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...

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.