0.1 Curvature
Before we get into curvature, we need to discuss some important topics and notation that will be used throughout the notes. As always, let’s get started with definition.
Definition 0.1.1
Given a vector-valued function \(\mathbf {r}\) of a smooth curve \(C\), the unit tangent vector to \(C\) is defined as the following. \[ \mathbf {T}(t) = \frac {\mathbf {r}'(t)}{\| \mathbf {r}'(t) \|} \]
As the name suggests, \(\| \mathbf {T}(T) \| = \frac {\| \mathbf {r}'(t) \|}{\| \mathbf {r}'(t) \|} = 1\). Also, it is worth noting that the unit tangent vector at a point of all smooth parametrizations of a curve are the same because the tangent vectors are normalized. Continuing, we can define another term.
Definition 0.1.2
Given a smooth parametrization \(\mathbf {r}\) of a curve and the unit tangent vector \(\mathbf {T}(t)\), the unit normal vector to the curve at \(t\) is defined as the following. \[ \mathbf {N}(t) = \frac {\mathbf {T}'(t)}{\| \mathbf {T}'(t) \|} \]
Notice that because \(\mathbf {T}'(t) \equiv 0\) if a curve is a straight line, \(\mathbf {N}(t)\) is not defined. Moreover, notice that because \(\mathbf {T}(t) \cdot \mathbf {T}(t) = 1\) and \(2\mathbf {T}'(t) \cdot \mathbf {T}(t) = 0\), \(\mathbf {N}(t)\) is orthogonal to \(\mathbf {T}(t)\) as the name suggests! However, there are different ways for \(\mathbf {N}\) to be orthogonal to \(\mathbf {T}\) geometrically. For instance, in 2-space, we have two directions and we have infinitely many possibilities in 3-space. To clarify the direction in which \(\mathbf {N}\) is orthogonal in 2-space, consider the following theorem.
Theorem 0.1.3
Given a smooth curve \(C\) in 2-space, the unit normal vector at a point on the curve is towards the concave side.
Proof.
Notice that to show that \(\mathbf {N}(t)\) point towards the concave side of the curve, it suffices to show that the curve bends towards the direction of \(\mathbf {T}'(t)\). Let \(\theta (t)\) be the angle of \(\mathbf {T}\) from the positive \(x\)-axis. Notice that because \(\mathbf {T}\) is a unit vector, \(\mathbf {T}\) can be written as \(\mathbf {T} = (\cos \theta (t), \sin \theta (t))\).
Continuing, \(\mathbf {T}'(t) = \theta '(t)(-\sin \theta (t), \cos \theta (t))\), which can be obtained by rotating \(\mathbf {T}\) by \(90^\circ \) and multiplying by \(\theta '(t)\). Note that if \(\theta '(t) > 0\), then it is pointing towards the left, or towards the concave side, of the curve. Similarly, if \(\theta '(t)\), then \(\mathbf {T}'(t)\) is pointing towards the left, or towards the concave side, of the curve. Because the normal vector is not defined when \(\theta '(t) = 0\), the theorem holds.
Before we get into curvature, here is one last definition.
Definition 0.1.4
Given a smooth curve in 3-space, the binormal vector to the curve at \(t\) is defined as \(\mathbf {B}(t) = \mathbf {T}(t) \times \mathbf {N}(t)\).
From the definition, we can infer that \(\mathbf {B}\) is orthogonal to both unit tangent vector and unit normal vector, and is also a unit vector. The coordinate system using \(\mathbf {N}\), \(\mathbf {T}\), and \(\mathbf {B}\) is known as Frenet frame or TNB-frame.
One more thing that we can notice from the definition is that \(\mathbf {B}(t)\) can be written as the following at \(t\) and \(s\) if the parametrization is \(\mathbf {r}(s)\). \begin{align*} \mathbf {B}(t) &= \frac {\mathbf {r}'(t) \times \mathbf {r}''(t)} {\| \mathbf {r}'(t) \times \mathbf {r}''(t) \|} \\ \mathbf {B}(s) &= \frac {\mathbf {r}'(s) \times \mathbf {r}''(s)} {\| \mathbf {r}'(s) \times \mathbf {r}''(s) \|} = \frac {\mathbf {r}'(s) \times \mathbf {r}''(s)} {\| \mathbf {r}''(s) \|} \end{align*}
Because the derivation is fairly straightforward with definition and \(\mathbf {r}' \times \mathbf {r}' = 0\), let’s skip the proof. Now let’s discuss about curvature!
Just for the sake of intuition, consider a circle and a square root function. From those two, the circle has larger curvature because it curves more, and we can represent that by the rate of changes in the direction of the tangent vector. Here is more rigorous definition.
Definition 0.1.5
Given an arc length parametrization of a curve, the curvature of the curve at \(s\) is defined as \(\kappa (s) = \| \mathbf {T}'(s) \|\).
In other words, curvature is the rate of change of the tangent vector with respect to its arc length. Notice that \(\kappa (s)\) is a function in terms of \(s\).
If the parametrization is not arc-length, then we write it as \(\kappa (t = t_0)\) for clarity and the following equation holds. \[ \kappa (t = t_0) = \frac {\| \mathbf {T_1}' (t_0) \|}{\| \mathbf {r_1}' (t_0) \|} \]
To see why this is true, consider the arc-length parametrization \(\mathbf {r_2}(s) = \mathbf {r_1}(g(s))\). Similarly, let \(\mathbf {T_1}\) and \(\mathbf {T_2}\) be the respective unit tangent vectors. Then, the following equations hold. \begin{align*} \kappa (t = t_0) &= \kappa (s = s_0) = \| \mathbf {T_2}'(s_0) \| = \| \mathbf {T_1}'(g(s_0)) \cdot g'(s_0) \| \\ &= \frac {\| \mathbf {T_1}'(g(s_0)) \cdot g'(s_0) \|} {\| \mathbf {r_2}'(s_0) \|} = \frac {\mathbf {T_1}'(g(s_0))}{\mathbf {r_1}'(g(s_0))} \end{align*}
Therefore, the equation holds.
Continuing, here is another interesting theorem. This is quite important because many times curvature is defined and used with this theorem, or formula, below.
Theorem 0.1.6
Let \(\mathbf {r}(t)\) be a smooth parametrization of a curve in 3-space. If \(\mathbf {r}''(t_0)\) is defined, then the following equation holds. \[ \kappa (t = t_0) = \frac { \| \mathbf {r}'(t_0) \times \mathbf {r}''(t_0) \| } {\| \mathbf {r}'(t_0) \|^3} \]
Proof.
Define \(\mathbf {r_1}(s) = \mathbf {r}(g(s))\) where \(g\) is an arc length change. Then, it suffices to show that \(\| \mathbf {r_1}''(s_0) \| = \frac { \| \mathbf {r}'(t_0) \times \mathbf {r}''(t_0) \| } {\| \mathbf {r}'(t_0) \|^3}\).
Notice that by construction, the following equations hold. \begin{align*} \mathbf {r_1}'(s) &= \mathbf {r}'(g(s)) g'(s) \\ \mathbf {r_1}''(s) &= \mathbf {r}''(g(s)) g'(s)^2 + \mathbf {r}'(g(s)) g''(s) \end{align*}
Therefore, the following equations hold. \begin{align*} \frac { \| \mathbf {r}'(t_0) \times \mathbf {r}''(t_0) \| } {\| \mathbf {r}'(t_0) \|^3} &= \frac { \| \mathbf {r}'(g(s_0)) \times \mathbf {r}''(g(s_0)) \| } {\| \mathbf {r}'(g(s_0)) \|^3} \\ &= \frac { \| \mathbf {r}'(g(s_0)) \times \mathbf {r}''(g(s_0)) \| | g'(s_0) |^3 }{\| \mathbf {r}'(g(s_0)) g'(s_0) \|^3} \\ &= \| \mathbf {r_1}'(s_0) \times \mathbf {r_1}''(s_0) \| \\ &= \| \mathbf {r_1}''(s_0) \| \end{align*}
Therefore, the equation holds.
Notice that if \(\mathbf {r}(t) = \langle x(t), y(t), 0 \rangle \), then this theorem also holds for 2-space.