...

..to the vertical at an angle with sine 4/5. Given that water is poured into the trough at the rate of 10 cubic feet per minute, how fast is the water level rising when the water is exactly 2 feet deep?

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### 1 Answer

The cross section of the trough is trapezoid. The bottom width is 3feet. So the top width is 3 +or- 2xsin(4/5) feet at a height x.

Case (i) the top width is 3+2xsin(4/5).

The area of the cross section = height* { top width and bottom width}/2 = x{3 + 3+2xsin(4/5)}/2 = x(3+xsin(4/5).

Therefore the volume v(x) of water when height is x = height {cross setional area)+length = x(3+xsin(4/5)}*12 = 12x(3+xsin(4)/5)} = 36x+x^2sin(4/5)...(1).

Therefore when the speed of water is 10 feet /m, the rate of increase in hieght per minute is required.

dv/dt = 10 feet.

But dv/dt =( dv/dx)dx/dt.

Therefore dx/dt = (dv/dt)/ (dv/dx) = 10/((dv/dx) .....(2)

We get dv/dx:

v(x) = 36x+36x^2sin(4/5) = 36x+62xsin(4/5).

v'(x) = 36+36*2xsin(4/5) .

When water height x = 2, v'(2) v'(x) = 36+36*2*2 = 36+ 144sin(4/5)sin(4/5) = 38.0106

Therefore we substitute v'(2) = dv/dx at x= 2 in eq(2):

dx/dt = 10/38.0106. = 0.2631 feet/ minute is the rate of increase in height / minute when water level is 2 feet.

Case (ii) the top width is 3-2sin(4/5):

v(x) = 36x -72x^2sin(4/5)

dx/dt = 10/v'(x) = 10/(36-144sin(4/5)) at x= 2.

dx/dt = 10/33.9894 = 0.2942 feet / minute.