Hot tubs and hyperbolic curves

[DaveC426913]

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.

 

[jack action]

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

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

​

 

[DaveC426913]

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.

 

[jack action]

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".

 

[I like Serena]

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 $h$:
$$h = \sqrt[3]{\frac{3Q}{C\pi}}\sqrt[3]t$$

 

[DaveC426913]

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?

 

[DaveC426913]

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.

 

[I like Serena]

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.

 

[jbriggs444]

[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.

 

[DaveC426913]

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 :