Length of curve $X_2(t) = (t\cos t, t\sin t)$ where $(0 \le t \le 2\pi)$

Question

Here I got $X_2(t) = (t\cos t, t\sin t)$ where $(0 \le t \le 2\pi)$

Let $X_2(t) = (x, y)$ then $x^2 + y^2 = t^2 \;\;\forall t\in[0, 2\pi]$

now will divide $[0,2\pi]$ into $n-identical$ segments.

then we get $[0, \frac{2\pi}{n},2\frac{2\pi}{n},..k\frac{2\pi}{n}, (k+1)\frac{2\pi}{n},...n\frac{2\pi}{n} ]$

then we take $kth$ and $(k+1)th$ term, and measure the distance between two:

$d(X_2(k\frac{2\pi}{n}),X_2((k+1)\frac{2\pi}{n})) = \sqrt{[k\frac{2\pi}{n}\cos k\frac{2\pi}{n}-(k+1)\frac{2\pi}{n}\cos((k+1)\frac{2\pi}{n})]^2+[k\frac{2\pi}{n}\sin k\frac{2\pi}{n}-(k+1)\frac{2\pi}{n}\sin((k+1)\frac{2\pi}{n})]^2}=\sqrt{k^2(\frac{2\pi}{n})^2+(k+1)^2(\frac{2\pi}{n})^2-2k(k+1)(\frac{2\pi}{n})^2[\cos k\frac{2\pi}{n}\cos (k+1)\frac{2\pi}{n}+\sin k\frac{2\pi}{n}\sin (k+1)\frac{2\pi}{n}]} $

then if we let $n\rightarrow \infty$, we can estimate that $[\cos k\frac{2\pi}{n}\cos (k+1)\frac{2\pi}{n}+\sin k\frac{2\pi}{n}\sin (k+1)\frac{2\pi}{n}] \rightarrow 1$ thus $d(X_2(k\frac{2\pi}{n}),X_2((k+1)\frac{2\pi}{n})) \rightarrow \frac{2\pi}{n}$ thus, total length would be $\frac{2\pi}{n}\cdot n = 2\pi$

I had developed my own conjecture as denoted above, and wants to be verified and get some correction for illogical reasoning.

Any harsh ones are welcomed.

Answer 1

The arc length is equal to

$$\int_0^{2\pi}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt$$

We have $\displaystyle \frac{dx}{dt}=\cos t-t\sin t$ and $\displaystyle \frac{dy}{dt}=\sin t+t\cos t$

$$\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2=(\cos t-t\sin t)^2+(\sin t+t\cos t)^2=1+t^2$$

Let $t=\tan\theta$. The arc length is

\begin{align} \int_0^{2\pi}\sqrt{1+t^2}dt&=\int_{\theta=0}^{\tan\theta=2\pi}\sec\theta d\tan\theta\\ &=\Big[\sec\theta\tan\theta\Big]_{\theta=0}^{\tan\theta=2\pi}-\int_{\theta=0}^{\tan\theta=2\pi}\sec\theta\tan^2\theta d\theta\\ &=2\pi\sqrt{4\pi^2+1 }-\int_{\theta=0}^{\tan\theta=2\pi}(\sec^3\theta-\sec\theta) d\theta\\ &=2\pi\sqrt{4\pi^2+1 }-\int_{\theta=0}^{\tan\theta=2\pi}\sec\theta d\tan\theta+\int_{\theta=0}^{\tan\theta=2\pi}\sec\theta d\theta\\ &=\pi\sqrt{4\pi^2+1 }+\frac{1}{2}\int_{\theta=0}^{\tan\theta=2\pi}\sec\theta d\theta\\ &=\pi\sqrt{4\pi^2+1 }+\frac{1}{2}\Big[\ln(\sec\theta+\tan\theta)\Big]_{\theta=0}^{\tan\theta=2\pi}\\ &=\pi\sqrt{4\pi^2+1 }+\frac{1}{2}\ln(\sqrt{4\pi^2+1}+2\pi) \end{align}

The estimated arc length is not correct.

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The estimate of OP is incorrect at one point

It is correct that

$d(X_2(k\frac{2\pi}{n}),X_2((k+1)\frac{2\pi}{n})) =\sqrt{k^2(\frac{2\pi}{n})^2+(k+1)^2(\frac{2\pi}{n})^2-2k(k+1)(\frac{2\pi}{n})^2[\cos k\frac{2\pi}{n}\cos (k+1)\frac{2\pi}{n}+\sin k\frac{2\pi}{n}\sin (k+1)\frac{2\pi}{n}]} $

Note that

$$\cos k\frac{2\pi}{n}\cos (k+1)\frac{2\pi}{n}+\sin k\frac{2\pi}{n}\sin (k+1)\frac{2\pi}{n}=\cos[(k+1)-k]\left(\frac{2\pi}{n}\right)=\cos\frac{2\pi}{n}$$

So,

$$d(X_2(k\frac{2\pi}{n}),X_2((k+1)\frac{2\pi}{n})) =\frac{2\pi}{n}\sqrt{k^2+(k+1)^2-2k(k+1)\cos\frac{2\pi}{n}} $$

The arc length is approximately $\displaystyle \sum_{k=0}^n\frac{2\pi}{n}\sqrt{k^2+(k+1)^2-2k(k+1)\cos\frac{2\pi}{n}}$.

We cannot replace $\sqrt{k^2+(k+1)^2-2k(k+1)\cos\frac{2\pi}{n}}$ by $\sqrt{k^2+(k+1)^2-2k(k+1)}$ as there are many terms. Although $\displaystyle \cos\frac{2\pi}{n}$ is close to $1$, the error cumulatives when there are many terms.

Answer 2

Although the idea is nice and the result is true, I see a problem in your argument. You say, that you like to approximate $$ \cos k\frac{2\pi}{n}\cos (k+1)\frac{2\pi}{n}+\sin k\frac{2\pi}{n}\sin (k+1)\frac{2\pi}{n} $$ with $1$ since $n$ goes to infinity. But this is not valid. An example where this argument fail: $$ e=\lim_{n\to\infty}\left(1+\frac1n\right)^n\neq \lim_{n\to\infty} 1^n=1 $$ although $1+\frac1n\to 1$ for $n\to\infty$.

But your argument can be fixed with a small change:

You can write $$ \sqrt{k^2(\frac{2\pi}{n})^2+(k+1)^2(\frac{2\pi}{n})^2-2k(k+1)(\frac{2\pi}{n})^2[\cos k\frac{2\pi}{n}\cos (k+1)\frac{2\pi}{n}+\sin k\frac{2\pi}{n}\sin (k+1)\frac{2\pi}{n}]} =\frac{2\pi}{n}\underbrace{\sqrt{k^2+(k+1)^2-2k(k+1)\left(\cos k\frac{2\pi}{n}\cos (k+1)\frac{2\pi}{n}+\sin k\frac{2\pi}{n}\sin (k+1)\frac{2\pi}{n}\right)}}_{=:f(n)} $$ Your approximated length is now $n\frac{2\pi}nf(n)=2\pi f(n)$.

Now it is easy to see, that $f(n)\to 1$ for $n\to\infty$ without problems.

Answer 3

You already recevied a nice answer from Mundron Schmidt.

Using you approach and simplifying as much as I could, we can arrive to $$d^2=\frac{4 \pi ^2}{n^2} \left(1+4 k (k+1) \sin ^2\left(\frac{\pi }{n}\right) \right)$$ which shows the limit and how it is approached.

Now, we can use Taylor series for infinitely large values of $n$ $$\sin\left(\frac{\pi }{n}\right)=\frac{\pi }{n}-\frac{\pi ^3}{6 n^3}+O\left(\frac{1}{n^5}\right)$$ $$\sin ^2\left(\frac{\pi }{n}\right)=\frac{\pi ^2}{n^2}-\frac{\pi ^4}{3 n^4}+O\left(\frac{1}{n^6}\right)$$ $$d^2=\frac{4 \pi ^2}{n^2} \left(1+\frac{4 \pi ^2 k (k+1)}{n^2}+O\left(\frac{1}{n^4}\right)\right)$$ $$d=\frac{2\pi}n \left(1+\frac{2 \pi ^2 k (k+1)}{n^2}+O\left(\frac{1}{n^4}\right)\right)$$