Having trouble with the 2nd part of this question, on proving that the radius of curvature is inversely proportional to the length of the normal intercepted between the point on the curve and y-axis.
From my understanding at point $P(x,y)$, the tractix has derivative $y'=\frac{dy}{dx} = \frac{\sqrt{1-x^2}}{x}$ for the upper half of the curve. The equation to the normal of the curve at $P$ given by $(n,m)$ is:
$$n-y_0=-1/y' (m-x)$$ The formula for the radius of curvature, $k$, should be: $$k=\frac{(1+y'^2)^{3/2}}{y''}$$
Now we need only find the length between the point $P$ and the intersection of the normal with the y-axis and compare it with $k$. But for some reason I can't get the result.
At a point $(x_0,y_0)$ on the curve the tangent line is $$\frac{y-y_0}{x-x_0}=y_0^{\prime}$$ Where the line intersects the $y$-axis, $x=0$, so $y=y_0-x_0y_0^{\prime}$ and the distance to the $y$-intercept (squared) is $$(x-x_0)^2+(y-y_0)^2=x_0^2+x_0^2y_0^{\prime2}=1^2=1$$ So the differential equation of the curve is indeed $$y^{\prime}=\pm\frac{\sqrt{1-x^2}}x$$ Taking the branch with $x>0$, $y^{\prime}>0$ we have $$\begin{align}y&=\int\frac{\sqrt{1-x^2}}xdx=\int\frac{\tanh\theta}{\text{sech}\,\theta}(-\text{sech}\,\theta\tanh\theta\,d\theta)\\ &=\int\left(\text{sech}^2\theta-1\right)d\theta=\tanh\theta-\theta+C\\ &=\sqrt{1-x^2}-\cosh^{-1}\left(\frac1x\right)+C\\ &=\sqrt{1-x^2}-\ln\left(\frac1x+\sqrt{\frac1{x^2}-1}\right)+C\\ &=\sqrt{1-x^2}+\ln\left(\frac x{1+\sqrt{1-x^2}}\right)+C\end{align}$$ Along the curve, $\vec r=\langle x,y,0\rangle$, $\vec v=\langle 1,y^{\prime},0\rangle=v\hat T$, and $\vec a=\langle 0,y^{\prime\prime},0\rangle=a_T\hat T+\frac{v^2}{\rho}\hat N$, so $v=\sqrt{1+y^{\prime2}}$, and $\vec v\times\vec a=\langle0,0,y^{\prime\prime}\rangle=\frac{v^3}{\rho}\hat B$, giving us the formula $$\rho=\left|\frac{v^3}{y^{\prime\prime}}\right|=\frac{\left(1+y^{\prime2}\right)^{3/2}}{|y^{\prime\prime}|}$$ Now,$$\begin{align}y^{\prime\prime}&=\frac12\frac{(-2x)}{\sqrt{1-x^2}}\frac1x-\frac{\sqrt{1-x^2}}{x^2}=\frac{-1}{x^2\sqrt{1-x^2}}\\ 1+y^{\prime2}&=1+\frac{1-x^2}{x^2}=\frac1{x^2}\end{align}$$ So $$\rho=\frac{\frac1{x^3}}{\frac1{x^2\sqrt{1-x^2}}}=\frac{\sqrt{1-x^2}}x$$ At a point $(x_0,y_0)$ along the curve the equation of the normal is $$\frac{y-y_0}{x-x_0}=-\frac1{y_0^{\prime}}$$ It intersect the $y$-axis at $x=0$ and $y=y_0+\frac{x_0}{y_0^{\prime}}$ so the (squared) distance to the $y$-intercept is $$(x-x_0)^2+(y-y_0)^2=x_0^2+\frac{x_0^2}{y_0^{\prime2}}=x_0^2+\frac{x_0^4}{1-x_0^2}=\frac{x_0^2}{1-x_0^2}$$ The distance to the intercept along the normal from point $(x,y)$ is thus $$\frac x{\sqrt{1-x^2}}=\frac1{\rho}$$ As claimed. For the arc length, $$ds=\sqrt{1+y^{\prime2}}dx=\frac{dx}x$$ So $s=\ln|x|+C_1$; $x=\pm e^{C_1}e^s=C_2e^s$. We can take $C_2=1$ if we start the arc from $x=1$. Then along the curve $$\begin{align}\vec r&=\langle x,y\rangle=\langle x,\sqrt{1-x^2}+\ln\left(\frac x{1+\sqrt{1-x^2}}\right)\rangle\\ &=\langle e^s,\sqrt{1-e^{2s}}+s-\ln\left(1+\sqrt{1-e^{2s}}\right)\rangle\end{align}$$ Where we have also started the arc at $y=0$.