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I've been given the the question

The length $L$ of a heavy cable hanging under gravity is given you the equation

$L=2\sinh x + 3\cosh x$

$L$ is given as $5$

I have working this out into a quadratic formula and found the answer.

I have been asked to check the answer using an alternative method. Could anyone help me out

Carl

amWhy
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C . Martin
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1 Answers1

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Method I \begin{align*} 2\sinh x+3\cosh x &= 5 \\ 2\left( \frac{e^{x}-e^{-x}}{2} \right)+ 3\left( \frac{e^{x}+e^{-x}}{2} \right) &= 5 \\ (2+3)e^{2x}+(3-2) &= 10e^{x} \\ 5e^{2x}-10e^{x}+1 &= 0 \\ e^{x} &= \frac{10\pm \sqrt{10^2-4(5)(1)}}{2(5)} \\ &= 1 \pm \frac{\sqrt{80}}{10} \\ &= 1 \pm \frac{2}{\sqrt{5}} \\ x &= \ln \left( 1\pm \frac{2}{\sqrt{5}} \right) \end{align*}

Method II \begin{align*} \sqrt{a^2-b^2} \cosh \left( x+\tanh^{-1} \frac{b}{a} \right) &= a\cosh x+b\sinh x \; , \qquad a>|b| \\ 2\sinh x+3\cosh x &= 5 \\ \sqrt{3^2-2^2} \cosh \left( x+\tanh^{-1} \frac{2}{3} \right) &= 5 \\ \cosh \left( x+\tanh^{-1} \frac{2}{3} \right) &= \sqrt{5} \\ x+\tanh^{-1} \frac{2}{3} &= \pm \cosh^{-1} \sqrt{5} \\ x &= \pm \cosh^{-1} \sqrt{5}-\tanh^{-1} \frac{2}{3} \\ &= \pm \ln (\sqrt{5}+\sqrt{5-1})- \ln \sqrt{\frac{1+\frac{2}{3}}{1-\frac{2}{3}}} \\ &= \pm \ln (\sqrt{5}+2)-\ln \sqrt{5} \\ &= \ln (\sqrt{5} \pm 2)-\ln \sqrt{5} \\ &= \ln \left( 1\pm \frac{2}{\sqrt{5}} \right) \end{align*}

Ng Chung Tak
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