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Differentiability (Real Parametric Curves)#

Definition: Differentiability

Let \(\gamma: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}^n\) be a parametric curve and let \(t \in \mathcal{D}\) be an accumulation point of \(\mathcal{D}\).

We say that \(\gamma\) is differentiable at \(t\) if the limit

\[\lim_{x \to t} \frac{\gamma(x) - \gamma(t)}{x - t}\]

exists. In this case, the value of this limit is known as \(\gamma\)'s derivative at \(t\).

Notation

\[\gamma'(t) \qquad \dot{\gamma}(t)\]

If \(\gamma\) is differentiable at each \(t \in S \subseteq \mathcal{D}\), then we say that \(\gamma\) is differentiable on \(S\). If there exists some finite \(E \subseteq \mathcal{D}\) such that \(\gamma\) is differentiable on \(S \setminus E\), then we say that \(\gamma\) is piecewise differentiable on \(S\). For \(S = \mathcal{D}\), we just say that \(\gamma\) is (piecewise) differentiable.

Theorem: Differentiability at Interior Points

A parametric curve \(\gamma: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}^n\) is differentiable at an interior point \(t_0\) of \(\mathcal{D}\) if and only if it is totally differentiable there. In this case, the derivative of \(\gamma\) is \(\gamma\)'s Jacobian matrix:

\[\gamma'(t_0) = J_{\gamma}(t_0)\]
Proof

We need to prove two things:

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

\[\begin{aligned}\lim_{t \to t_0} \frac{||\gamma(t) - \gamma(t_0) - T(t - t_0)||}{||t- t_0||} &= \lim_{t \to t_0} \frac{||\gamma(t) - \gamma(t_0) - (t - t_0)\mathbf{L}||}{|t- t_0|} \\ & = \lim_{t \to t_0} \frac{ |t-t_0| \cdot \left|\left|\frac{\gamma(t) - \gamma(t_0)}{t-t_0} - \mathbf{L}\right|\right|}{|t- t_0|} \\ & = \lim_{t \to t_0} \left|\left|\frac{\gamma(t) - \gamma(t_0)}{t-t_0} - \mathbf{L}\right|\right| \\ & = \left|\left| \lim_{t \to t_0} \left( \frac{\gamma(t) - \gamma(t_0)}{t - t_0} - \mathbf{L}\right) \right|\right| \\ & = \left|\left|\lim_{t \to t_0} \left( \frac{\gamma(t) - \gamma(t_0)}{t - t_0} - \frac{\gamma(t) - \gamma(t_0)}{t-t_0}\right)\right|\right| \\ & = ||\mathbf{0}|| = 0 \end{aligned}\]

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,

\[\lim_{t \to t_0} \frac{||\gamma(t) - \gamma(t_0) - \mathbf{T}\begin{bmatrix}t-t_0\end{bmatrix}||}{|t- t_0|} = \lim_{t \to t_0} \frac{||\gamma(t) - \gamma(t_0) - (t-t_0)\mathbf{T}||}{|t- t_0|} = 0\]

Factor out \(t-t_0\) from the numerator and move it outside the magnitude.

\[\begin{aligned}\lim_{t \to t_0} \frac{||\gamma(t) - \gamma(t_0) - (t-t_0)\mathbf{T}||}{|t- t_0|} &= \lim_{t - t_0} \frac{|t-t_0|\cdot \left|\left| \frac{\gamma(t) - \gamma(t_0)}{t - t_0} - \mathbf{T} \right|\right|}{|t-t_0|} \\ & = \lim_{t \to t_0} \left|\left| \frac{\gamma(t) - \gamma(t_0)}{t - t_0} - \mathbf{T} \right|\right| \\ & = \left|\left| \lim_{t \to t_0} \left( \frac{\gamma(t) - \gamma(t_0)}{t - t_0} - \mathbf{T} \right) \right|\right| = 0\end{aligned}\]

Therefore,

\[\begin{aligned}& \lim_{t \to t_0} \left(\frac{\gamma(t) - \gamma(t_0)}{t - t_0} - \mathbf{T}\right) = \mathbf{0} \\ & \lim_{t \to t_0} \left(\frac{\gamma(t) - \gamma(t_0)}{t - t_0}\right) - \mathbf{T} = \mathbf{0} \\ & \lim_{t \to t_0} \frac{\gamma(t) - \gamma(t_0)}{t - t_0} = \mathbf{T}\end{aligned}\]

Theorem: Differentiability \(\implies\) Continuity

Let \(\gamma: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}^n\) be a parametric curve and let \(t \in \mathcal{D}\) be a limit point of \(\mathcal{D}\).

If \(\gamma\) is differentiable at \(t\), then it is also continuous there.

Proof

For \(\gamma\) to be continuous at \(t\), we need to prove the following limit:

\[\lim_{x \to t} \gamma(x) = \gamma(t)\]

We have:

\[\lim_{x \to t} \gamma(x) = \gamma(t) \iff \left(\lim_{x \to t} \gamma(x)\right) - \gamma(t) = \boldsymbol{0} \iff \lim_{x \to t} \left(\gamma(x) - \gamma (t)\right) = \boldsymbol{0}\]

We now examine the last limit:

\[\lim_{x \to t} \left(\gamma(x) - \gamma (t)\right) = \lim_{x \to t} \left(\frac{\gamma(x) - \gamma (t)}{x - t} (x-t)\right)\]

Since \(\gamma\) is differentiable at \(t\), we have:

\[\begin{aligned}\lim_{x \to t} \left(\frac{\gamma(x) - \gamma (t)}{x - t} (x-t)\right) = & \left(\lim_{x \to t} \frac{\gamma(x) - \gamma (t)}{x - t}\right) \left(\lim_{x \to t} (x-t) \right) \\ & = \gamma'(t) \cdot 0 \\ & = \boldsymbol{0}\end{aligned}\]

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 \(\mu,\lambda \in \mathbb{R}\):

\[ (\lambda\,\mathbf{u}(t) + \mu\,\mathbf{v}(t))' = \lambda\,\mathbf{u}'(t) + \mu\,\mathbf{v}'(t) \]
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

\[ (\mathbf{r} \circ f)(t)' = f'(t)\,\mathbf{r}'(f(t)) \]
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

\[ (f(t)\mathbf{r}(t))' = f'(t)\mathbf{r}(t) + f(t)\mathbf{r}'(t) \]
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

\[ (\mathbf{u}(t)\cdot\mathbf{v}(t))' = \mathbf{u}'(t)\cdot \mathbf{v}(t) + \mathbf{u}(t)\cdot\mathbf{v}'(t) \]
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

\[ (\mathbf{u}(t)\times \mathbf{v}(t))' = \mathbf{u}'(t)\times\mathbf{v}(t) + \mathbf{u}(t)\times\mathbf{v}'(t) \]
Proof

TODO

Orientation#

Theorem: Colinearity of Tangent Vectors

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

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

\[ \gamma'(t) = \lambda \, \varphi'(s) \]
Proof

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

\[ \varphi(s) = \gamma(h(s)) \]

Differentiate both sides and apply the chain rule:

\[ \varphi'(s) = h'(s) \gamma'(h(s)) \]

Since \(h\) is regular, we know that \(h'(s) \ne 0\) and so

\[ \gamma'(h(s)) = \frac{1}{h'(s)} \varphi'(s) \]

Let \(\lambda = \frac{1}{h'(s)}\) and substitute back \(t = h(s)\) and the proof is complete:

\[ \gamma'(t) = \lambda \, \varphi'(s) \]

Important: Unit Tangent Vectors

It immediately follows that the unit tangent vectors of \(\gamma\) and \(\varphi\) are either always equal or always opposite.

Definition: Orientation of Parametrizations

We say that \(\gamma\) and \(\varphi\) have:

  • the same orientation if their unit tangent vectors are always equal. In this case, we also say that \(\gamma\) and \(\varphi\) 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 \(\varphi\) are equivalent up to an orientation-reversing reparametrization.