A water trough is 5 m long and has a cross-section in the shape of an isosceles trapezoid that is 30 cm wide at the bottom, 70 cm wide at the top, and has height 40 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 20 cm deep?

Answer 1

Q = incoming flow rate = .1 m^3/min

dh/ dt = Q A
where
A = surface area
= length * width at 20 cm depth which depth is (1/2) height

width = 30 + 1/2(70-30) = 30+20 = 50 cm = .5 m wide water surface
so
A = 5 * .50 =2.5 m^3
so finally
dh/dt = .1 * 2.5 = .25 m/min = 25 cm/min

Answer 2

when the water has depth x, the cross-section is a trapezoid with bases
30 and 30+x. So the volume of water at depth x is

v = (60+x)/2 * x * 500 cm^3
= 250x^2 + 15000x

so, knowing that

dv/dt = (500x + 15000) dx/dt

just solve that for dx/dt when x=20

Answer 3

LOL - Guess which of us is the mathematician and which is the Engineer :)