So I know there is the method of using the parametric equation for the arc length of a curve which is derived geometrically, is there a similar geometric derivation for polar arc length.

Currently I attempt to do this by considering a segment of a curve and approximating the arc length as a straight line then using the cosine rule to calculate it's length. However this just cancels and produces L = R which is clearly wrong.


If you have a curve in polar coordinates given by $r=f(\theta)$ for $ a\leq\theta\leq b$, you can parameterize the curve as $x = f(\theta)\cos{\theta}$ and $y=f(\theta)\sin{\theta}$.

The differential arclength is: $ds=\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}d\theta$

If you substitute in for $x$ and $y$ this simplifies to: $ds=\sqrt{\left( f'(\theta)\right) ^2+\left( f(\theta)\right) ^2}d\theta$

  • $\begingroup$ Ok I understand this derivation but is there a more intuitive directly geometric method of finding this result? $\endgroup$ – Class May 9 '18 at 17:08

Let $C$ be a curve in the plane given by its polar representation $\rho(\theta)$, where $\theta\in[a,b]\subset(-\pi,\pi]$.

Then the position vector for any point along the curve is given by

$$\vec{r}(\theta) = \rho(\theta)\begin{pmatrix}\cos\theta \\ \sin\theta\end{pmatrix}$$

We approximate the arc length of the curve by breaking it up into $N$ segments of equal length, and approximating $\vec{r}(\theta)$ by lines along each of these segments.

Let $a=x_0<x_1<\text{ }...\text{ }<x_N = b$ be an equidistant partition of $[a,b]$ with $h_N$ being the distance between any two endpoints, that is $h_N =x_{k+1}-x_k= \frac{a-b}{N}$.

Then, the line segment best approximating the curve over an Interval $I_k:=[x_{k},x_{k+1})$ is given by

$$\vec{l_k}(t) = \vec{r}(x_k) + t\vec{r'}(x_k)$$

where $t\in[0,h_N]$. Its length is given by

$$l_k = \frac{b-a}{N}\left|\vec{r'}(x_k)\right|$$

Thus, in approximation, the arc length of the curve $C$ is given by the sum over all these lengths of line-segments $l_k$:

$$L_N =\sum_{k=0}^{N-1}l_k = \frac{b-a}{N}\sum_{k=0}^{N-1}\left|\vec{r'}(x_k)\right|$$

We now compute the tangent vector $\vec{r'}$:

\begin{align} \vec{r'} = \rho'(\theta)\begin{pmatrix}\cos\theta \\ \sin\theta\end{pmatrix} + \rho(\theta)\begin{pmatrix}-\sin\theta \\ \cos\theta\end{pmatrix} \end{align}

\begin{align} \left|\vec{r'}\right|&=\sqrt{(\rho'(\theta)\cos\theta - \rho(\theta)\sin\theta)^2+(\rho'(\theta)\sin\theta + \rho(\theta)\cos\theta)^2}\\ &=\sqrt{(\rho'(\theta))^2+(\rho(\theta))^2} \end{align}

Thus, we have:

$$L_N = \frac{b-a}{N}\sum_{k=0}^{N-1}\sqrt{(\rho'(x_k))^2+(\rho(x_k))^2}$$

This, however, is the left Riemann sum for the function $f(\theta) = \sqrt{(\rho'(\theta))^2+(\rho(\theta))^2}$ over the interval $[a,b]$. Taking the limit as $N\rightarrow\infty$, we obtain:

$$L := \lim\limits_{N\rightarrow\infty}L_N = \int_a^b\sqrt{(\rho'(\theta))^2+(\rho(\theta))^2}d\theta$$

which retrieves the arc length formula for polar curves.


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