Curvature at a Point on a Curve

We are now going to look at a very important property of a curve $C$ defined by a space curve known as the curvature at a point on the curve. In essence, the curvature at a point on $C$ is defined to be the rate at which the unit tangent vector $\hat{T}$ changes direction.

Of course, this is difficult to define since $\vec{r}(t) = (x(t), y(t), z(t))$ with the parameterization $x = x(t)$, $y = y(t)$ and $z = z(t)$ may trace the curve $C$ "faster" or "slower" depending on what parameterization we use. For example, a car that turns at a sharp corner slowly and a car that turns at a sharp corner quickly will have different rates of change for their respective unit tangent vectors. For that reason, we will measure the curvature at a point as the rate of change of the unit tangent vector with respect to its arc length. Recall that the arc length of $C$ is independent of its parameterization, so then the curvature of a point on $C$ defined by arc length will be independent of its parameterization as well.

Definition: Let $\vec{r}(s) = (x(s), y(s), z(s))$ be a vector-valued function parameterized in terms of the arc length of a smooth curve $C$ that is traced out by $\vec{r}$. Then the Curvature of $C$ at the point $\vec{r}(s)$ denoted $\kappa (s) = \biggr \| \frac{d\hat{T}(s)}{ds} \biggr \| = \biggr \| \frac{d\hat{T}(s) / dt}{ds / dt} \biggr \|$ is the length/magnitude of derivative of $\frac{d\hat{T}}{ds}$.

Often times it can be a nuissance to determine the curvature at a point on a curve with the parameter $s$, and so for that reason, we will look at two other formulas for computing the curvature at a point without parameterization before we look at any examples.

Theorem 1: Let $\vec{r}(t) = (x(t), y(t), z(t))$ be a vector-valued function that traces the smooth curve $C$. Then the curvature of $C$ is also given by the following formula: $\kappa (t) = \frac{\| \hat{T'}(t) \|}{\| \vec{r'}(t) \|}$.

• Proof: By the definition of curvature, we have that $\kappa = \biggr \| \frac{d\hat{T} / dt}{ds / dt} \biggr \|$. Now we note that $s(t) = \int_a^t \| \vec{r'}(u) \| \: du$, and so by the Fundamental Theorem of Calculus, $\frac{ds}{dt} = \| \vec{r'}(t) \|$. Substituting this into the curvature equation and we get that:

(1)

\begin{align} \kappa (t) = \biggr \| \frac{d\hat{T} / dt}{ds/dt} \biggr \| = \frac{\| \hat{T'}(t) \|}{\| \vec{r'}(t) \|} \quad \blacksquare \end{align}

Theorem 2: Let $\vec{r}(t) = (x(t), y(t), z(t))$ be a vector-valued function that traces the smooth curve $C$. Then the curvature of $C$ at the point $\vec{r}(t)$ is also given by the formula: $\kappa (t) = \frac{\| \vec{r'}(t) \times \vec{r''}(t) \|}{\| \vec{r'}(t) \|^3}$.

• Proof: To prove theorem 2, we will first find $\vec{r'}(t)$ and $\vec{r''}(t)$ and then compute their cross product.

• We note that the unit tangent vector $\hat{T}(t)$ can be rewritten as $\hat{T}(t) = \frac{\vec{r'}(t)}{\| \vec{r'}(t) \|}$, and so rearranging this formula we have that $\vec{r'}(t) = \hat{T} (t) \| \vec{r'}(t) \|$. Furthermore since $s(t) = \int_a^t \| \vec{r'}(u) \| \: du$, then by the Fundamental Theorem of Calculus, $\frac{ds}{dt} = \| \vec{r'}(t) \|$, so substituting this into the earlier formula we get:

(2)

\begin{align} \vec{r'}(t) = \hat{T} (t) \| \vec{r'}(t) \| = \hat{T} (t) \frac{ds}{dt} \end{align}

• So we have a formula for $\vec{r'}(t)$. Now we need a formula for $\vec{r''}(t)$. If we differentiate both sides of this equation by applying the product rule, then:

(3)

\begin{align} \vec{r''}(t) = \hat{T}(t) \frac{d^2s}{dt^2} + \hat{T'}(t) \frac{ds}{dt} \end{align}

• So now we have a formula for $\vec{r''}(t)$. Taking the cross product $\vec{r'}(t) \times \vec{r''}(t)$ we get that:

(4)

\begin{align} \quad \quad \vec{r'}(t) \times \vec{r''}(t) = \left (\hat{T}(t) \frac{ds}{dt} \right ) \times \left (\hat{T}(t) \frac{d^2s}{dt^2} + \hat{T'}(t) \frac{ds}{dt} \right ) \\ \quad \quad \vec{r'}(t) \times \vec{r''}(t) = \left [ \hat{T}(t) \frac{ds}{dt} \right ] \times \left [ \hat{T}(t) \frac{d^2s}{dt^2} \right ] + \left [ \hat{T}(t) \frac{ds}{dt} \right ] \times \left [ \hat{T'}(t) \frac{ds}{dt} \right ] \\ \quad \quad \vec{r'}(t) \times \vec{r''}(t) = \left ( \frac{ds}{dt} \right ) \left ( \frac{d^2s}{dt^2} \right) [ \hat{T}(t) \times \hat{T}(t) ] + \left ( \frac{ds}{dt} \right)^2 [ \hat{T}(t) \times \hat{T'}(t) ] \end{align}

• Recall that for any vector $\vec{u} \in \mathbb{R}^3$ that $\vec{u} \times \vec{u} = 0$, and so $\hat{T}(t) \times \hat{T}(t) = 0$:

(5)

\begin{align} \quad \quad \vec{r'}(t) \times \vec{r''}(t) = \left ( \frac{ds}{dt} \right)^2 [ \hat{T}(t) \times \hat{T'}(t) ] \end{align}

• Now we take the norm of both sides to get:

(6)

\begin{align} \quad \quad \| \vec{r'}(t) \times \vec{r''}(t) \| = \left ( \frac{ds}{dt} \right)^2 \| \hat{T}(t) \times \hat{T'}(t) \| \end{align}

• Now recall that for any two vectors $\vec{u}, \vec{v} \in \mathbb{R}^3$ that $\| \vec{u} \times \vec{v} \| = \| \vec{u} \| \| \vec{v} \| \sin \theta$ where $\theta$ is the angle between $\vec{u}$ and $\vec{v}$. So $\| \hat{T}(t) \times \hat{T'}(t) \| = \| \hat{T}(t) \| \| \hat{T'}(t) \| \sin \theta$ where $\theta$ is the angle between $\hat{T}(t)$ and $\hat{T'}(t)$. But notice that $\| \hat{T}(t) \| = 1$ is a constant (since $\| \hat{T}(t) \|$ is the unit tangent vector), and so by a theorem on the Derivative Rules for Vector-Valued Functions page, this implies that $\hat{T}(t) \perp \hat{T'}(t)$ and so $\theta = \frac{\pi}{2}$ and $\sin \theta = 1$, therefore:

(7)

\begin{align} \quad \quad \| \vec{r'}(t) \times \vec{r''}(t) \| = \left ( \frac{ds}{dt} \right)^2 \| \hat{T}(t) \| \| \hat{T'}(t) \| \sin \theta \\ \quad \quad \| \vec{r'}(t) \times \vec{r''}(t) \| = \left ( \frac{ds}{dt} \right)^2 \| \hat{T'}(t) \| \\ \end{align}

• Rearranging this equation and replacing $\left ( \frac{ds}{dt} \right )^2 = \| \vec{r'}(t) \| ^2$ we get that $\| \hat{T'}(t) \| = \frac{\| \vec{r'}(t) \times \vec{r''}(t) \|}{\| \vec{r'}(t) \|^2}$. Then divide both sides by $\| \vec{r'}(t) \|$:

(8)

\begin{align} \quad \kappa (t)= \frac{\| \hat{T'}(t) \|}{\| \vec{r'}(t) \|} = \frac{ \| \vec{r'}(t) \times \vec{r''}(t) \|}{\| \vec{r'}(t) \|^3} \quad \blacksquare \end{align}