Explanation for geometric proof of arc length cosh

Question

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 $\sqrt{y(N)^2-1}$ ?

Answer 1

First of all:

If a planar curve in $\mathbb{R}^2$ is defined by the equation $\text{y}=\text{f}\left(x\right)$ where $\text{f}$ is continuously differentiable, then it is simply a special case of a parametric equation where $x=t$ and $\text{y}=\text{f}\left(t\right)$, the arc length is given by:

$$\mathcal{S}\left(\text{a},\text{b}\right):=\int_\text{a}^\text{b}\sqrt{1+\left(\frac{\text{d}\text{y}\left(x\right)}{\text{d}x}\right)^2}\space\text{d}x\tag1$$

For the derivation, look at Wikipedia.

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In your case, we have:

$$\mathcal{S}\left(\text{a},\text{b}\right)=\int_\text{a}^\text{b}\sqrt{1+\left(\frac{\text{d}}{\text{d}x}\left(\cosh\left(x\right)\right)\right)^2}\space\text{d}x=\int_\text{a}^\text{b}\sqrt{1+\sinh^2\left(x\right)}\space\text{d}x=$$ $$\int_\text{a}^\text{b}\cosh\left(x\right)\space\text{d}x=\left[\sinh\left(x\right)\right]_\text{a}^\text{b}=\sinh\left(\text{b}\right)-\sinh\left(\text{a}\right)\tag2$$

Answer 2

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 $\cosh(x)$.