Find the curves for which Rcosvarphi = k (k>0 constant)…

Find the curves for which $R\cos\varphi = k \qquad (k>0\ \text{constant}),$ where $R$ is the radius of curvature and $\varphi$ is the angle the tangent makes with the $x$-axis ($\tan\varphi=y'$, $\cos\varphi=1/\sqrt{1+y'^2}$) — i.e. $R$ is inversely proportional to $\cos\varphi$.

Step 1: Setting up the equation

Since $R=(1+y'^2)^{3/2}/|y''|$ and $\cos\varphi=1/\sqrt{1+y'^2}$, the condition $R\cos\varphi=k$ reduces to $(1+y'^2)/|y''|=k$, i.e.

\[ y'' = \pm\frac{1+y'^2}{k}. \tag{1} \]

Step 2: Reduce the order — $p=y'$

Equation (1) has no $y$, so with $p=y'$, $y''=p'$, it separates directly: \[\frac{dp}{1+p^2} = \pm\frac{dx}{k} \qquad\Longrightarrow\qquad \arctan p = \pm\frac{x}{k}+C_1.\]

Step 3: Second integration

Take the $+$ sign (the convex branch, $y''>0$): $y'=\tan\!\left(\dfrac{x}{k}+C_1\right)$. Integrating with $\int\tan u\,du=-\ln|\cos u|$: \[y = -k\ln\left|\cos\!\left(\frac{x}{k}+C_1\right)\right| + C_2. \tag{2}\]

Step 4: General solution

where $C_1'=-kC_1$ and $C_2$ are arbitrary constants. This curve is known as the catenary of equal resistance (or lintearia): the curve of uniform strength for a hanging cable of variable cross-section.

Step 5: Verification

Let $\alpha=x/k+C_1$, so $y=-k\ln|\cos\alpha|+C_2$. Then

\[ y' = \tan\alpha, \qquad y'' = \frac{\sec^2\alpha}{k}. \]

\[ R = \frac{(1+\tan^2\alpha)^{3/2}}{\sec^2\alpha/k} = k\sec\alpha, \qquad \cos\varphi = \frac{1}{\sqrt{1+\tan^2\alpha}} = \cos\alpha. \]

Hence $R\cos\varphi = k\sec\alpha\cos\alpha = k$, confirming $R$ is inversely proportional to $\cos\varphi$. $\checkmark$