# Thread: related rates, water running into trough

1. ## 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?

2. Originally Posted by oblixps
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?
the 3-4-5 triangles on both end sides of the trapezoid.

trapezoidal cross-section of water in the trough ...

upper base = $5+2x$

lower base = $5$

height = $h$

relationship between x and h ... $\frac{x}{h} = \frac{3}{4}$

$x = \frac{3h}{4}$

area of the trapezoidal cross-section of water ...

$A = \frac{h}{2}\left[(5+2x) + 5\right]$

$A = \frac{h}{2}\left[\left(5+ \frac{3h}{2}\right) + 5\right]$

can you take it from here?