I find this video, showing a way to find curve length of $cosh(x)$.
https://www.youtube.com/watch?v=0i1vecdN_pw
However, I fail to understand how exactly this works. Why is the length of the origin to intersection of circle with x-axis the curve length?
Isn't it via pythagoras $\sqrt{y(N)^2-1}$ ?
Best Answer
For reference, take a look at this demonstration of $\cosh(x)$ and the definitions of $\cosh(x)$ and $\sinh(x)$ below:
$$ \begin{array}{l} \sinh x:=\frac{e^{x}-e^{-x}}{2} \\ \cosh x:=\frac{e^{x}+e^{-x}}{2} \end{array} $$
From the demonstration, we see that the arclength $MN$ originates at $(0,1)$ and terminates at $(x,r)$ as per the definitions above. Notice that $r=\frac{e^{x}+e^{-x}}{2}=\cosh(x)$ and is the radius of the circle. To answer your question, it must be shown that the arclength of $\cosh(x)$ is the same as the bottom leg of the right triangle shown in the demonstration. Notice that this triangle has a vertical (smaller) leg of $1$ and a hypotenuse of $r$. From simple trigonometry, the long leg, or the distance along the $x$-axis, is $\sqrt{r^2-1}$.
Mathematically, this can be summarized as (using the arclength formula):
$$ \int_0^x \sqrt{1+(\frac{d}{du}\cosh(u))^2}\;du=\sqrt{r^2-1} $$
Taking the arclength expression first, note that $\frac{d}{dx}\cosh(x)=\sinh(x)$. This makes the argument of the square root $1+\sinh^2(x)$. To continue integrating this, this argument can be simplified to $\cosh^2(x)=1+\sinh^2(x)$ from the identity $\cosh^2(x)-\sinh^2(x)=1$. Our integral then becomes:
$$ \int_0^x \sqrt{\cosh^2(u)}\;du=\int_0^x \cosh(u)\;du $$
From inspection, also note that $\int \cosh(x)=\sinh(x)+C$. Thus, our result is:
$$ \sinh(x)-\sinh(0)=\sinh(x) $$
Now, we must show that $\sinh(x)=\sqrt{r^2-1}$:
$$ \begin{align*} r^2-1&=\frac{e^{2x}+2e^xe^{-x}+e^{-2x}}{4}-\frac{4}{4}\\ &=\frac{e^{2x}+e^{-2x}-2}{4} \end{align*} $$
$$ \begin{align*} \sinh^2(x)&=\frac{e^{2x}-2e^xe^{-x}+e^{-2x}}{4}\\ &=\frac{e^{2x}+e^{-2x}-2}{4} \end{align*} $$
$$\therefore r^2-1=\sinh^2(x)\implies \boxed{\sqrt{r^2-1}=\sinh(x)}$$
This shows that the length of the segment along the $x$-axis intersecting the circle ($\sqrt{r^2-1}$) is the same as the arclength of $\cosh(x)$.