Find the curves for which, at every point, R = 2N, where R…

Find the curves for which, at every point, $R = 2N,$ where $R$ is the radius of curvature and $N=y\sqrt{1+y'^2}$ is the length of the normal segment intercepted between the curve and the $x$-axis. Consider separately: $1^\circ$ the curve convex toward the $x$-axis ($y''>0$); $2^\circ$ concave toward the $x$-axis ($y''<0$).

Step 1: Setting up the equation

Since $R=(1+y'^2)^{3/2}/|y''|$, dividing the condition $R=2N$ by $\sqrt{1+y'^2}$ leaves $|y''|=(1+y'^2)/(2y)$, i.e.

\[ y'' = \frac{1+y'^2}{2y} \ \ (1^\circ), \qquad\qquad y'' = -\frac{1+y'^2}{2y} \ \ (2^\circ). \]

Step 2: Reduce the order — $y'=p(y)$

With $y'=p$, $y''=p\,dp/dy$, both cases separate as

\[ \frac{p\,dp}{1+p^2} = \frac{dy}{2y} \ \ (1^\circ), \qquad\qquad \frac{p\,dp}{1+p^2} = -\frac{dy}{2y} \ \ (2^\circ), \]

and integrating once, \[1+y'^2 = C_1^2\,y \ \ (1^\circ), \qquad\qquad 1+y'^2 = \frac{C_1^2}{y} \ \ (2^\circ).\]

Step 3: Case 1° — parabolas

Here $y'^2=C_1^2y-1$, so $dy/\sqrt{C_1^2y-1}=\pm dx$. Integrating ($u=C_1^2y-1$) and squaring: \[\frac{2\sqrt{C_1^2y-1}}{C_1^2} = \pm(x-C_2) \qquad\Longrightarrow\qquad C_1^2y = 1+\frac{C_1^4(x-C_2)^2}{4}.\]

Writing $a=C_1^2/4$:

$y = a(x-C_2)^2+\frac{1$, a family of parabolas with vertical axis, vertex $\left(C_2,\frac1{4a}\right)$, opening upward.

Step 4: Case 2° — cycloids

Here $y'^2=(C_1^2-y)/y$, so $\sqrt y\,dy/\sqrt{C_1^2-y}=\pm dx$. With $y=C_1^2\sin^2t$ (so $\sqrt y=C_1\sin t$, $\sqrt{C_1^2-y}=C_1\cos t$): \[\int\frac{\sqrt y\,dy}{\sqrt{C_1^2-y}} = C_1^2\!\left(t-\tfrac12\sin2t\right) = C_1^2\arcsin\frac{\sqrt y}{C_1} - \sqrt{y(C_1^2-y)}.\]

Renaming $C_1\to a$, the implicit general solution is $a^2\arcsin\frac{\sqrt y$.

Setting $y=\tfrac{a^2}{2}(1-\cos t)$ (so $\sqrt y/a=\sin(t/2)$) turns this into the familiar parametric form $x = \frac{a^2$.

a cycloid traced by a circle of radius $a^2/2$ rolling on the $x$-axis.

Step 5: Verification, Case 1°

For $y=a(x-C_2)^2+\frac1{4a}$: $y'=2a(x-C_2)$, $y''=2a$. The condition $R=2N$ is equivalent to $1+y'^2=4ay$, and indeed

\[ 4ay = 4a\!\left[a(x-C_2)^2+\frac{1}{4a}\right] = 4a^2(x-C_2)^2+1 = 1+y'^2.\ \checkmark \]

Step 6: Verification, Case 2°

For the cycloid, $\dot x=\frac{a^2}{2}(1-\cos t)$, $\dot y=\frac{a^2}{2}\sin t$, so $y'=\dot y/\dot x=\cot(t/2)$ and $1+y'^2=\csc^2(t/2)$. Then

\[ y'' = \frac{d(y')/dt}{\dot x} = \frac{-\frac12\csc^2(t/2)}{a^2\sin^2(t/2)} = -\frac{1}{2a^2\sin^4(t/2)}. \]

\[ N = y\sqrt{1+y'^2} = \frac{a^2}{2}(1-\cos t)\csc\frac t2 = a^2\sin\frac t2, \qquad R = \frac{(1+y'^2)^{3/2}}{|y''|} = 2a^2\sin\frac t2. \] So $R=2a^2\sin(t/2)=2N$. $\checkmark$