# Thread: Area Under an Inverted Catenary

1. ## Area Under an Inverted Catenary

Greetings all, this is my first post on these forums.

I'm attempting to construct an inverted catenary that will intersect the x-axis at y=0 and y=1 and have an area of 1 bound by the x-axis and the function.

I started off with $\displaystyle f(x)=cosh[1.5]-cosh[3x-1.5]$ and that satisfies the intersection points, but does not have an area of 1. So to get the area closer to one, I kept manipulating the numbers and integrating, until I came to $\displaystyle f(x)=cosh[1.543405]-cosh[3.08681*x-1.543405]$

This gives an area of 1.0000 under the curve bound by the x-axis, but I'm sure it's not actually equalling 1, I've just reached the limit of my program's (winplot) decimal display.

My goal is to find a more systematic approach, since repeatedly adjusting the numbers isn't exact. I wonder if there could be some sort of series representation for the function. I don't recognize the number 1.543405 as anything special, but perhaps it could be respresented exactly, rather than a decimal (eg. to say pi rather than 3.141592) and yield the proper results.

Thanks in advance for any help!
Cheers.

2. Originally Posted by yfronto
Greetings all, this is my first post on these forums.

I'm attempting to construct an inverted catenary that will intersect the x-axis at y=0 and y=1 and have an area of 1 bound by the x-axis and the function.

I started off with $\displaystyle f(x)=cosh[1.5]-cosh[3x-1.5]$ and that satisfies the intersection points, but does not have an area of 1. So to get the area closer to one, I kept manipulating the numbers and integrating, until I came to $\displaystyle f(x)=cosh[1.543405]-cosh[3.08681*x-1.543405]$

This gives an area of 1.0000 under the curve bound by the x-axis, but I'm sure it's not actually equalling 1, I've just reached the limit of my program's (winplot) decimal display.

My goal is to find a more systematic approach, since repeatedly adjusting the numbers isn't exact. I wonder if there could be some sort of series representation for the function. I don't recognize the number 1.543405 as anything special, but perhaps it could be respresented exactly, rather than a decimal (eg. to say pi rather than 3.141592) and yield the proper results.

Thanks in advance for any help!
Cheers.
Clearly, as you noted, $\displaystyle f(x)=\cosh(a)-\cosh\left(2ax-a\right)$ will do. So we need to solve $\displaystyle \int_0^{1}\left\{\cosh(a)-\cosh\left(2ax-a\right)\right\}dx=1\implies \cosh(a)-\frac{\sinh(a)}{a}=1$ this has no closed form solution though. - Wolfram|Alpha. If you want to get close numerical solutions to this use the fact that $\displaystyle \cosh(a)=\sum_{n=0}^{\infty}\frac{a^{2n}}{(2n)!},\ sinh(a)=\sum_{n=0}^{\infty}\frac{a^{2n+1}}{(2n+1)! }$. If you only take the first, say 4, terms of these series you can find an approximate closed form solution.

3. Originally Posted by Drexel28
Clearly, as you noted, $\displaystyle f(x)=\cosh(a)-\cosh\left(2ax-a\right)$ will do. So we need to solve $\displaystyle \int_0^{1}\left\{\cosh(a)-\cosh\left(2ax-a\right)\right\}dx=1\implies \cosh(a)-\frac{\sinh(a)}{a}=1$ this has no closed form solution though. - Wolfram|Alpha. If you want to get close numerical solutions to this use the fact that $\displaystyle \cosh(a)=sum_{n=0}^{\infty}\frac{a^{2n}}{(2n)!},\s inh(a)=\sum_{n=0}^{\infty}\frac{a^{2n+1}}{(2n+1)!}$.
That makes sense. I was just having trouble connecting the dots. Thanks!