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B Hot tubs and hyperbolic curves

  1. Nov 25, 2017 #1

    DaveC426913

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    I am filling my circular hottub, and charting the water level height. Its sides have a small, constant slope from vertical - i.e. it is a truncated, inverted cone.

    Imagining an ideal hottub of unlimited height*, the water level will always be increasing - but at a decreasing rate - it will level off more and more over time.

    So it can't be an asymptote, since it will never reach a limit. Then again, it will also never stop curving.

    What is this curve?


    *OK, that's not an ideal hottub in my eyes. Easy to drown.
     
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  3. Nov 25, 2017 #2

    jack action

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    Isn't it ##y = x^2##?

    By mixing ideal hot tubs and physics, I thought this thread would've been about time travelling:

    tMWNhMC00ZTA1LTkwYTgtMmQ2ZjVhY2I5YjhjXkEyXkFqcGdeQXVyNTIzOTk5ODM@._V1_SY1000_CR0,0,1332,1000_AL_.jpg
     
  4. Nov 25, 2017 #3

    DaveC426913

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    Definitely not.
    The rate of height increase decreases. (i.e. it will tend toward horizontal over time)

    When the water level is a mere one foot, the surface area is quite small, so with constant inflow, the height increases rapidly.
    When the water level is a hundred feet, the hot tub has a surface area much larger, thus a constant flow of water takes a long time to raise the level even a little bit. So the height increases very slowly.

    The rate of height increase will drop off over time. But it never zeroes out.

    hottub.png
     
    Last edited: Nov 25, 2017
  5. Nov 25, 2017 #4

    jack action

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    I misunderstood the question, I'll try again with the square root instead:

    $$y = \max\left(0; \sqrt{x} - \frac{a}{2} \right)$$

    Where ##a## is the width of the base of your truncated "cone".
     
  6. Nov 25, 2017 #5

    I like Serena

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    Hey Dave,

    After an infinitesimal time dt, the level will rise by dh with surface area of ##A=C\pi h^2##, where ##C## is some constant.
    The change in volume is ##dV=A\,dh##.
    Therefore the volumetric flow rate ##Q## that is presumably constant, is:
    $$Q = \frac{dV}{dt} = \frac{Adh}{dt}=C\pi h^2 \frac{dh}{dt}$$
    Integrate to find:
    $$Qt=V=\frac {C\pi}{3}h^3$$
    Solving for ##h##:
    $$h = \sqrt[3]{\frac{3Q}{C\pi}}\sqrt[3]t$$
     
  7. Nov 25, 2017 #6

    DaveC426913

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    Wouldn't you sort of need to know the angle of the walls to have a meaningful calculation?
    Highly vertical walls would have a very different rate of leveling than nearly horizontal walls, so it's gotta be a factor.

    Or was that intended as a proportional formula?
     
  8. Nov 25, 2017 #7

    DaveC426913

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    So, what kind of curve is it?
    Logarithmic?

    I guess it's that cube root that dictates the curve's nature.
     
  9. Nov 25, 2017 #8

    I like Serena

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    It's a third root curve ##h \propto \sqrt[3] t##.
    It rises faster than logarithmic.
    In a semi-log plot, we'll get a straight line.
     
  10. Nov 25, 2017 #9

    jbriggs444

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    [Lots of traffic between when I started to compose this and when I hit post]
    Let us begin by transforming the problem slightly. Instead of a truncated cone, use a sharp-tipped cone. Now the volume of the fluid in the cone is proportional to the cube of the height and is also a linear multiple of elapsed time. If you insist on setting t=0 to the time when the water level hits some predetermined truncation point, that is a simple offset to the result.$$h(t)=k\sqrt[3]{t}$$ for some constant k that depends on cone angle and fill rate.$$h'(t)=kt^{-\frac{2}{3}}$$ for some [different] constant k [a factor of three smaller] that depends on cone angle and fill rate.
     
  11. Nov 25, 2017 #10

    DaveC426913

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    Oh.

    *click*

    A linear dimension will vary as the cube root of a volume.
    So, height is rising as the cube root of the volume of the filled portion of the tub.

    The angle of the slope simply adds a multiplier. (If angle is 0, multiplier is one.)

    That's so ... basic.

    : ancient gears start to grind :
     
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