Related Rates: water pumped in at 0.2 m/min; find rate level

[joeluver53]

Water is being pumped into a trough at a rate of 0.2 meters per minute. THe trough is 10 meters long and has cross-section in the shape of an isosceles trapezoid that is 30 centimeters wide at the bottom, 80 centimeters wide at the top, and 50 centimeters high. Find the rate at which the water level is rising when the depth of water is 30 centimeters

 

[BigGlenntheHeavy]

Re: Related Rates

Given: dV/dt = .2m/min.
The volume of the trough is the cross-sectional area of the trough times the length of the trough, so V = (10A).

The area of the cross-sectional of the trough is A =h/2(a+b). When the height of the water is .3m, a = .3m and b = .3m+2b so
V =10/2(.3m+.3m+2a)h. By similiar triangles a/h =.25/.5= 1/2,so 2a=h.
Ergo. V = 5(.6m+h)h = 3h+5h^2. now dY/dt =(dV/dh)(dh/dt) -> .2m=(3+10h)dh/dt ->.2m = (3+10h)dh/dt ->dh/dt = .2m/(3+10h). When h =.3m, dh/dt =.2/(6)m/min = 10/3cm/min.
Therefore when the trough is being filled at 20cm/min, when the trough has a depth of 30cm, the height of water in the trough is increasing at a rate of 10/3cm/min.

Note: I revised my answer as I mistool a variable for a constant. This answer rings true.