# Homework Help: Rate of change of water in a trough

1. Mar 15, 2009

### Emethyst

1. The problem statement, all variables and given/known data
The cross section of a trough for holding water is an inverted equilateral triangle. The trough is 6m long and 50cm deep. If the water is flowing in a t a rate of 0.6m^3/min, find the rate at which the height is changing when the water is 40cm deep.

2. Relevant equations
implicit differentation

3. The attempt at a solution
I started with the equation V=1/2(bh)l and the given information of dV/dt=0.6. From the equation and all information given, I can see that I am not given b as either a static number or as a rate, and thus need to somehow get rid of it by finding another relationship. My problem is I can't seem to find that relationship, as the only thing that seems to make sense to me is to use similar triangles to get a relationship, but I used this method I ended up with 3 variables instead of 2. Could anyone be of assistance and point me in the right direction for this question, as I have no idea where to find the correct equation for differentation to solve for the answer. Thanks in advance.

2. Mar 15, 2009

### Dick

An equilateral triangle has a very definite relation between base and height. Can you find it? Use trigonometry! Or just the Pythagorean theorem!

3. Mar 15, 2009

### Emethyst

Well I can't remember my earlier lessons on trig :tongue: so I used the Pythagorean Theorem and came up with B=(2/sqrt3)h, where B is the base and h is the height. This brings up another question as there seems to be 2 heights here, one that is the actual trough height and the other that is the height of the water at a specific time. Will there not be 2 seperate heights invovled in the equation, or will one simply be ignored in the solving process?

4. Mar 15, 2009

### Staff: Mentor

The overall width of the tank is not 2/sqrt(3) cm. By my calculations it is roughly 60 cm wide. Yes, there are two heights in this problem, but you need to work with only the height of the water at a particular time (when h = 60 cm.).

5. Mar 15, 2009

### Emethyst

60cm? How would I come up with such an answer? What I did was go B^2=((1/2)B)^2+h^2 and solved the expression for B. For me this made sense because the answer is 8/sqrt3. EDIT: After trying to solve it though I can now see it doesn't work, so that must be where I am going wrong, along with being unsure about the heights.

6. Mar 15, 2009

### Dick

Emethyst said that the width of the cross section is 2*h/sqrt(3). I concur with that. And you want the width at h=40cm. What's your problem with that? And yes, you don't care about the 50cm number, you care about the 40cm number. Where did you come up with h=60cm?

7. Mar 15, 2009

### Emethyst

Ok I think I see how to solve the question now. The only problem I seem to be having is getting the correct answer. What I did was:

V=(1/2)Bhw
V=(1/2)((2/sqrt3)h)hw
V=((1/sqrt3)h^2)(6)
V=(6/sqrt3)h^2
dV/dt=(6/sqrt3)(2h)(dh/dt)
0.6=(6/sqrt3)(2(0.4))(dh/dt)
0.6=(4.8/sqrt3)(dh/dt)
0.6/(4.8/sqrt3)=dh/dt
sqrt3/0.125=dh/dt

This is not the answer that the sheet says (the correct answer is sqrt3/8). I know that 4.8/0.6=8 though, so I know I must have gone wrong somewhere in my calculations. If any of you see the mistake please point it out.

8. Mar 16, 2009

### Staff: Mentor

My mistake. In doing mental calculations, I was thinking I was getting half the width, but on reflection it was the full width, which is roughly 30 cm. And I missed her h.

9. Mar 16, 2009

### Dick

0.6/(4.8/sqrt(3))=sqrt(3)*(0.6/4.8)=sqrt(3)*(1/8). You somehow flipped the 0.6/4.8 part upside down.

10. Mar 16, 2009

### Emethyst

Ahh I see my mistake now, thanks for the help Dick, what I did was invert 0.6/(4.8/sqrt3) to get 0.6*(sqrt3/4.8) in order to get rid of the fraction, I just forgot to seperate the sqrt3 then multiply it by the 1/8 by the looks of it. Thanks for all of that help, it has been greatly appreciated.