How to find the maximum arc length of this equation?

In summary, the conversation is about attempting to graph the relationship between ##x^x + y^y = r^r##, particularly when the values of x, y, and r are infinite. The first attempt was to use the Lambert W function to define the equation and graph it, but this was unsuccessful due to technical difficulties. The next approach was to find the arc length of the equation and use the derivative to find the maximum arc length, but this was also hindered by difficulty in converting the equation to polar coordinates. The conversation ends with a request for alternative methods or help in solving the equation.
  • #1
Saracen Rue
150
10
TL;DR Summary
How do you find the maximum arc length of ##{^{\infty}x}+{^{\infty}y}={^{\infty}r}## and the value of ##r## at which it occurs?
After seeing a discussion about graphs of the relationship ##x^x + y^y = r^r##, it got me interested in attempting to see what the graphical appearance of ##{^{\infty}x}+{^{\infty}y}={^{\infty}r}## would look like. The first step I did was use the relationship of ##{^{\infty}n}=-\frac{W(-\ln(n))}{\ln(n)}## to give me the equation of:
$$-\frac{W(-\ln(x))}{\ln(x)} -\frac{W(-\ln(y))}{\ln(y)} = -\frac{W(-\ln(r))}{\ln(r)} $$

Where the range of possible real values for ##r## is ##e^{−e} < r < e^{1/e}##.
Using this I attempted to define the Lambert W function inside of Desmos using
$$\text{}$$
$$W\left(x\right)=-\frac{2}{\pi}\int_{0}^{\pi}\frac{\sin\left(\frac{t}{2}\right)\left(\sin\left(\frac{3t}{2}\right)+e^{\cos\left(t\right)}x\sin\left(\frac{5t}{2}-\sin\left(t\right)\right)\right)}{1+e^{2\cos\left(t\right)}x^{2}+2e^{\cos\left(t\right)}x\cos\left(t-\sin\left(t\right)\right)}dt\left\{-\frac{1}{e}<x<e\right\}$$
$$\text{}$$

and from there try to graph ##-\frac{W(-\ln(x))}{\ln(x)} -\frac{W(-\ln(y))}{\ln(y)} = -\frac{W(-\ln(r))}{\ln(r)} ##, in the hopes of being able to get a rough idea of what value of ##r## gives the greatest arc length. However, Desmos was extremely laggy and any change in the value of ##r## would take over a minute to be reflected by the graph.

This would take far too long to reasonably do so instead I decided to try to find a way to define a new function as being the arc length of the equation ##-\frac{W(-\ln(x))}{\ln(x)} -\frac{W(-\ln(y))}{\ln(y)} = -\frac{W(-\ln(r))}{\ln(r)} ##, get the derivative of said new function and solve for when the derivative equals zero to get the value of ##r## that produces the maximum arc length.

However, at this point I encountered another problem. The only way I know how to calculate the length of a curve is by using the formula ##\int_a^b \sqrt{1+(\frac{dy}{dx})^2} dx## for Cartesian coordinates and ##\int_{\theta_1}^{\theta_2} \sqrt{r^2+(\frac{dr}{d\theta})^2} d\theta## for polar coordinates, but the prior isn't useful without rearranging for ##y## to be the subject and the latter isn't useful as I'm unsure of how to rearrange into the form ##r=f(\theta)## after converting this particular equation to polar coordinates.

So, to summarize, I would very much appreciate if someone could help by telling me if there's a way to make ##y## the subject in ##-\frac{W(-\ln(x))}{\ln(x)} -\frac{W(-\ln(y))}{\ln(y)} = -\frac{W(-\ln(r))}{\ln(r)} ## or if there's an alternate method I could use to evaluate the arc length of this curve.
 
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  • #2
Whatever your notation means, the only alternative I see is to rectify the curve and sum up the polygon lengths.
 
  • #3
Saracen Rue said:
Summary:: How do you find the maximum arc length of ##{^{\infty}x}+{^{\infty}y}={^{\infty}r}## and the value of ##r## at which it occurs?
What is the notation ##^{\infty}x## supposed to mean? Keep in mind that ##\infty## can't be used in arithmetic or algebraic expressions.
 
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Likes jim mcnamara
  • #4
Mark44 said:
What is the notation ##^{\infty}x## supposed to mean? Keep in mind that ##\infty## can't be used in arithmetic or algebraic expressions.
The notation being used is Tetration; another way of expressing Kunths double up arrow notation: ##{ ^{\infty}n}=n \uparrow \uparrow \infty =## x^x^x^{.}^{.}^{.} (An infinitely tall power tower).

You can read more about tetration here: https://en.m.wikipedia.org/wiki/Tetration

And you can read more about power towers here: http://mathworld.wolfram.com/PowerTower.html
 

Related to How to find the maximum arc length of this equation?

1. What is the formula for finding the maximum arc length of an equation?

The formula for finding the maximum arc length of an equation is given by the integral of the square root of 1 plus the square of the derivative of the equation, with respect to the variable of integration. This can be expressed as:
∫ √(1 + (dy/dx)^2) dx

2. How do you determine the limits of integration for finding the maximum arc length?

The limits of integration for finding the maximum arc length are determined by the range of the independent variable in the given equation. For example, if the equation is defined from x = 0 to x = 5, then the limits of integration would be from 0 to 5.

3. Can the maximum arc length of an equation be negative?

No, the maximum arc length of an equation cannot be negative. The arc length represents the distance along the curve and therefore cannot be negative.

4. Is it possible to find the maximum arc length of an equation analytically?

Yes, it is possible to find the maximum arc length of an equation analytically by using the formula mentioned in the first question. However, in some cases, it may be difficult or impossible to find an exact analytical solution and numerical methods may need to be used instead.

5. Can the maximum arc length of an equation be greater than the length of the curve itself?

Yes, it is possible for the maximum arc length of an equation to be greater than the length of the curve itself. This can happen if the curve has multiple loops or if it is a closed curve, in which case the maximum arc length would be the circumference of the curve.

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