Looking for a special spiral!

[Milosz]

My long day of music production culminated in thinking about spirals. Go figure.

I'm looking for a spiral that satisfies a seemingly simple condition.

Consider an arbitrary point u on the spiral. Let the Euclidean distance between the origin and the point u be d. Let the arc length of the spiral between the origin and u be l. Then l = 2^d.

The moral of the story is that if I cut across the arms of the spiral, I'll get to my destination exponentially faster than by following the rules of the road and riding along the arms.

Does this beast have a name? Has it already been under the scrutiny of countless mathematicians? Does it have a "nice" formula? Or perhaps have I uncovered something so ugly that no one wants to look at it?

The furthest I got in determining the formula for this thing was to write down this awful parametric differential equation with parameter u:

2^{\sqrt{{x(u)}^2+{y(u)}^2}} = \int_0^u\sqrt{{\left(\frac{dx(t)}{dt}\right)}^2 + {\left(\frac{dy(t)}{dt}\right)}^2} dt

Trouble is, I never did learn how to solve these things, and I have no idea where to start. Perhaps parametric is the wrong way to go about this? I can't really figure out how to write the above in polar notation either though. I just need a nudge in the right direction, but won't complain if I get a treatise

[tomtom2357]

Well, you can replace sqrt((x(u)^2)+(y(u)^2)) with r, and (dx(t)/dt)^2+(dy/dt)^2 with (dr/dt)^2, and after that it decomposes into 2^rln2*r^'(t)=r^'(t), which transforms into 2^rln2=1, so I don't think that your spiral exists.

[Proginoskes]

Well, you can replace sqrt((x(u)^2)+(y(u)^2)) with r, and (dx(t)/dt)^2+(dy/dt)^2 with (dr/dt)^2,

Not quite.

\left( dr \over dt \right)^2 = \left( x \over r \right)^2 \left( dx \over dt \right)^2 +2\left( x \over r \right) \left( y \over r \right) \left( dx \over dt \right)\left( dy \over dt \right)+ \left( y \over r \right)^2 \left( dy \over dt \right)^2

[Proginoskes]

In polar coordinates, arc length is given by the formula

\int_a^b \sqrt{r\,^2 + (r\,')^2}\,dt.

You want

2^r = \int_0^r \sqrt{r(t)^2 + (r\,'(t))^2}\,dt.

Taking derivatives of both sides and then squaring yields the non-linear differential equation

4^r (\ln 2)^2 = r\,^2 + (r\,')^2.

Maple has trouble with this one, so I doubt there's a closed form. (r being in the exponent and a base suggests that the Lambert-W function is involved in any "finite" form.)

[tomtom2357]

Well, you can replace sqrt((x(u)^2)+(y(u)^2)) with r, and (dx(t)/dt)^2+(dy/dt)^2 with (dr/dt)^2,

Not quite.

\left( dr \over dt \right)^2 = \left( x \over r \right)^2 \left( dx \over dt \right)^2 +2\left( x \over r \right) \left( y \over r \right) \left( dx \over dt \right)\left( dy \over dt \right)+ \left( y \over r \right)^2 \left( dy \over dt \right)^2

Oops, sorry.

In polar coordinates, arc length is given by the formula

\int_a^b \sqrt{r\,^2 + (r\,')^2}\,dt.

You want

2^r = \int_0^r \sqrt{r(t)^2 + (r\,'(t))^2}\,dt.

Taking derivatives of both sides and then squaring yields the non-linear differential equation

4^r (\ln 2)^2 = r\,^2 + (r\,')^2.

Maple has trouble with this one, so I doubt there's a closed form. (r being in the exponent and a base suggests that the Lambert-W function is involved in any "finite" form.)

Wolfram Alpha doesn't have a solution either, it is probably an unsolvable equation.

[gmalivuk]

Careful how you word that. "Unsolvable" suggests there is no solution, while what you mean to say is simply that there's no nice closed form solution.

[quantropy]

4^r (\ln 2)^2 = r\,^2 + (r\,')^2.

Something doesn't seem right here - or maybe I'm confused
This can be rewritten as

r\,'=\sqrt{4^r (\ln 2)^2 - r\,^2}

Which I think is positive for r>0, and gets large pretty quickly. Hence the curve is soon zooming away from the origin. But this doesn't seem to match the curve described by the OP, which would need to be ever more tightly wound as r increases.

[++$_]

In polar coordinates, arc length is given by the formula

\int_a^b \sqrt{r\,^2 + (r\,')^2}\,dt.

You want

2^r = \int_0^r \sqrt{r(t)^2 + (r\,'(t))^2}\,dt.

Taking derivatives of both sides and then squaring yields the non-linear differential equation

4^r (\ln 2)^2 = r\,^2 + (r\,')^2.

Maple has trouble with this one, so I doubt there's a closed form. (r being in the exponent and a base suggests that the Lambert-W function is involved in any "finite" form.)

This makes no sense. (As quantropy points out, you get something that grows at an increasing speed, which isn't right.)

The upper bound on your integral is wrong because what you want to do is go from 0 to x, not from 0 to r(x).

Anyway, making this change, we can then differentiate both sides with respect to x, getting

4^{r(x)}(\ln 2)^2r'(x)^2 = r(x)^2 + r'(x)^2

This is separable. Nice.

[Dark Avorian]

I don't think this definition actually makes sense. If d=0 then l=1, so when it's at the origin it's arc length from the origin is 1, so I don't think we can actually do this.

[t1mm01994]

When the arclength is 0, it's at distance 1 from the origin, so it starts on the circle with radius 1 around the origin.

[++$_]

When the arclength is 0, it's at distance 1 from the origin, so it starts on the circle with radius 1 around the origin.

No, that's backwards -- but the definition still makes sense on the complement of any neighborhood of 0, right?

[gmalivuk]

Yeah, I believe so. Set the initial conditions at some point other than the origin, and it should be doable, and just spiral around a whole hell of a lot as you get toward the origin.

[moiraemachy]

I think it won't. The arclength can't be expected to grow slower than d, which is what happens close to 0.

[Sagekilla]

Your spiral looks pretty interesting. I solved it in polar form where the angular portion (dw/dt) was constant.

Here's some nice plots for r(0) = 0.6:

dw/dt = 1

Spoiler:

dw/dt = 5

Spoiler:

dw/dt = 9

Spoiler:

Obviously, you can pick dw/dt to be anything you want. But it looks nice for constant angular velocity.

Very close to the origin the arclength growth drops exponentially to a minimum at around t = 0.2, then grows linearly
from t = 0.2 to about t = 3 and logarithmically thereafter.

Change in arclength w.r.t. time:

Spoiler:

The smooth purple curve is the actual ds/dt, where s is arclength. The blue curve is my piecewise fit using
exponential, linear, and logarithmic least-squares fit curves to the given regions.


I'll post pictures of that in a few minutes.

[tomtom2357]

Very cool spiral!

[Proginoskes]

In polar coordinates, arc length is given by the formula

\int_a^b \sqrt{r\,^2 + (r\,')^2}\,dt.

You want

2^r = \int_0^r \sqrt{r(t)^2 + (r\,'(t))^2}\,dt.

Taking derivatives of both sides and then squaring yields the non-linear differential equation

4^r (\ln 2)^2 = r\,^2 + (r\,')^2.

Maple has trouble with this one, so I doubt there's a closed form. (r being in the exponent and a base suggests that the Lambert-W function is involved in any "finite" form.)

This makes no sense. (As quantropy points out, you get something that grows at an increasing speed, which isn't right.)

The upper bound on your integral is wrong because what you want to do is go from 0 to x, not from 0 to r(x).

Yeah, I realized this about half an hour ago, when I was going over the problem again in my mind.

Anyway, making this change, we can then differentiate both sides with respect to x, getting

4^{r(x)}(\ln 2)^2r\,'(x)^2 = r(x)^2 + r\,'(x)^2

This is separable. Nice.

Yes, but not integrable, and you'd also have to invert the function if you could integrate it. I got the equation

\theta = \int {~\sqrt{4^r (\ln 2)^2 - 1}~~ \over r}\,dr

It suggests a way to construct spirals with certain nice properties, though.

[quantropy]

Note r must be greater than (-ln(ln 2))/(ln 2), that is about 0.53, so that can be thought of as where the curve starts. At that point r' is infinite, so the curve starts off by going directly away from the origin.

[Sagekilla]

Well, if you're restricted to the real line then yes. r = 0 is a unstable equilibrium.
If you allow complex values a small real starting point will move away from the origin
and oscillate in a fairly unpredictable way about the imaginary axis.

[gmalivuk]

Obviously, you can pick dw/dt to be anything you want. But it looks nice for constant angular velocity.

How are all those pictures valid solutions? Surely when r = 6 the arclength of the spiral up to that point has to be 32, while in your three pictures it would seem to have 3 distinct values depending on dw/dt.

I think only your one for dw/dt=1 is actually a solution to the differential equation, with whatever you come up with for dw/dt = k is actually a solution for l = 2^{k d}

[Sagekilla]

Obviously, you can pick dw/dt to be anything you want. But it looks nice for constant angular velocity.

How are all those pictures valid solutions? Surely when r = 6 the arclength of the spiral up to that point has to be 32, while in your three pictures it would seem to have 3 distinct values depending on dw/dt.

I think only your one for dw/dt=1 is actually a solution to the differential equation, with whatever you come up with for dw/dt = k is actually a solution for l = 2^{k d}

Oops. Yeah you're right. I actually did it the right way the first time around where constant angular velocity kicks
in the constant k into the radial portion as a prefactor.

I'll update those plots later on to reflect that.

[moiraemachy]

The derivative of the arclength with respect to the distance must be at least 1, since considering two points in the spiral, the length of the arc that connects them is at least as big as a the difference between their distances to the center. In the proposed spiral, this derivative is ln(2)*(2^d), so when you get to the point where d is smaller than (-ln(ln(2))/ln(2)) you can't go further. The thing is unsolveable then unless you force the apropriate initial value for the arclen at this point. (which is, essentially, changing the spiral to arclen = 2^d - k)

The spiral arclen = exp(d) - 1, however, will start at the origin with no problems. I ended in something quite similar to what Proginoskes got. (angular velocity ω = sqrt( exp(d)² - 1)/d.

What I'm really curious is about the spirals arclen = k^d, where k>e. I'm not sure if, by starting at an point outside the origin, the arclength will tend to 0 as it approaches the origin.

[MartianInvader]

I don't know what you guys are talking about. The OP is clearly looking for the constant curve r=0.

[Milosz]

I don't know what you guys are talking about. The OP is clearly looking for the constant curve r=0.

Haha, but then we have arc length 2^d = 0 which doesn't make sense. However, that does force me to reconsider the equation for my spiral. Under the current equation, we have a bit of a situation in the neighborhood of the origin: as the distance d to the origin becomes vanishingly small (which I suppose we've renamed to the radius r at this point), the arc length 2^d approaches 1. This implies that near the origin, the spiral is wrapping around an infinite number of times to allow for the relatively gargantuan arc length. At least that's my intuition.

A few of you guys seem to be coming up with the lower bound \frac{-\ln(\ln(2))}{\ln(2)} for r. I'm curious how this is coming about? Forgive me, but I'm not very good at differential equations or differential geometry (only had brushes with them in my academic career so far).

Sagekilla, would you mind sharing the code you used to graph the first spiral? I think it would really help me understand what's going on. Was your derivation similar to that which was come up with in the rest of the thread, culminating in

\begin{equation*}
4^{r(x)} {(\ln2)}^2 {r^\prime(x)}^2 = {r(x)}^2 + {r^\prime(x)^2} \text{ ?}
\end{equation*}

[gmalivuk]

A few of you guys seem to be coming up with the lower bound \frac{-\ln(\ln(2))}{\ln(2)} for r. I'm curious how this is coming about?

Below that point, the OP's equation would have arclength growing more slowly than distance from the origin. In other words, going from r to r+dr would have the arclength increase by less than dr, which would be like trying to connect two points a meter apart with an unstretchable string that's less than a meter long.

Was your derivation similar to that which was come up with in the rest of the thread, culminating in

\begin{equation*}
4^{r(x)} {(\ln2)}^2 {r^\prime(x)}^2 = {r(x)}^2 + {r^\prime(x)^2} \text{ ?}
\end{equation*}

Yeah, that's what it looks like to me. (I independently came up with that same equation and graphed its numerical solution as a PolarPlot in Mathematica, and I'm pretty sure it's identical to the first spiral Sagekilla posted.) Note that Sagekilla started with r = 0.6, which is slightly above the lower bound of \frac{-\ln(\ln(2))}{\ln(2)}. If you tell most numerical differentiators to start right at that point, they'll choke on the infinite derivative there, so best to start a bit above that.

[Sagekilla]

My complete code can be more or less summed up as:

Code: Select all

x[t_] := r[t] Cos[t]
y[t_] := r[t] Sin[t]
eq = Power[2, Sqrt[x[t]^2 + y[t]^2]] ==
   Integrate[Sqrt[x'[s]^2 + y'[s]^2], {s, 0, t}];
de = r'[t] == (r'[t] /.
     Last@Solve[Simplify[D[eq, t] , {r[t] > 0, t > 0}], r'[t]]);

tmin = 0;
tmax = 200;
r0 = 0.6;
sol = r[t] /. First@NDSolve[{de, r[tmin] == r0},
     r[t], {t, tmin, tmax}];


The first part is merely for verification purposes. I solved for the differential form of the spiral
by hand and I got the same thing that Mathematica told me.

sol just contains the solution plotted from t = 0 to 200. The code really isn't useful unless you have Mathematica.

The starting value of r(0) = 0.6 I picked from, as mentioned above, a starting point slightly higher than the singularity:

Code: Select all

Solve[4^r Log[2]^2 - 1 == 0, r] (* r -> -2 Log[Log[2]] / Log[4] = 0.528766 *)