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D.E. Word Problem (Extruded Volume, constant decrease in volume)

  1. Sep 15, 2009 #1
    This was the problem before the one I am asking about (backstory maybe?) - Maybe relevant to my current problem

    [tex]
    \frac{dh}{dt}=-\frac{A_{h}}{A_{w}}\sqrt{2gh}
    [/tex]

    Problem statement:

    It seems that, by placing marks on the tank corresponding to the water level at intervals of, say, one hour one should be able to create a reasonably accurate clock. The idea is distinctly not new. Such clocks, known later as clepsydra (from the Greek kleptein, to steal, and hydro, water) are believed to have appeared in China around 4000 BCE. The known versions all have the rather inelegant property that, as [itex]\frac{dh}{dt}[/itex]is not constant, the distances between successive hourly marks decreases with time. Your task here is to use the formula (??), with [itex]A_{w}=A_{w}(h)[/itex]
    UAB_MA252_September_2009_HW1_02b.jpg
    now assumed to depend on h, to design a 12-hour clepsydra with the dimensions shown above and shaped like the surface obtained by roataing the curve [itex]x=g(y)[/itex] around the y-axis. You are to determine a formula for the function g describing the shape of your clepsydra, together with the radius of the circular bottom hole, in order that the water level [itex]h=h(t)[/itex] will fall at the constant rate of 4 inches per hour (be careful with the units here).

    Here is my work:

    [tex]
    \frac{dh}{dt}=\frac{4 in}{1 hr} = \frac{\frac{4}{12} ft}{60 s}
    [/tex]

    [tex]
    \frac{dh}{dt}=-0.00\bar{55}
    [/tex]

    [tex]
    h(t)=-0.00\bar{55}t+C_{0}
    [/tex]

    find: [itex]h(t)[/itex] for [itex]t=0, h=4[/itex]

    [tex]
    4=-0.00\bar{55}(0)+C_{0}
    [/tex]

    [tex]
    C_{0}=4
    [/tex]

    [tex]
    h(t) \rightarrow h(t)=0.00\bar{55}t+4
    [/tex]

    find: [itex]t[/itex] for [itex]h(t)=0[/itex]

    [tex]
    0=0.00\bar{55}t+4
    [/tex]

    [tex]t=\frac{4}{0.00\bar{55}}\rightarrow720\left(s\right)[/tex]

    or

    [tex]12\left(h\right)[/tex]

    Now, here is where I am stuck:

    I have tried and tried to find an equation for the line that should be [itex]x=g(y)[/itex]

    I know that the eqn of a circle is involved.

    It seems that the radius of the "circles" (stacked to form the volume of the paraboloid) should vary with time as well. ([itex]dx[/itex]?)

    I even tried to deduce the eqn from the formula for a paraboloid ([itex]\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=z[/itex]), but obviously, other than being able to plot the eqn in Maple, this did me little good.

    Someone throw me a bone, please?
     
    Last edited: Sep 15, 2009
  2. jcsd
  3. Sep 15, 2009 #2

    tiny-tim

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    Hi EtherealMonke! :smile:

    What is the equation for the rate of flow when the height is h, and the diameter of the hole is x? :wink:
     
  4. Sep 15, 2009 #3
    Honestly, I am that lost.

    I thought that the rate of flow for this particular problem is constant.

    I did try to calculate the area of the drain hole using the initial area and [itex]\frac{dh}{ht}[/itex] for [itex]t=0[/itex].

    But, then I focused on finding the equation of the "hull" to be extruded about y.

    I am nicht so gut at the word problems...
     
  5. Sep 15, 2009 #4

    tiny-tim

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    The equation will involve pressure and cross-section areas. :smile:
     
  6. Sep 16, 2009 #5
    Okay, I have an idea about this...

    Using: [itex]\frac{dh}{dt}=-\frac{A_{h}}{A_{w}}\sqrt{2gh}[/itex]

    1. Solve [itex]A_{w}[/itex]

    2. Use [itex]\frac{dh}{dt}[/itex] found in my first post, [itex]A_{w}[/itex] from #1 and [itex]h(t)[/itex] to find [itex]\frac{d}{dt}A_{h}[/itex]
     
  7. Sep 16, 2009 #6
    Well, that seemed like a good idea, but...

    I have:

    [tex]
    \frac{d}{dt}A_{h}=\frac{1.32212e^{-8}}{\sqrt{h(t)}}
    [/tex]

    Which doesn't look too good.
     
  8. Sep 16, 2009 #7

    tiny-tim

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    eugh! :yuck: never use itex for fractions!!

    use tex, and it looks like this :wink:
    ok, you're losing sight of the original problem …

    you don't want to integrate d/dt of anything, you only want dh/dt = -.0055 …

    so use that, and your equation dh/dt = -(Ah/Aw)√(2gh) :smile:
     
  9. Sep 16, 2009 #8
    Okay, I will try that.

    BTW, thank you Tim.

    I am not trying to be dense. I really am having difficulty with this.

    This assignment was due last Wednesday, and I promise that I have been trying to work it out ever since then (well, before then too, but not soon enough apparently - should have gone straight home and started then).

    Whole assignment posted here:
    https://www.physicsforums.com/showpost.php?p=2341102&postcount=9"

    ---

    I still have a question though...

    Why should it be obvious to me (as I hope it will be soon), that I do not need to integrate anything else?

    I mean, [tex]\frac{dh}{dt}[/tex] is giving me the change in "height" for y (actually z as I see it), so what is the reason that the differential radius can be "ignored."

    Is it because of the approximation of "area" under the curve [tex]x = g(y)[/tex] is the value being given for [tex]\frac{dh}{dt}[/tex]? (actually, I just had this thought while writing a different question that I disregarded when this popped to my mind, so is that the reason, and if not; what is?)

    Thanks again
     
    Last edited by a moderator: Apr 24, 2017
  10. Sep 16, 2009 #9

    tiny-tim

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    So -(Ah/Aw)√(2gh) = -.0055,

    so Aw = … ? :smile:
     
  11. Sep 16, 2009 #10
    [tex]A_{w}=0.059ft[/tex]

    or

    [tex]A_{w}=0.7in[/tex]
     
  12. Sep 16, 2009 #11

    tiny-tim

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    Aw(h) is a function of h …

    what happened to h? :confused:
     
  13. Sep 16, 2009 #12
    I am sorry.

    That was the value for [tex]A_{h}[/tex]

    (calculated using [itex]h=4[/itex] and [itex]A_{w}[/itex] when [itex]r = 1[/itex])

    ---

    [tex]A_{w}=\left(-0.00\bar{55}t+4\right)*\pi r^{2}[/tex]

    (r=x)

    [edit]
    Sorry about that (again...), I mangled my reply on the first post and then just now fixed it (I think)

    LaTex is interesting...
     
    Last edited: Sep 16, 2009
  14. Sep 16, 2009 #13
    Okay, so I guess that was wrong. Please do not be afraid by my mistakes. I have too much heart to quit...

    Anyway,

    Maybe this is better?

    [tex]r_{w}=\sqrt{\frac{A_{h} * \sqrt{2gh}}{\pi*0.00\bar{55}}}[/tex]
     
    Last edited: Sep 16, 2009
  15. Sep 16, 2009 #14
    Actually, I just checked the equation above for h=4 to see if r would = 1 and it does...

    Someone please verify that I have finally figured it out?

    Thank you
     
  16. Sep 16, 2009 #15
    Actually, I was looking for the radius (because I kept thinking that I needed it to "extrude" the "hull" about y).

    This is the equation that I found which satisfies all values I tested and exhibits a shape similar to that depicted in the problem statement:

    [tex]1.571041571*\sqrt{h}[/tex]

    Thank you Tim for your patience.

    I wish you might have been more explicit with me, but never-the-less, I still appreciate your effort, because the last post you made was the one that put me on the path to a solution.

    Thank you again.
     
    Last edited: Sep 16, 2009
  17. Sep 16, 2009 #16

    tiny-tim

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    Hi EtherealMonke! :smile:

    (I've been out all afternoon)
    That's right! :smile:
    ah … PF rules (unlike some other websites) say we mustn't help you too much …

    it's much better if we help you get the answer yourself

    (and it also means that professors don't mind their students using PF :wink:)
     
  18. Sep 16, 2009 #17
    I understand. I may ask inane questions, but I honestly have no expectation of being spoon fed the solution.

    I don't believe my degree would be worth much if I satisfied the requirements by proxy :wink:

    And, considering the choice of my major (Mechanical Engineering), I don't think it would serve any of us very well if I was not able to demonstrate the skills required to perform the job.

    Thanks again (I really can't thank you enough - that problem was about to cause emotional trauma... :smile: )
     
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