B Hot tubs and hyperbolic curves

AI Thread Summary
The discussion centers on the mathematical modeling of water height in a circular hot tub with a truncated cone shape. As water fills the tub, the height increases at a decreasing rate due to the larger surface area at higher levels, leading to a third root relationship where height is proportional to the cube root of time. The participants explore various mathematical equations, concluding that the rate of height increase is not linear but follows a curve dictated by the cone's geometry. The conversation also touches on the importance of the cone's angle in determining the rate of height change. Ultimately, the height of the water rises as the cube root of the volume, influenced by the fill rate and cone angle.
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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Isn't it ##y = x^2##?

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

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jack action said:
Isn't it ##y = x^2##?
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
 

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Last edited:
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".
 
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##:
$$h = \sqrt[3]{\frac{3Q}{C\pi}}\sqrt[3]t$$
 
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jack action said:
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".
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 got to be a factor.

Or was that intended as a proportional formula?
 
I like Serena said:
Hey Dave,

After an infinitesimal time dt, the level will rise by dh with surface area of
So, what kind of curve is it?
Logarithmic?

I guess it's that cube root that dictates the curve's nature.
 
DaveC426913 said:
So, what kind of curve is it?
Logarithmic?
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.
 
[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.
 
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jbriggs444 said:
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.
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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