- #1
Saracen Rue
- 150
- 10
- TL;DR
- 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.
$$-\frac{W(-\ln(x))}{\ln(x)} -\frac{W(-\ln(y))}{\ln(y)} = -\frac{W(-\ln(r))}{\ln(r)} $$
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.