Differentiability of Parametric Curves

Theorem: Differentiability of Parametric Curves

Let $\gamma :I\subseteq \mathbb{R}\to {\mathbb{R}}^n$ be a parametric curve and let $t$ be an accumulation point of $I$.

Then $\gamma$ is differentiable at $t_0\in I$ if and only if the following limit exists

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

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: Derivative of a Parametric Curve

The derivative of $\gamma$ at $t_0$ is the limit

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

provided that it exists.

If there is no specific $t_0\in I$ mentioned, then “derivative” refers to the function which to each $t^{\ast }$ assigns the aforementioned limit (if said limit exists).

NOTATION

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

Note: Derivative Terminology

Referring to this limit as the “derivative” is technically a misnomer, since the derivative of a vector function is a linear transformation and the aforementioned limit is only a matrix representation of said transformation. However, in the case of parametric curves, using the matrix representation of the derivative is very common. This is why, in this context, the term “derivative” is usually used for the matrix representation of the transformation instead of the transformation itself.

Tip: (Continuous) Differentiability of Curve Parameterisations

A curve parameterisation $\gamma$ is $k$-times (continuously) differentiable if its $k$-th order derivative exists (and is continuous).

Tip: Piecewise (Continuous) Differentiability of Curve Parameterisations

A curve parameterisation $\gamma :I\subseteq \mathbb{R}\to {\mathbb{R}}^n$ is $k$-times piecewise (continuously) differentiable if $I$ can be expressed as a disjoint union $I=I_1\cup \cdots \cup I_n$ such that the restrictions $\gamma {|}_{I_1},\cdots ,\gamma {|}_{I_n}$ are $k$-times continuously differentiable.

Tip: Smoothness of Curve Parameterisations

A curve parameterisation is smooth if it is $k$-times continuously differentiable for all $k\in \mathbb{N}$.

Tip: Piecewise Smoothness of Curve Parameterisations

A curve parameterisation $\gamma :I\subseteq \mathbb{R}\to {\mathbb{R}}^n$ is piecewise smooth if $I$ can be expressed as a union of a finite collection of intervals $I=I_1\cup \cdots \cup I_p$ such that the restrictions $\gamma {|}_{I_1},\dots ,\gamma {|}_{I_n}$ are smooth.