Curvature at a Point on a Single Variable Real Valued Function

We first learned about curves in the form of a single-variable function $y = f(x)$ in $\mathbb{R}^2$, which takes every $x \in D(f)$ and maps it to an element $f(x) \in R(f) \subseteq C(f)$. We will now look at a theorem which will tell us the curvature at a point on a curve of this form, provided that the curve is twice differentiable.

Theorem 1: Let $y = f(x)$ be a real-valued twice-differentiable function that traces the curve $C$. Then the curvature at the point $x \in D(f)$ is given by the formula: $\kappa (x) = \frac{\mid f''(x) \mid}{(1 + (f'(x))^2)^{3/2}}$.

• Proof: Let $y = f(x)$ be a twice differentiable function. Then this function can rewritten as the set of parametric equations $\left\{\begin{matrix} x = x \\ y = f(x) \\ z = 0 \end{matrix}\right.$.

• Now recall from the Curvature at a Point on a Curve page that $\kappa (x) = \frac{ \| \vec{r'}(x) \times \vec{r''}(x) \|}{\| \vec{r'}(x) \|^3}$.

• So we have that $\vec{r'}(x) = (1, f'(x), 0)$ and $\vec{r''}(x) = (0, f''(x), 0)$. We now want to compute the cross product $\vec{r'}(x) \times \vec{r''}(x)$ as follows:

(1)

\begin{align} \quad \quad \vec{r'}(x) \times \vec{r''}(x) = [(f'(x))(0) - (0)(f''(x))]\vec{i} - [(1)(0) + (0)(0)] \vec{j} + [(1)(f''(x)) - 0(f'(x))] \vec{k} \\ \quad \quad \vec{r'}(x) \times \vec{r''}(x) = (0, 0, f''(x)) \end{align}

• Therefore $\| \vec{r'}(x) \times \vec{r''}(x) \| = \sqrt{(f''(x))^2} = \mid f''(x) \mid$. So we have computed the numerator to our above equation and so now we need to compute the denominator.

• We have that $\| \vec{r'}(x) \| = \sqrt{1 + (f'(x))^2}$, and so $\| \vec{r'}(x) \|^3 = (1 + (f'(x))^2)^{3/2}$. Putting this all together we see that:

(2)

\begin{align} \quad \kappa = \frac{ \| \vec{r'}(x) \times \vec{r''}(x) \|}{\| \vec{r'}(x) \|^3} = \frac{\mid f''(x) \mid}{(1 + (f'(x))^2)^{3/2}} \quad \blacksquare \end{align}

Let's look at some examples applying theorem 1.

Example 1

Let $f(x) = 4x^2 - 2x + 1$. Compute the curvature of $f$ at $x = 2$.

Let's first compute the first and second derivatives of $f$. $f'(x) = 8x - 2$, and $f''(x) = 8$. Therefore we have that:

(3)

\begin{align} \kappa (x) = \frac{\mid 8 \mid}{(1 + (8x - 2)^2)^{3/2}} \end{align}

Plugging in $x = 2$ and:

(4)

\begin{align} \kappa (2) = \frac{8}{(197)^{3/2}} \end{align}