The Curvature of Plane Polar Curves

The following theorem will give us a method of obtaining the curvature of a plane polar curve $r = f(\theta)$ at a point $(r, \theta)$.

Theorem 1: Suppose that $r = f(\theta)$ is a plane polar curve. Then the curvature at $(r, \theta)$ is given by the formula $\kappa (\theta) = \frac{\mid 2 (f'(\theta))^2 + (f(\theta))^2 - f(\theta)f''(\theta) \mid}{\left [(f'(\theta))^2 + (f(\theta))^2 \right ]^{3/2}}$.

• Proof: Let $r = f(\theta)$ be a plane polar curve. This curve can be parameterized as $\vec{r}(\theta) = (r \cos \theta, r \sin \theta, 0) = (f(\theta) \cos \theta, f(\theta) \sin \theta, 0)$. To prove Theorem 1, we will compute the necessary components and plug them into the formula for curvature.

• We first compute $\vec{r'}(\theta)$ as:

(1)

\begin{align} \quad \vec{r'}(\theta) = (-f(\theta) \sin \theta + f'(\theta) \cos \theta, f(\theta)\cos \theta + f'(\theta)\sin \theta, 0 ) \end{align}

• We then compute $\vec{r''}(\theta)$ as:

(2)

\begin{align} \quad \vec{r''}(\theta) = (-f(\theta)\cos(\theta) -f'(\theta)\sin \theta - f'(\theta) \sin \theta + f''(\theta)\cos \theta, -f(\theta)\sin \theta + f'(\theta)\cos \theta + f'(\theta)\cos (\theta) + f''(\theta) \sin \theta, 0) \\ \quad \vec{r''}(\theta) = (f''(\theta) \cos \theta - 2f'(\theta)\sin \theta - f(\theta) \cos \theta, f''(\theta)\sin \theta + 2f'(\theta) \cos \theta - f(\theta) \sin \theta, 0) \end{align}

• We now compute the cross product $\vec{r'}(\theta) \times \vec{r''}(\theta)$ as follows:

(3)

\begin{align} \quad \vec{r'}(\theta) \times \vec{r''}(\theta) = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -f(\theta)\sin \theta + f'(\theta)\cos \theta & f(\theta) \cos \theta + f'(\theta) \sin \theta & 0\\ f''(\theta) \cos \theta - 2f'(\theta)\sin \theta - f(\theta) \cos \theta & f''(\theta)\sin \theta + 2f'(\theta) \cos \theta - f(\theta) \sin \theta & 0 \end{vmatrix} = (0, 0, 2(f'(\theta))^2 + (f(\theta))^2 - f(\theta)f''(\theta)) \end{align}

• Therefore we have that $\| \vec{r'}(\theta) \times \vec{r''}(\theta) \|$ is:

(4)

\begin{align} \quad \| \vec{r'}(\theta) \times \vec{r''}(\theta) \| = \sqrt{ \left [ 2(f'(\theta))^2 + (f(\theta))^2 - f(\theta)f''(\theta) \right ]^2} = \mid 2(f'(\theta))^2 + (f(\theta))^2 - f(\theta)f''(\theta) \mid \end{align}

• Lastly we compute $\| \vec{r'}(\theta) \|$ as:

(5)

\begin{align} \quad \| \vec{r'}(\theta) \| = \sqrt{(f'(\theta))^2 + (f(\theta))^2} = \left [ (f'(\theta))^2 + (f(\theta))^2 \right]^{1/2} \end{align}

• Now we will plug $\| \vec{r'}(\theta) \times \vec{r''}(\theta) \|$ and $\| \vec{r'} (\theta) \|$ into the formula for curvature, $\kappa (\theta) = \frac{\| \vec{r'}(\theta) \times \vec{r''}(\theta) \|}{\| \vec{r'}(\theta) \|^{3}}$ to get that:

(6)

\begin{align} \quad \kappa (\theta) = \frac{\mid 2(f'(\theta))^2 + (f(\theta))^2 - f(\theta)f''(\theta) \mid}{\left [ (f'(\theta))^2 + (f(\theta))^2 \right]^{3/2}} \quad \blacksquare \end{align}

Let's look at an example of applying Theorem 1.

Example 1

Find the curvature of $f(\theta) = \sin \theta$ at any point on the curve.

We should first note that $f(\theta) = \sin \theta$ represents a circle with radius $0.5$ centered at $(0, 0.5)$. We thus should get that $\kappa (\theta) = 2$ for all $\theta$.

We note that $f'(\theta) = \cos \theta$ and $f''(\theta) = -\sin \theta$. Applying the formula in Theorem 1 and we get that:

(7)

\begin{align} \kappa (\theta) = \frac{\mid 2(f'(\theta))^2 + (f(\theta))^2 - f(\theta)f''(\theta) \mid}{\left [ (f'(\theta))^2 + (f(\theta))^2 \right]^{3/2}} = \frac{\mid 2(\cos \theta)^2 + (\sin \theta)^2 + \sin \theta \sin \theta \mid}{\left [ (\cos \theta)^2 + (\sin \theta)^2 \right]^{3/2}} = \frac{2}{1} = 2 \end{align}

Hence the curvature $\kappa (\theta) = 2$ as expected.