Optimizing V=SEH: Solving Really Hard Related Rates Problem | dH/dt = -7/36

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Homework Statement


Homework Equations


The Attempt at a Solution


V=20 S=6 E=2 H=20/12 (from V=SEH?)

[itex]\frac{dS}{dt}=-.2[/itex] <--- + OR -?

[itex]\frac{dE}{dt}=.3[/itex] <--- + OR -?

V=SEH

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

[itex]0=SE\frac{dH}{dt}+EH(-.2)+SH(.3)[/itex]

[itex]-EH(-.2)-SH(.3)=SE\frac{dH}{dt}[/itex]

[itex]\frac{-EH(-.2)-SH(.3)}{SE}=\frac{dH}{dt}[/itex]

[itex]\frac{-(2)(\frac{20}{12})(-.2)-(6)(\frac{20}{12})(.3)}{SE}=\frac{dH}{dt}[/itex]

[itex]\frac{-7}{36}=\frac{dH}{dt}[/itex]

Negative, therefore decreasing.
 
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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?
 
It means that they're getting smaller, therefore negative right? so that's why dS/dt is negative and not dE/dt?
 
Dick said:
If the side walls are getting closer it's the end walls that are getting shorter, right?

OH I see, one sec let me retry this.
 
Ended up getting -1/12 = dH/dt is this correct? So therefore the depth of the liquid is decreasing.
 
iRaid said:
Ended up getting -1/12 = dH/dt is this correct? So therefore the depth of the liquid is decreasing.

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

What did you use for dS/dt and dE/dt?
 
That's what I thought I did wrong, thanks for all your help.

Much appreciated.