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I've the following parametric equations for a curve: $$\begin{cases}x(t)=a\cdot \operatorname{sech} (t) \\ y(t)=a\cdot(t-\tanh(t))\end{cases}$$ Now let $\theta(t)=-\arctan(\sinh(t))$ how does the parametric form change? I need to substitute $\theta$ in above equations...I'm not good at simplifying these tricky expressions involving even inverse hyperbolic functions. Any help would be greatly appreciated.

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  • $\begingroup$ Since $\theta(t)=-\arctan(\sinh(t))$ you have $t=\operatorname{arcsinh}(\tan(-\theta(t)))$, which you can subsitute for $t$. I'd start from there and try to simplify the thus generated terms. $\endgroup$ – dinosaur Oct 11 '14 at 17:28
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Hints:

You can use that $t=-\sinh^{-1}(\tan\theta)$, along with the identities that

$\text{sech} (-x)=\text{sech}(x), \;\;\text{sech}(\sinh^{-1}t)=\frac{1}{\sqrt{t^2+1}},\;\;\tan^{2}\theta+1=\sec^{2}\theta$, and

$\sinh^{-1}x=\ln(x+\sqrt{x^2+1}), \;\;\tanh(-x)=-\tanh (x),\;\;\tanh(\sinh^{-1}t)=\frac{t}{\sqrt{t^2+1}}$


Notice that $\displaystyle\text{sech}(\sinh^{-1}(t))=\frac{1}{\cosh(\sinh^{-1}(t))}$ and $\displaystyle\tanh(\sinh^{-1}(t))=\frac{\sinh(\sinh^{-1}(t))}{\cosh(\sinh^{-1}(t))}$

where $\cosh(\sinh^{-1}(t))=\sqrt{t^2+1}$ and $\sinh(\sinh^{-1}(t))=t$.

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