Can anyone please tell me where to go with this problem.

A chain hangs in a shape called a catenary, the equatio of which is;

f(x) = a*cosh(x/a)

Determine value ox when f(x)=48 and a=35.

Thanks in advance.

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- Aug 9th 2009, 10:50 AMady72Determine value of x
Can anyone please tell me where to go with this problem.

A chain hangs in a shape called a catenary, the equatio of which is;

f(x) = a*cosh(x/a)

Determine value ox when f(x)=48 and a=35.

Thanks in advance. - Aug 9th 2009, 10:55 AMRandom Variable
$\displaystyle \cosh^{-1} x = \ln (x+ \sqrt{x^{2}-1} ) \ \text{for} \ x \ge 1$

- Aug 9th 2009, 11:09 AMady72
I'm still none the wiser, I need to know how to get x on it's own.

I'm struggling with this one. - Aug 9th 2009, 11:14 AMRandom Variable
$\displaystyle 48 = 35 \cosh \Big(\frac{x}{35}\Big) $

$\displaystyle \frac{48}{35} = \cosh \Big(\frac{x}{35}\Big) $

$\displaystyle \cosh^{-1} \Big(\frac{48}{35}\Big) = \frac{x}{35} $

$\displaystyle x = 35 \cosh^{-1} \Big(\frac{48}{35}\Big) $ - Aug 9th 2009, 11:17 AMady72
Hi, I have tried that but when I place the value of x back into original formula it doesn't come back to f(x)=48.

- Aug 9th 2009, 11:26 AMRandom Variable
- Aug 9th 2009, 11:27 AMady72
Sorry, I,ve got it now.

I wasn't putting it to calculator correctly.

Thanks for your advice!!