Find the curves for which R = k r, where R is the radius of…

Find the curves for which $R = k\,r,$ where $R$ is the radius of curvature, $r=\sqrt{x^2+y^2}$ is the radius vector of the point, and $k>0$ is a constant.

Step 1: Pass to polar coordinates

With $x=r\cos\theta$, $y=r\sin\theta$, and $r=r(\theta)$, the radius of curvature in polar form is

\[ R = \frac{(r^2+r'^2)^{3/2}}{|r^2+2r'^2-r\,r''|}, \qquad r'=\frac{dr}{d\theta}, \] so the condition $R=kr$ becomes

\[ \frac{(r^2+r'^2)^{3/2}}{|r^2+2r'^2-r\,r''|} = k\,r. \tag{1} \]

Step 2: Simplify with $u=\ln r$

Let $u=\ln r$, so $r=e^u$, $r'=u'e^u$, $r''=(u''+u'^2)e^u$. Then $r^2+r'^2=e^{2u}(1+u'^2)$ and $r^2+2r'^2-rr''=e^{2u}(1+u'^2-u'')$, and (1) collapses to

\[ \frac{(1+u'^2)^{3/2}}{|1+u'^2-u''|} = k. \tag{2} \]

Step 3: Reduce the order — $p=u'=\tan\psi$

Equation (2) does not contain $u$ itself, so with $p=u'=du/d\theta$ it reads $u''=(1+p^2)\!\left[1\mp\dfrac{\sqrt{1+p^2}}{k}\right]$. Substituting $p=\tan\psi$ (so $1+p^2=\sec^2\psi$):

\[ \sec^2\psi\,\frac{d\psi}{d\theta} = \sec^2\psi\!\left(1\mp\frac{\sec\psi}{k}\right) \qquad\Longrightarrow\qquad \frac{d\psi}{d\theta} = 1\mp\frac{1}{k\cos\psi}. \tag{3} \]

This separates as

\[ \frac{k\cos\psi\,d\psi}{k\cos\psi\mp1} = d\theta. \tag{4} \]

Step 4: Integrate (4)

Adding and subtracting $1$ in the numerator splits the integrand: \[\left(1\pm\frac{1}{k\cos\psi\mp1}\right)d\psi = d\theta \qquad\Longrightarrow\qquad \psi\pm\int\frac{d\psi}{k\cos\psi\mp1} = \theta+C_1. \tag{5}\]

The remaining integral is evaluated with the Weierstrass substitution $t=\tan(\psi/2)$, $\cos\psi=\dfrac{1-t^2}{1+t^2}$, $d\psi=\dfrac{2\,dt}{1+t^2}$; for the $-$ branch it becomes $\displaystyle\int\frac{2\,dt}{(k-1)-(k+1)t^2}$, a rational integral whose closed form depends on the sign of $k-1$:

Step 5: Recover $r(\theta)$

Once $\psi(\theta)$ is known from (5), $u=\ln r$ follows from $du/d\theta=\tan\psi$: $\ln r = \int\tan\psi\,d\theta + C_2.$

Step 6: General solution

with the integral evaluated as in Step 4 according to whether $k>1$, $k=1$, or $k<1$, and $C_1$, $C_2$ arbitrary constants.

Step 7: Simplest particular solution

If $p=u'=0$ (i.e. $r'=0$), equation (1) gives $R=r^2/|r|=r$, so $k=1$ and $r = C$.

— the family of circles centred at the origin.

Step 8: Verification (particular solution)

For $r=C$ (constant), $r'=0$ and $r''=0$, so directly from the polar curvature formula in Step 1: \[R = \frac{(C^2+0)^{3/2}}{|C^2+0-0|} = \frac{C^3}{C^2} = C = r.\]

Hence $R=1\cdot r$, confirming $R=kr$ with $k=1$. $\checkmark$