# Horizontal Eavestrough Problem

1. Apr 5, 2009

### ghostanime2001

1. The problem statement, all variables and given/known data
A horizontal eavestrough 3m long has a triangular cross section 10 cm across the top and 10 cm deep. During a rainstorm the water in the trough is rising at the rate of 1 cm/min when the depth is 5 cm.

a) How fast is the volume of water in the trough increasing?
b) After the rain stopped, the water drained out of the trough at the rate of 0.06 m3/min. How fast is the surface of the water falling when the depth is 1 cm?

3. The attempt at a solution
V =$$\frac{1}{2}$$ bhl

Solving in terms of b:
b 5
-- = --
h 10

Last edited: Apr 5, 2009
2. Apr 5, 2009

### Hootenanny

Staff Emeritus
I'm not entirely sure what you're doing here. You're given dl and dh at a certain time and you're asked to determine dV.

3. Apr 5, 2009

### ghostanime2001

I know this formula that relates Volume with base, height and lenght of the water trough. The formula is V=1/2 bhl

From similiar triangles:
5/10 = b/h
5h=10b
b=5/10 h
b=1/2h

Then substituting 1/2h for b into the formula we get:
V=1/2(1/2 h)hl
V=1/2(1/2 h2 l)
V=1/4 h2 l
Now differentiating h we get:
dV/dt = 2(1/4)h dh/dt l
dV/dt = 1/2h dh/dt l
Now submitting in all the values we get:
dV/dt = 1/2(5 cm)(1 cm/min)(300 cm)
dV/dt = 750 cm3/min

However the answer is 1500 cm3/min. Now my question is WHAT HAPPENED!??!?! i did everything correct from everything ive learned but i still dont get the correct answer.... :( please help me understand thank you.

as for b) ill wait until i get the first part correct.

4. Apr 5, 2009

### Dick

When the trough is full b=10, and h=10. When the trough is half full b=5 and h=5, right? The correct relation is b=h, not b=(1/2)h. I would rethink the logic that lead you to that.