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anonymous

  • one year ago

Question) Show that Torricelli's Law may be solved by separation of variables. State Assumptions

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  1. anonymous
    • one year ago
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    can you post Toricelli's law

  2. anonymous
    • one year ago
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    \[A(h)\frac{ dh }{ dt } = -k \sqrt{h}\]

  3. anonymous
    • one year ago
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    not sure what A(h) stands for

  4. anonymous
    • one year ago
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    Toricelli’s Law. Suppose that a water tank has a hole with area \( a\) at its bottom and that water is draining from the hole. Let y(t) (in feet) and V(t ) (in cubic feet) denote the depth and the volume of water in the tank at time t (in seconds). Then (under ideal conditions) the velocity of the stream of water exiting the tank will be \( v = \sqrt{ 2g y }\)

  5. anonymous
    • one year ago
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    yeah, I get that part. And I know how velocity in this case is also derived. I'm just curious on the conditions required to make this 'equation' work.

  6. IrishBoy123
    • one year ago
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    \[-\frac{A(h)}{\sqrt{h}} \, dh = -k \, dt \] i posted something on this yesterday, will try find a link, it goes into more detail

  7. anonymous
    • one year ago
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    here is a pdf on the separation of variable, and an intro to it

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  8. IrishBoy123
    • one year ago
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    down the bottom of this http://openstudy.com/users/irishboy123#/updates/55fffe4ae4b0ed58e276cd99

  9. anonymous
    • one year ago
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    Aright, Much appreciated

  10. anonymous
    • one year ago
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    @IrishBoy123 can you explain this part in the pdf i posted http://prntscr.com/8j6f7s It follows that V =∫ A(y)*dy <---- this is my reasoning then dV/dt = d/dt ( ∫ A(y)*dy ) = A(t) ?

  11. anonymous
    • one year ago
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    oh i think it follows from chain rule

  12. anonymous
    • one year ago
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    V =∫ A(y)*dy on [ 0, k] where k is the height of the vessel. dV/dt = d/dt ( ∫ A(y)*dy ) = d/dy ( ∫ A(y)*dy ) * dy/dt = A(y) * dy/dt I am not sure exactly what rule this is. AL

  13. IrishBoy123
    • one year ago
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    yeah, chain rule that pdf is a good find :-)

  14. anonymous
    • one year ago
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    i hate it when they skip steps,. this does not seem obvious to me, lol

  15. anonymous
    • one year ago
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    I noticed your post had a more interesting argument using volume and washers . but this is pretty cool too . i think i made a proof

  16. anonymous
    • one year ago
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    Want to make sure the limits make sense here. Let \( A(y) \) be the cross sectional area at height \( y \) of vessel. $$ { \large V(t) = \int_{0}^{y(t)}A(y)~ dy \\ \frac{dV}{dt}= \frac{d}{dt} \left( \int_{0}^{y(t)}A(y) dy \right)= \frac{d}{dy} \left( \int_{0}^{y(t)}A(y) dy \right)\cdot \frac{dy}{dt}= A(y) \cdot \frac{dy}{dt} } $$

  17. anonymous
    • one year ago
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    wikipedia has a nice argument https://en.wikipedia.org/wiki/Torricelli's_law#Application_for_time_to_empty_the_container

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