Arc length of the curve $x^1=2a\cos t$, $x^2=sint$.

Question

Using the metric Euclidean tensor for polar coordinates, calculate the arc length for the curve:

$x^1=2a\cos t$

$x^2=\sin t$ where $0\leq t\leq2\pi$

If I use the formula:

$L=\int_{a} ^{b} \sqrt{\left(\cfrac{dx^1} {dt} \right)^2+\left(\cfrac{dx^2} {dt}\right) ^2}=\sqrt{\left| \delta_{ij} \cfrac{dx^1} {dt}\cfrac{dx^2} {dt}\right|} $

I get an elliptic integral, can anyone help me to find the arc lenght? and How could you interpret the result geometrically?

Thanks in advanced.

Answer 1

As you probably noticed, the result is $$L=2 \left(E\left(-4 a^2\right)+\sqrt{4 a^2+1} E\left(\frac{4 a^2}{4 a^2+1}\right)\right)$$ What you can do is to approximate $L$ using $[2n,2n]$ Padé approximants built around $a=0$.

For example $$P_2=2\pi\, \frac {1+\frac{159 }{44}a^2+\frac{453 }{176}a^4 } { 1+\frac{115 }{44}a^2+\frac{125 }{176}a^4}$$ $$P_3=2\pi\, \frac {1+\frac{39575 }{7116}a^2+\frac{20621}{2372}a^4+\frac{129235 }{37952}a^6} {1+\frac{32459 }{7116}a^2+\frac{34741 }{7116}a^4+\frac{79037 }{113856}a^6 } $$ $$P_4=2\pi\, \frac {1+\frac{133542997 }{17725732}a^2+\frac{1325913585 }{70902928}a^4+\frac{1210596065 }{70902928}a^6+\frac{4808786003 }{1134446848}a^8 } {1+\frac{115817265 }{17725732}a^2+\frac{915821721 }{70902928}a^4+\frac{553597479 }{70902928}a^6+\frac{777708891 }{1134446848}a^8 }$$

The table shows how good (or bad) are these approximations $$\left( \begin{array}{ccccc} a & P_2 & P_3 & P_4 & \text{exact} \\ 0.00 & 6.283185307 & 6.283185307 & 6.283185307 & 6.283185307 \\ 0.25 & 6.659164725 & 6.659167214 & 6.659167222 & 6.659167222 \\ 0.50 & 7.639438651 & 7.640368153 & 7.640394783 & 7.640395578 \\ 0.75 & 8.969481052 & 8.985328348 & 8.986601081 & 8.986713343 \\ 1.00 & 10.44445389 & 10.52687833 & 10.53873159 & 10.54073433 \\ 1.25 & 11.91631116 & 12.15523595 & 12.20520464 & 12.21838971 \\ 1.50 & 13.29114874 & 13.78900721 & 13.92449946 & 13.97441783 \\ 1.75 & 14.52204955 & 15.36792068 & 15.64828398 & 15.78301361 \\ 2.00 & 15.59531652 & 16.85165250 & 17.33739745 & 17.62855113 \end{array} \right)$$

Edit

To better see the gain, consider the norm $$\Phi_n=\int^2_0 \big[L-P_n\big]^2\,da$$ which leads to $$\Phi_2=0.983 \qquad \Phi_3=0.114 \qquad \Phi_4=0.013$$ that is to say almost a factor of $10$ when $n \to n+1$.