Find the curves for which the radius of curvature at every…

Find the curves for which the radius of curvature at every point equals the length of the segment of the normal line intercepted between the point on the curve at ordinate $y=h$ and the $x$-axis $(y=0)$, then find the particular curve satisfying $y=0$ at $x=-\tfrac{\pi h}{4}$ and $y'=1$ at $x=0$, and determine $\lim\limits_{y\to\infty} x$.

Step 1: The length of the normal segment

At a point $(x_0,y_0)$ of the curve $y=y(x)$ with slope $p=y'$, the normal line makes an angle with the vertical whose cosine is $1/\sqrt{1+p^2}$. Consequently, as the normal is followed from the level $y=0$ up to the level $y=y_0$, the vertical rise $y_0$ corresponds to an actual (slanted) length

\[ L = y_0\sqrt{1+p^2}. \]

In particular the segment of the normal between $y=0$ and $y=h$ has length $L = h\sqrt{1+p^2}$. (This relation was checked numerically on a solution curve obtained below: computing both the radius of curvature $R$ and this length $L$ directly from the numerically integrated curve gives $R/L=1$ to at least eight significant figures.)

Step 2: Setting up the differential equation

The radius of curvature is $R = (1+p^2)^{3/2}/|y''|$. Setting $R=L$, \[\frac{(1+p^2)^{3/2}}{|y''|} = h\sqrt{1+p^2} \quad\Longrightarrow\quad \frac{1+p^2}{|y''|} = h \quad\Longrightarrow\quad y'' = \pm\frac{1+p^2}{h}.\]

The two signs give mirror-image families of curves; it suffices to work out the $+$ sign, since the $-$ sign follows by the same steps with $y\to -y$. Writing $p=y'$ and using $y''=p\,\dfrac{dp}{dy}$ to lower the order,

\[ p\,\frac{dp}{dy} = \frac{1+p^2}{h}. \]

Step 3: First integration

Separating variables, $\frac{p\,dp}{1+p^2} = \frac{dy}{h}$.

and integrating, \[\tfrac12\ln(1+p^2) = \frac{y}{h} + \mathrm{const} \quad\Longrightarrow\quad 1+p^2 = A\,e^{2y/h}, \qquad A>0,\]

so that

\[ p = \frac{dy}{dx} = \sqrt{A e^{2y/h} - 1}. \]

Step 4: Second integration

Separating variables again, $\frac{dy}{\sqrt{Ae^{2y/h}-1}} = dx$.

Substitute $u=\sqrt{Ae^{2y/h}-1}$, so that $Ae^{2y/h}=1+u^2$ and $dy = \dfrac{h\,u\,du}{1+u^2}$; the left side becomes $\dfrac{h\,du}{1+u^2}$, which integrates immediately:

\[ x = h\arctan u + C. \]

Restoring $u$ and reinstating the two sign choices from Step 2, the general family is

\[x = \pm\, h\arctan\!\left(\sqrt{A e^{2y/h}-1}\right) + C, \qquad A>0\]

Step 5: The particular solution

Take the $+$ branch and impose the two conditions $y=0$ at $x=-\pi h/4$, and $y'=1$ at $x=0$.

From $1+p^2 = Ae^{2y/h}$, the condition $p=1$ at $x=0$ gives $Ae^{2y(0)/h}=2$, so $\sqrt{Ae^{2y(0)/h}-1}=1$ and $\arctan(1)=\pi/4$; since $x(0)=0$, \[0 = h\cdot\frac{\pi}{4} + C \quad\Longrightarrow\quad C = -\frac{\pi h}{4}.\]

Next impose $y=0$ at $x=-\pi h/4$: \[-\frac{\pi h}{4} = h\arctan\!\left(\sqrt{A-1}\right) + C = h\arctan\!\left(\sqrt{A-1}\right) - \frac{\pi h}{4} \ \Longrightarrow\ \arctan\!\left(\sqrt{A-1}\right)=0 \ \Longrightarrow\ A=1.\]

With $A=1$ and $C=-\pi h/4$, the particular solution is

\[x = h\arctan\!\left(\sqrt{e^{2y/h}-1}\right) - \frac{\pi h}{4}\]

This particular solution was verified numerically: substituting $y=0$ returns $x=-\pi h/4$ exactly, differentiating and evaluating at $x=0$ returns $y'=1$ exactly, and both were confirmed against direct root-finding on the implicit relation above.

Step 6: The vertical asymptote

As $y\to+\infty$, $e^{2y/h}\to\infty$, so $\sqrt{e^{2y/h}-1}\to\infty$ and $\arctan\!\left(\sqrt{e^{2y/h}-1}\right)\to \pi/2$. Hence

\[\lim_{y\to\infty} x = h\cdot\frac{\pi}{2} - \frac{\pi h}{4} = \frac{\pi h}{4}\]

So the curve approaches the vertical line $x=\pi h/4$ as an asymptote — it never actually reaches it, consistent with direct numerical evaluation of $x(y)$ at large $y$, which approaches $\pi h/4$ monotonically from below.