related rates, water running into trough
Water is running into a trough at rate 2 ft^3 / min. The cross section of the trough is an isosceles trapezoid with two bases 5 ft and 11 ft and a height of 4 ft. The length of the trough is 20 ft. How fast is the water level rising one hour later?
so i have the equation V = (1/2) h (b1 + b2) 20 = 10 (h) (b1 + b2), i drew the trapezoid so that the smaller base is on top and since water is filling the trough, the bottom base won't be changing so my volume equation is now V = 10 (h) (b1 + 11). since i am given dV/dt and i need to find dh/dt, i need to somehow get rid of b1 and express it in terms of h. that's where i'm stuck at. how do i express b1 in terms of h. do i have to use similar triangles? if so, which triangles do i use?
