# Determine value of x

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• Aug 9th 2009, 11:50 AM
ady72
Determine 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, 11:55 AM
Random Variable
$\cosh^{-1} x = \ln (x+ \sqrt{x^{2}-1} ) \ \text{for} \ x \ge 1$
• Aug 9th 2009, 12:09 PM
ady72
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, 12:14 PM
Random Variable
$48 = 35 \cosh \Big(\frac{x}{35}\Big)$

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

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

$x = 35 \cosh^{-1} \Big(\frac{48}{35}\Big)$
• Aug 9th 2009, 12:17 PM
ady72
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, 12:26 PM
Random Variable
Quote:

Originally Posted by ady72
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.

$35 \cosh \Big(35 \frac{\cosh^{-1} \Big(\frac{48}{35}\Big)}{35} \Big)$ $= 35 \cosh \Big(\cosh^{-1} \frac{48}{35}\Big) = 35 \Big(\frac{48}{35}\Big) = 48$
• Aug 9th 2009, 12:27 PM
ady72
Sorry, I,ve got it now.
I wasn't putting it to calculator correctly.
Thanks for your advice!!