Arc length from $\exp(x)$ , $x$ from $0$ to $t$

Question

How do I find $t$ such that arc length of $\exp(x)$ from $x=0$ to $t$ is $2\pi$? https://en.wikipedia.org/wiki/Arc_length I know that it would be equal to $$\int_0^{t} \sqrt{\exp(2x)+1}\ dx$$

Answer 1

Well, first of all, let's look at the integral:

$$\mathcal{I}_{\space\text{n}}\left(t\right):=\int_0^t\sqrt{1+\exp\left(\text{n}\cdot x\right)}\space\text{d}x\tag1$$

Now, let's do a couple substitutions:

1. Substitute $\text{u}:=\text{n}\cdot x$
2. Substitute $\text{s}:=\exp\left(\text{u}\right)$
3. Substitute $\text{p}:=\sqrt{1+\text{s}}$

So, then we get:

$$\mathcal{I}_{\space\text{n}}\left(t\right)=\frac{2}{\text{n}}\cdot\int_\sqrt{2}^{\sqrt{1+\exp\left(\text{n}\cdot t\right)}}\frac{\text{p}^2}{\text{p}^2-1}\space\text{d}\text{p}\tag2$$

Using long divison, we can write:

$$\frac{\text{p}^2}{\text{p}^2-1}=1+\frac{1}{2}\cdot\frac{1}{\text{p}-1}-\frac{1}{2}\cdot\frac{1}{\text{p}+1}\tag3$$

Now, let's do a couple substitutions again:

1. Substitute $\text{w}:=\text{p}+1$
2. Substitute $\text{v}:=\text{p}-1$

So, we get:

$$\mathcal{I}_{\space\text{n}}\left(t\right)=\frac{2}{\text{n}}\cdot\int_\sqrt{2}^{\sqrt{1+\exp\left(\text{n}\cdot t\right)}}1\space\text{d}\text{p}+\frac{1}{\text{n}}\cdot\int_{\sqrt{2}-1}^{\sqrt{1+\exp\left(\text{n}\cdot t\right)}-1}\frac{1}{\text{v}}\space\text{d}\text{v}-\frac{1}{\text{n}}\cdot\int_{1+\sqrt{2}}^{1+\sqrt{1+\exp\left(\text{n}\cdot t\right)}}\frac{1}{\text{w}}\space\text{d}\text{w}=$$ $$\frac{2}{\text{n}}\cdot\left(\sqrt{1+\exp\left(\text{n}\cdot t\right)}-\sqrt{2}\right)+\frac{1}{\text{n}}\cdot\ln\left|\frac{\sqrt{1+\exp\left(\text{n}\cdot t\right)}-1}{\sqrt{2}-1}\right|-\frac{1}{\text{n}}\cdot\ln\left|\frac{1+\sqrt{1+\exp\left(\text{n}\cdot t\right)}}{1+\sqrt{2}}\right|=$$ $$\frac{2}{\text{n}}\cdot\left(\sqrt{1+\exp\left(\text{n}\cdot t\right)}-\sqrt{2}\right)+\frac{1}{\text{n}}\cdot\left(\ln\left(3+2\sqrt{2}\right)+\ln\left|\frac{\sqrt{1+\exp\left(\text{n}\cdot t\right)}-1}{\sqrt{1+\exp\left(\text{n}\cdot t\right)}+1}\right|\right)\tag4$$

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For your case, Let $\text{n}=2$, so we want to solve:

$$\frac{2}{2}\cdot\left(\sqrt{1+\exp\left(2\cdot t\right)}-\sqrt{2}\right)+\frac{1}{2}\cdot\left(\ln\left(3+2\sqrt{2}\right)+\ln\left|\frac{\sqrt{1+\exp\left(2\cdot t\right)}-1}{\sqrt{1+\exp\left(2\cdot t\right)}+1}\right|\right)=2\pi\tag5$$

Using Mathematica, I found:

So:

$$t\approx1.92985096273712170694334669601\tag6$$

Answer 2

Taking into account the antiderivative given by Wolfram Alpha, you want to solve $$\sqrt{e^{2 t}+1}-\tanh ^{-1}\left(\sqrt{e^{2 t}+1}\right)-\sqrt{2}+\tanh ^{-1}\left(\sqrt{2}\right)=2\pi$$ and, as said in comments, only numerical methods will do the job.

So, consider that you look for the zero of function $$f(t)=\sqrt{e^{2 t}+1}-\tanh ^{-1}\left(\sqrt{e^{2 t}+1}\right)-\left(2\pi+\sqrt{2}-\tanh ^{-1}\left(\sqrt{2}\right)\right)$$ $$f'(t)=\sqrt{e^{2 t}+1}\qquad \text{and}\qquad f''(t)=\frac{e^{2 x}}{\sqrt{e^{2 x}+1}}> 0 \,\,\, \forall t$$ and use Newton method which, starting from a guess $t_0$ will update it according to $$t_{k+1}=t_k-\frac{f(t_k)}{f(t_k)}$$

Being lazy, start using $t_0=1$ to get the following iterates $$\left( \begin{array}{cc} k & t_k \\ 0 & 1.00000 \\ 1 & 2.47760 \\ 2 & 2.05477 \\ 3 & 1.93721 \\ 4 & 1.92988 \\ 5 & 1.92985 \end{array} \right)$$ You can notice that we face a serious overshoot of the solution because, at $t_0=1$ we have $f(t_0)\times f''(t_0) <0$ (this is Darboux-Fourier theorem).

Using instead $t_0=3$, the iterates would be $$\left( \begin{array}{cc} k & t_k \\ 0 & 3.00000 \\ 1 & 2.34140 \\ 2 & 2.00324 \\ 3 & 1.93243 \\ 4 & 1.92985 \end{array} \right)$$

Edit after Jan's answer

For the most general case of $I_n(t)=L$, equation $(4)$ in Jan's answer write $$2 \left(k-\sqrt{2}\right)+\log \left(\left|\frac{k-1}{k+1}\right|\right)+\log \left(3+2 \sqrt{2}\right)= nL$$ Assuming that $k$ is large, this can be approximated as $$k-\frac 1k=a\qquad \text{with} \qquad a=\frac{1}{2} \left(nL +2 \sqrt{2}-\log \left(3+2 \sqrt{2}\right)\right)$$ and then a first estimate $$k^{(0)}_{est}=\frac{1}{2} \left(a+\sqrt{a^2+4}\right)\implies t^{(0)}_{est}=\frac 1 n{\log \left(\left[k^{(0)}_{est}\right]^2-1\right)} $$ A second estimate can be generated using one iteration of Newton method to get $$k^{(1)}_{est}=k^{(0)}_{est}-\frac {\left[k^{(0)}_{est}\right]^2-1 } {\left[k^{(0)}_{est}\right]^2 }\left(k^{(0)}_{est}+\frac 12 \log\left(\frac {k^{(0)}_{est}-1}{k^{(0)}_{est}+1}\right) \right)\implies t^{(1)}_{est}=\frac 1 n{\log \left(\left[k^{(1)}_{est}\right]^2-1\right)}$$

To show how good is Jan's solution, I produced a table for a few values of $n$ for $L=2 \pi$. $$\left( \begin{array}{ccc} n & t^{(0)}_{est} &t^{(1)}_{est} & t_{exact} \\ 1 & 2.669771799 & 2.672681287 & 2.672681561 \\ 2 & 1.929707113 & 1.929850962 & 1.929850962 \\ 3 & 1.535537276 & 1.535559127 & 1.535559127 \\ 4 & 1.287720377 & 1.287725928 & 1.287725928 \\ 5 & 1.115764696 & 1.115766588 & 1.115766588 \\ 6 & 0.988563624 & 0.988564405 & 0.988564405 \\ 7 & 0.890161014 & 0.890161382 & 0.890161382 \\ 8 & 0.811477041 & 0.811477232 & 0.811477232 \\ 9 & 0.746939084 & 0.746939191 & 0.746939191 \\ 10 & 0.692924541 & 0.692924605 & 0.692924604 \end{array} \right)$$