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

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Homework Help Overview

The discussion revolves around a related rates problem involving the volume of a shape defined by the equation V=SEH, where S, E, and H represent dimensions that are changing over time. Participants are analyzing how the rates of change of these dimensions affect the overall volume and the rate of change of height (dH/dt).

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of the product rule for three variables and question the signs of the rates of change (dS/dt and dE/dt) based on the physical scenario described. There is uncertainty about whether certain rates should be positive or negative, leading to further exploration of the relationships between the dimensions.

Discussion Status

There is an ongoing examination of the assumptions regarding the rates of change of the dimensions. Some participants have proposed different values for dS/dt and dE/dt, leading to conflicting results for dH/dt. The discussion reflects a collaborative effort to clarify these assumptions and their implications on the calculations.

Contextual Notes

Participants are grappling with the interpretation of the problem's physical setup, particularly how the movement of the side walls affects the dimensions and their rates of change. There is a lack of consensus on the correct signs for the rates of change, which is central to resolving the problem.

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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?
 
iRaid said:
Does it mean that dS/dt=-.4? Not sure?

If the side walls are getting closer it's the end walls that are getting shorter, right?
 
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?
 
  • #10
dS/dt=-.3
dE/dt=+.2
 
  • #11
iRaid said:
dS/dt=-.3
dE/dt=+.2

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

Much appreciated.
 

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