The curvature of a plane curve with polar coordinate

Question

let a plane curve with polar coordinates ( $\theta$ , $\rho (\theta )$) and let $k(\theta)$ be it's curvature one can preuve that

$k(\theta) = \frac{2({\rho }')^{2}-\rho {\rho }''+ \rho ^{2}}{({\rho }'^{2}+\rho ^{2})^{\frac{3}{2}}} $

we can do this with some calculation ,but i need method with less calculation possible.

thank you for help .

Answer 1

The equation for curvature is somewhat simpler in complex variables. Namely, if $z=r(\theta)e^{i\theta}$, then

$$\kappa(\theta)=\frac{\mathfrak{Im}\{\dot z^*\ddot z\}}{|\dot z|^3}$$

where $\dot z=dz/d\theta~$ and $()^*$ is the conjugate. Of course, in the final analysis both must be the same, but if you're doing this in a program, then this is much simpler.

Answer 2

@RobertLewis has requested a derivation of the expression for curvature that I gave in my other answer. This is sufficiently more involved to warrant a separate answer.

There are several things to understand before we can do the derivation proper:

1. The argument of a complex number. Consider that

$$z=|z|e^{i\theta},\quad z^*=|z|e^{-i\theta}$$

then

$$\frac{z}{z^*}=e^{2i\theta}\quad \text{or}\quad e^{i\theta}=\sqrt{\frac{z}{z^*}}\quad \text{or}\quad \theta=\frac{1}{2i}\ln\frac{z}{z^*}$$

You can verify this for yourself. Start with $\ln z=\ln |z|+i\theta$.

2. The curvature, $\kappa$ is defined in terms of the tangent angle, $\tau$ and the arc length, $s$ by Euler's integral solution for plane curves,

$$\tau =\int{\kappa \left( s \right)\,ds;\,\,\,\,\,\,\,\,{d\tau }/{ds=\kappa \left( s \right)}\;}\\ \kappa(\theta)=\frac{d\tau}{d\theta}\frac{d\theta}{ds}$$

where $\theta$ is the polar angle. We also know that for any parametric representation, the arc length is given by

$$ s=\int |\dot z|du\\ \frac{ds}{du}=|\dot z|=\sqrt{\dot z\dot z^*} $$

3. We can now pull this all together to derive the curvature in terms of the polar angle. First, however, we must realize that the tangent angle is equal to the argument of the derivative of the curve, i.e.,

$$ \begin{align} \tau(\theta) &=\frac{d\tau}{d\theta}\frac{d\theta}{ds}\\ &=\frac{1}{2i}\ln\frac{\dot z}{\dot z^*} \end{align} $$

Then

$$ \begin{align} \tau(\theta) &=\arg \dot z\\ \kappa(\theta) &=\frac{1}{2i}\frac{d}{d\theta}\left(\ln\frac{\dot z}{\dot z^*}\right)\frac{1}{\sqrt{\dot z\dot z^*}}\\ &=\frac{1}{2i}\frac{d}{d\theta}(\ln\dot z-\ln\dot z^*)\frac{1}{\sqrt{\dot z\dot z^*}}\\ &=\frac{1}{2i}\left(\frac{\ddot z}{\dot z}-\frac{\ddot z^*}{\dot z^*}\right)\frac{1}{\sqrt{\dot z\dot z^*}}\\ &=\frac{1}{2i}\frac{\dot z^*\ddot z-\dot z\ddot z^*}{(\dot z\dot z^*)^{3/2}}\\ &=\frac{1}{2i}\frac{\dot z^*\ddot z-(\dot z^*\ddot z)^*}{(\dot z\dot z^*)^{3/2}}\\ &=\frac{1}{2i}\frac{2i\mathfrak{Im}\{\dot z^*\ddot z\}}{(\dot z\dot z^*)^{3/2}}\quad \text{since}\quad Z-Z^*=2i\mathfrak{Im}\{Z\}\\ &=\frac{\mathfrak{Im}\{\dot z^*\ddot z\}}{|\dot z|^3} \end{align} $$

This completes the derivation.

Answer 3

The trick is not to be scared of making long calculations.

You may have encountered the following proposition

The curvature of a plane curve $\alpha(t)=(x(t),y(t))$ is given by the $$\kappa(t)=\dfrac{x'y''-x''y'}{((x')^2+(y')^2)^{3/2}}$$

Consider the polar curve $\rho = \rho(\theta)$ and your parametrization in cartesian coordinates $\alpha(\theta)=\rho(\theta)\left(\cos\theta,\sin\theta\right)$ from that we get \begin{align*} \alpha'(\theta)&=(x'(\theta),y'(\theta))=\rho'(\cos\theta,\sin\theta)+\rho(-\sin\theta,\cos\theta)\\ \alpha''(\theta)&=(x''(\theta),y''(\theta))=(\rho''-\rho)(\cos\theta,\sin\theta)+2\rho'(-\sin\theta,\cos\theta) \end{align*} In this way,

\begin{align*} x'(\theta)y''(\theta)&=(\rho'\cos\theta-\rho\sin\theta)[(\rho''-\rho)\sin\theta+2\rho'\cos\theta]\\ &=2(\rho')^2\cos^2\theta+[\rho'(\rho''-\rho)-2\rho\rho']\sin\theta\cos\theta-\rho(\rho''-\rho)\sin^2\theta \end{align*}

and

\begin{align*} x''(\theta)y'(\theta)&=[(\rho''-\rho)\cos\theta-2\rho'\sin\theta](\rho'\sin\theta+\rho\cos\theta)\\ &=\rho(\rho''-\rho)\cos^2\theta+[\rho'(\rho''-\rho)-2\rho\rho']\sin\theta\cos\theta-2(\rho')^2\sin^2\theta \end{align*}

Using the proposition stated in the beginning the curvature of $\rho(\theta)$ is given by

\begin{align*} \kappa(\theta)&=\dfrac{x'(\theta)y''(\theta)-x''(\theta)y'(\theta)}{((x')^2+(y')^2)^{3/2}}=\dfrac{2(\rho')^2-\rho(\rho''-\rho)}{\left((\rho')^2+\rho^2\right)^{3/2}}=\dfrac{2(\rho')^2-\rho\rho''+\rho^2}{\left(\rho^2+(\rho')^2\right)^{3/2}} \end{align*}