# calculus

A water trough is 8 m long and has a cross-section in the shape of an isosceles trapezoid that is 40 cm wide at the bottom, 100 cm wide at the top, and has height 60 cm. If the trough is being filled with water at the rate of 0.1 m3/min how fast is the water level rising when the water is 30 cm deep?

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1. The real question is how wide is the tank when it is half full?
I draw a rectangle in the cross section of the tank, 40 wide and 60 high.
them at the top and bottom the sides stick out (100-40)/2 = 30cm on each side
halfway up they stick out 15 on each side.
so the width halfway up is 40 + 30 = 70 cm
Now
the area of the surface is 70cm * 800 cm = .7 m * 8 m = 5.6 m^2
now finally calculus
dV/dt = A * dh /dt
0.1 m^3/min = 5.6 m^2 * dh/dt
dh/dt = 0.1 / 5.6 meters/ minute = 0.0179 meters/minute = 1.79 cm/minute

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Damon

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