Orientation of Parametric Curves

Differentiation

In the case of vector-valued functions of the form $\gamma :\mathcal{D}\subseteq \mathbb{R}\to {\mathbb{R}}^n$, the definition of differentiability reduces to the following.

Definition: Differentiability of Parametric Curves

Let $\gamma :\mathcal{D}\subseteq \mathbb{R}\to {\mathbb{R}}^n$ be a vector-valued function on an open subset and let $t_0\in \mathcal{D}$.

We say that $\gamma$ is differentiable at $t_0$ if and only if the following limit exists:

$$\underset{t\to t_0}{\lim }\frac{\gamma (t)-\gamma (t_0)}{t-t_0}$$

If this limit exists, then its value is the Jacobian matrix of $\gamma$ and so we call it the derivative of $\gamma$ at $t_0$.

NOTATION

$$\frac{\mathrm{d}\gamma }{\mathrm{d}t}(t_0){\frac{\mathrm{d}\gamma }{\mathrm{d}t}|}_{t_0}\frac{\mathrm{d}}{\mathrm{d}t}\gamma (t_0){\gamma }^{\prime }(t_0)\dot{\gamma }(t_0)$$

We say that $\gamma$ is $k$-times (continuously) differentiable at $t_0$ if its $k$-order derivative exists.

PROOF

We need to prove two things:

• (I) If $\underset{t\to t_0}{\lim }\frac{\gamma (t)-\gamma (t_0)}{t-t_0}$ exists, then there is a linear transformation $T:I\to {\mathbb{R}}^n$ such that

$$\underset{t\to t_0}{\lim }\frac{||\gamma (t)-\gamma (t_0)-T(t-t_0)||}{|t-t_0|}=0$$

• (II) If there is a linear transformation $T:I\to {\mathbb{R}}^n$ such that $\underset{t\to t_0}{\lim }\frac{||\gamma (t)-\gamma (t_0)-T(t-t_0)||}{||t-t_0||}=0$, then $\underset{t\to t_0}{\lim }\frac{\gamma (t)-\gamma (t_0)}{t-t_0}$ exists.

Proof of (I):

Let $\mathbf{L}={\lim }_{t\to t_0}\frac{\gamma (t)-\gamma (t_0)}{t-t_0}$. Define $T:I\to {\mathbb{R}}^n$ as

$$T(x)=x\mathbf{L}$$

We can easily check that $T$ is a linear transformation:

$$T(\alpha x+\beta y)=(\alpha x+\beta y)\mathbf{L}=\alpha x\mathbf{L}+\beta y\mathbf{L}=\alpha T(x)+\beta T(y)$$

Then

$$\begin{array}{rl}\underset{t\to t_0}{\lim }\frac{||\gamma (t)-\gamma (t_0)-T(t-t_0)||}{||t-t_0||}& =\underset{t\to t_0}{\lim }\frac{||\gamma (t)-\gamma (t_0)-(t-t_0)\mathbf{L}||}{|t-t_0|}\\ & =\underset{t\to t_0}{\lim }\frac{|t-t_0|\cdot \left|\left|\frac{\gamma (t)-\gamma (t_0)}{t-t_0}-\mathbf{L}\right|\right|}{|t-t_0|}\\ & =\underset{t\to t_0}{\lim }\left|\left|\frac{\gamma (t)-\gamma (t_0)}{t-t_0}-\mathbf{L}\right|\right|\\ & =\left|\left|\underset{t\to t_0}{\lim }\left(\frac{\gamma (t)-\gamma (t_0)}{t-t_0}-\mathbf{L}\right)\right|\right|\\ & =\left|\left|\underset{t\to t_0}{\lim }\left(\frac{\gamma (t)-\gamma (t_0)}{t-t_0}-\frac{\gamma (t)-\gamma (t_0)}{t-t_0}\right)\right|\right|\\ & =||0||=0\end{array}$$

Proof of (II):

Let $\mathbf{T}$ be the matrix representation of $T$ with respect to the standard bases of $\mathbb{R}$ and ${\mathbb{R}}^n$. Then,

$$\underset{t\to t_0}{\lim }\frac{||\gamma (t)-\gamma (t_0)-\mathbf{T}\left[\begin{array}{c}t-t_0\end{array}\right]||}{|t-t_0|}=\underset{t\to t_0}{\lim }\frac{||\gamma (t)-\gamma (t_0)-(t-t_0)\mathbf{T}||}{|t-t_0|}=0$$

Factor out $t-t_0$ from the numerator and move it outside the magnitude.

$$\begin{array}{rl}\underset{t\to t_0}{\lim }\frac{||\gamma (t)-\gamma (t_0)-(t-t_0)\mathbf{T}||}{|t-t_0|}& =\underset{t-t_0}{\lim }\frac{|t-t_0|\cdot \left|\left|\frac{\gamma (t)-\gamma (t_0)}{t-t_0}-\mathbf{T}\right|\right|}{|t-t_0|}\\ & =\underset{t\to t_0}{\lim }\left|\left|\frac{\gamma (t)-\gamma (t_0)}{t-t_0}-\mathbf{T}\right|\right|\\ & =\left|\left|\underset{t\to t_0}{\lim }\left(\frac{\gamma (t)-\gamma (t_0)}{t-t_0}-\mathbf{T}\right)\right|\right|=0\end{array}$$

Therefore,

$$\begin{array}{rl}& \underset{t\to t_0}{\lim }\left(\frac{\gamma (t)-\gamma (t_0)}{t-t_0}-\mathbf{T}\right)=0\\ & \underset{t\to t_0}{\lim }\left(\frac{\gamma (t)-\gamma (t_0)}{t-t_0}\right)-\mathbf{T}=0\\ & \underset{t\to t_0}{\lim }\frac{\gamma (t)-\gamma (t_0)}{t-t_0}=\mathbf{T}\end{array}$$

Definition: Length

Let $\gamma :\mathcal{D}\subseteq \mathbb{R}\to \mathbb{R}$ be a differentiable vector-valued function.

The length of $\gamma$ is the integral of the magnitude of its derivative, if it exists:

$${\int }_{\mathcal{D}}||\dot{\gamma }||$$

Definition: Regularity

A vector-valued function $\gamma :\mathcal{D}\subseteq \mathbb{R}\to {\mathbb{R}}^n$ is regular if it is differentiable on $\mathcal{D}$ and its derivative is never zero.

Theorem: Linearity of Differentiation

If $\mathbf{u}$ and $\mathbf{v}$ are differentiable at some $t$, then for all $\mu ,\lambda \in \mathbb{R}$:

$$(\lambda \mathbf{u}(t)+\mu \mathbf{v}(t){)}^{\prime }=\lambda {\mathbf{u}}^{\prime }(t)+\mu {\mathbf{v}}^{\prime }(t)$$

PROOF

TODO

Theorem: Chain Rule

Let $f:{\mathcal{D}}_f\subseteq \mathbb{R}\to \mathbb{R}$ be a real function and let $\mathbf{r}:{\mathcal{D}}_{\mathbf{r}}\subseteq \mathbb{R}\to {\mathbb{R}}^n$ be vector-valued.

If $f$ is differentiable at some $t\in {\mathcal{D}}_f$ and $\mathbf{r}$ is differentiable at $f(t)$, then their composition is also differentiable at $t$ with

$$(\mathbf{r}\circ f)(t{)}^{\prime }=f^{\prime }(t){\mathbf{r}}^{\prime }(f(t))$$

PROOF

TODO

Theorem: Product Rule

Let $f:{\mathcal{D}}_f\subseteq \mathbb{R}\to \mathbb{R}$ be a real function and let $\mathbf{r}:{\mathcal{D}}_{\mathbf{r}}\subseteq \mathbb{R}\to {\mathbb{R}}^n$ be vector-valued.

If $f$ is differentiable at some $t\in {\mathcal{D}}_f$ and $\mathbf{r}$ is differentiable at $f(t)$, then

$$(f(t)\mathbf{r}(t){)}^{\prime }=f^{\prime }(t)\mathbf{r}(t)+f(t){\mathbf{r}}^{\prime }(t)$$

PROOF

TODO

Theorem: Dot Product Rule

Let $\mathbf{u}:{\mathcal{D}}_{\mathbf{u}}\subseteq \mathbb{R}\to {\mathbb{R}}^n$ and $\mathbf{v}:{\mathcal{D}}_{\mathbf{v}}\subseteq \mathbb{R}\to {\mathbb{R}}^n$ be vector-valued.

If $\mathbf{u}$ and $\mathbf{v}$ are both differentiable at some $t\in {\mathcal{D}}_{\mathbf{u}}\cap {\mathcal{D}}_{\mathbf{v}}$, then their dot product is also differentiable at $t$ with

$$(\mathbf{u}(t)\cdot \mathbf{v}(t){)}^{\prime }={\mathbf{u}}^{\prime }(t)\cdot \mathbf{v}(t)+\mathbf{u}(t)\cdot {\mathbf{v}}^{\prime }(t)$$

PROOF

TODO

Theorem: Cross Product Rule

Let $\mathbf{u}:{\mathcal{D}}_{\mathbf{u}}\subseteq \mathbb{R}\to {\mathbb{R}}^3$ and $\mathbf{v}:{\mathcal{D}}_{\mathbf{v}}\subseteq \mathbb{R}\to {\mathbb{R}}^3$ be vector-valued.

If $\mathbf{u}$ and $\mathbf{v}$ are both differentiable at some $t\in {\mathcal{D}}_{\mathbf{u}}\cap {\mathcal{D}}_{\mathbf{v}}$, then their cross product is also differentiable at $t$ with

$$(\mathbf{u}(t)\times \mathbf{v}(t){)}^{\prime }={\mathbf{u}}^{\prime }(t)\times \mathbf{v}(t)+\mathbf{u}(t)\times {\mathbf{v}}^{\prime }(t)$$

PROOF

TODO

Frenet–Serret Basis

Definition: Tangent Vector (Velocity)

Let $\gamma :\mathcal{D}\subseteq \mathbb{R}\to {\mathbb{R}}^n$ be a vector-valued function on an open subset and let $t_0\in \mathcal{D}$.

If $\gamma$ is differentiable at $t_0$, then the tangent vector or velocity of $\gamma$ at $t_0$ is its derivative there.

We also talk of the tangent vector as a function which to each $t\in \mathcal{D}$ assigns the value of $\gamma$‘s derivative, if it exists.

Definition: Speed

The Euclidean norm of $\gamma$‘s velocity is known as $\gamma$‘s speed.

Definition: Unit Tangent Vector

The normalization of $\gamma$‘s tangent vector is known as $\gamma$‘s unit tangent vector:

$$\frac{1}{||\dot{\gamma }(t)||}\dot{\gamma }(t)$$

NOTATION

$$\mathbf{T}(t)$$

Definition: Normal Vector

Let $\gamma :\mathcal{D}\subseteq \mathbb{R}\to {\mathbb{R}}^n$ be a vector-valued function on an open subset and let $t_0\in \mathcal{D}$.

If $\gamma$ and its tangent vector are both differentiable at $t_0$, then the derivative of the latter is known as the normal vector of $\gamma$ at $t_0$.

Definition: Unit Normal Vector

The normalization of $\gamma$‘s normal vector is known as $\gamma$‘s unit tangent vector:

$$\frac{1}{||\dot{\mathbf{T}}(t)||}\dot{\mathbf{T}}(t)$$

NOTATION

$$\mathbf{N}(t)$$

Definition: Binormal Vector

Let $\gamma :\mathcal{D}\subseteq \mathbb{R}\to {\mathbb{R}}^n$ be a vector-valued function on an open subset and let $t\in \mathcal{D}$.

If $\gamma$ is twice differentiable at $t$, then the binormal vector of $\gamma$ at $t$ is the cross product of its unit tangent vector and its unit normal vector:

$$\mathbf{T}(t)\times \mathbf{N}(t)$$

Notation

$$\mathbf{B}(t)$$

Orientation

Theorem: Colinearity of Tangent Vectors

Let $\gamma :{\mathcal{D}}_{\gamma }\subseteq \mathbb{R}\to {\mathbb{R}}^n$ and $\phi :{\mathcal{D}}_{\phi }\subseteq \mathbb{R}\to {\mathbb{R}}^n$ be differentiable vector-valued functions with the same image $\mathcal{C}\subseteq {\mathbb{R}}^n$.

If $\gamma$ and $\phi$ are equivalent up to a regular reparametrization, then for each $t\in {\mathcal{D}}_{\gamma }$ and each $s\in {\mathcal{D}}_{\phi }$ with $\gamma (t)=\phi (s)$ there exists some $\lambda \in \mathbb{R}$ such that

$${\gamma }^{\prime }(t)=\lambda {\phi }^{\prime }(s)$$

PROOF

Let’s call this reparametrization $h$, i.e $t=h(s)$ and so

$$\phi (s)=\gamma (h(s))$$

Differentiate both sides and apply the chain rule:

$${\phi }^{\prime }(s)=h^{\prime }(s){\gamma }^{\prime }(h(s))$$

Since $h$ is regular, we know that $h^{\prime }(s)\ne 0$ and so

$${\gamma }^{\prime }(h(s))=\frac{1}{h^{\prime }(s)}{\phi }^{\prime }(s)$$

Let $\lambda =\frac{1}{h^{\prime }(s)}$ and substitute back $t=h(s)$ and the proof is complete:

$${\gamma }^{\prime }(t)=\lambda {\phi }^{\prime }(s)$$

Important: Unit Tangent Vectors

It immediately follows that the unit tangent vectors of $\gamma$ and $\phi$ are either always equal or always opposite.

Definition: Orientation of Parametrizations

We say that $\gamma$ and $\phi$ have:

• the same orientation if their unit tangent vectors are always equal. In this case, we also say that $\gamma$ and $\phi$ are equivalent up to an orientation-perserving reparametrization.
• opposite orientations if their unit tangent vectors are always opposite. In this case, we also say that $\gamma$ and $\phi$ are equivalent up to an orientation-reversing reparametrization.