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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$\displaystyle \cosh^{-1} x = \ln (x+ \sqrt{x^{2}-1} ) \ \text{for} \ x \ge 1$
I'm still none the wiser, I need to know how to get x on it's own. I'm struggling with this one.
$\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) $
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.
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. $\displaystyle 35 \cosh \Big(35 \frac{\cosh^{-1} \Big(\frac{48}{35}\Big)}{35} \Big)$ $\displaystyle = 35 \cosh \Big(\cosh^{-1} \frac{48}{35}\Big) = 35 \Big(\frac{48}{35}\Big) = 48 $
Sorry, I,ve got it now. I wasn't putting it to calculator correctly. Thanks for your advice!!
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