# Thread: Writing an equation to catenary curve.

1. ## Writing an equation to catenary curve.

A catenary curve is an idealised hanging chain or cable. The curve is the graph of a hyperbolic cosine function and is similar in appearance to a parabola. The equation of a catenary can be expressed in two ways: y = c cosh (x/c) or y = 2/a (e^bx + e^-bx)

To find the equation of this catenary curve two points much be chosen, and their coordinates must be substituted into the equation to find the values of a and b.
The points which are used to find the equation are (0,2) and (5,16)
Substituting (0,2) into the equation:

y = 2/a (e^bx + e^-bx)
2 = 2/a (e^b0 + e^-b0)
2 = 2/a (e^0 + e^-0)
2 = 2/a (1 + 1)
2 = 2/a * 2
2 = a

The value of a is 2.
Therefore the equation becomes:

y = e^bx + e^-bx

By substituting (5,16) into the equation you can find the value of b:
y = ebx + e-bx
16 = e5b + e-5b
loge16 = loge e5b + logee-5b
2.773 = 5b -5b

I have to get an equation so is my 'a' value correct? And how do I get my b value because I have them cancelling out. Any help on where I've gone wrong or what I do?

2. Check the equation of catenary in Wikipedia.

3. Originally Posted by eriiin
...
16 = e5b + e-5b
loge16 = loge e5b + logee-5b <---
...

Where did you learn this step?

$16=e^{5b} + e^{-5b}~\implies~16 e^{5b} = \left(e^{5b}\right)^2+1$

This is a quadratic in $e^{5b}$. Use substitution to solve the equation.

4. $log(x+ y)\ne log(x)+ log(y)$