Really hard related reates problem

1. Feb 2, 2012

iRaid

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
V=20 S=6 E=2 H=20/12 (from V=SEH?)

$\frac{dS}{dt}=-.2$ <--- + OR -?

$\frac{dE}{dt}=.3$ <--- + OR -?

V=SEH

Product rule for 3 variables:
$\frac{dV}{dt}=SE\frac{dH}{dt}+EH\frac{dS}{dt}+SH \frac{dE}{dt}$

$0=SE\frac{dH}{dt}+EH(-.2)+SH(.3)$

$-EH(-.2)-SH(.3)=SE\frac{dH}{dt}$

$\frac{-EH(-.2)-SH(.3)}{SE}=\frac{dH}{dt}$

$\frac{-(2)(\frac{20}{12})(-.2)-(6)(\frac{20}{12})(.3)}{SE}=\frac{dH}{dt}$

$\frac{-7}{36}=\frac{dH}{dt}$

Negative, therefore decreasing.

Last edited: Feb 2, 2012
2. Feb 2, 2012

Dick

The method of solution is fine. But picture the problem a little more carefully. It tells you the side walls are moving together at .2 m/s. This doesn't tell you dS/dt=(-.2). What does it tell you?

3. Feb 2, 2012

iRaid

It means that they're getting smaller, therefore negative right? so that's why dS/dt is negative and not dE/dt?

4. Feb 2, 2012

Dick

If the side walls are getting closer it's the end walls that are getting shorter, right?

5. Feb 2, 2012

iRaid

OH I see, one sec let me retry this.

6. Feb 2, 2012

iRaid

Ended up getting -1/12 = dH/dt is this correct? So therefore the depth of the liquid is decreasing.

7. Feb 2, 2012

Dick

Well, I got +1/12=dH/dt. One of us made a mistake.

8. Feb 2, 2012

iRaid

Um I think I made the mistake because I said dS/dt is negative, should be positive since the sides are getting bigger and the ends are getting smaller... But I'm not positive..

9. Feb 2, 2012

Dick

What did you use for dS/dt and dE/dt?

10. Feb 2, 2012

iRaid

dS/dt=-.3
dE/dt=+.2

11. Feb 2, 2012

Dick

Now that's not right, is it? If the side walls are getting closer the end walls are getting shorter. So dE/dt should be negative.

12. Feb 2, 2012

iRaid

That's what I thought I did wrong, thanks for all your help.

Much appreciated.