How to solve tractrix equation for x?

Question

Tractrix curve is defined by the following equation:

$y= asech^{-1}(x/a)-sqrt(a^2-x^2)$

where a is a constant.

I would like to apply this equation for an algorithm where y is a known value. Is it possible to solve it for x?

Answer 1

I haven't been able to fully solve it, but I'll share what I've got so far: $$y = a{sech}^{-1}\Bigl(\frac{x}{a}\Bigr)-\sqrt{a^2-x^2}$$ Since ${sech}^{-1}(x) = {cosh}^{-1}\bigl(\frac{1}{x}\bigr)$: $$y = a{cosh}^{-1}\Bigl(\frac{a}{x}\Bigr)-\sqrt{a^2-x^2} \\ \sqrt{a^2-x^2} = a{cosh}^{-1}\Bigl(\frac{a}{x}\Bigr)-y \\ a^2-x^2 = a^2\biggr({cosh}^{-1}\Bigr(\frac{a}{x}\Bigl)\biggl)^2 + y^2 -2a{cosh}^{-1}\Bigl(\frac{a}{x}\Bigr)y$$ With the substitution of ${cosh}^{-1}\bigr(\frac{a}{x}\bigl) := w$ and $\frac{a}{cosh(w)} = x$ we get: $$a^2 - \frac{a^2}{cosh^{2}(w)} = a^2w^2 + y^2 + 2awy \\ 0 = a^2w^2 + 2awy + \frac{a^2}{cosh^{2}(w)} - a^2 + y^2$$ I'll use the quadratic formula for $w$: $$w_{1,2} = \frac{-2ay \pm \sqrt{4a^2y^2-4a^2\Bigl(\frac{a^2}{{cosh}^{2}(w)}-a^2+y^2\Bigr)}}{2a^2}$$ We can fortunately simplify with $2$ and $a$: $$w_{1,2} = \frac{-y \pm \sqrt{ay^2-a\Bigl(\frac{a^2}{{cosh}^{2}(w)}-a^2+y^2\Bigr)}}{a} \\ w_{1,2} = \frac{-y \pm \sqrt{a^2\Bigl(1-\frac{1}{{cosh^{2}(w)}}\Bigr)}}{a} \\ w_{1,2} = \frac{-y \pm \sqrt{a^2({tanh}^{2}(w))}}{a} \\ w_{1,2} = \frac{-y \pm atanh(w)}{a}$$ $$aw \mp atanh(w) = -y$$ $$w \pm tanh(w) = -\frac{y}{a}$$ From here I wasn't able to continue, however we can deduce the following and approximate $x$: $$\lim_{w \to \infty}(w \pm tanh(w)) = w \pm 1$$ So for large $w$ values, this is a good approximation: $$w \pm 1 \approx -\frac{y}{a}$$ $${cosh}^{-1}\Bigr(\frac{a}{x}\Bigl) \pm 1 \approx -\frac{y}{a}$$ $${cosh}^{-1}\Bigr(\frac{a}{x}\Bigl)\approx -\frac{y}{a} \mp 1 $$ $$\frac{a}{x}\approx cosh\Bigr(-\frac{y}{a} \mp 1\Bigl)$$ $$x \approx \frac{a}{cosh\Bigr(1 \pm \frac{y}{a}\Bigl)}$$ What are large $w$ values? $w$ is large, if $x$ is small, since: $$\lim_{w \to \infty}(w \pm tanh(w)) = w \pm 1$$ $$\lim_{{cosh}^{-1}(\frac{a}{x}) \to \infty}\Bigl({cosh}^{-1}\Bigr(\frac{a}{x}\Bigr) \pm tanh\Bigl({cosh}^{-1}\Bigl(\frac{a}{x}\Bigr)\Bigr)\Bigr) = {cosh}^{-1}\Bigr(\frac{a}{x}\Bigr) \pm 1$$ $$\lim_{\frac{a}{x} \to \infty}\Bigl( \pm tanh\Bigl({cosh}^{-1}\Bigl(\frac{a}{x}\Bigr)\Bigr)\Bigr) = \pm 1$$ $$\lim_{x \to 0}\left( \frac{\sqrt{\Bigl(\frac{a}{x}\Bigr)^{2}-1}}{\frac{a}{x}}\right) = 1$$ $$\lim_{x \to 0}\left(\sqrt{\Bigl(\frac{a}{x}\Bigr)^{2}-1}\right) = \lim_{x \to 0}\left(\frac{a}{x}\right)$$ Which is true for all $x \in \mathbb{R}.$

SUMMARY: If $x$ is small ($x \approx 0$), then a good approximation is: $$x_{1,2} \approx \frac{a}{cosh\Bigl(1 \pm \frac{y}{a}\Bigr)}$$