Parametric Curve

A parametric curve is a differentiable map that constructs a curve from as a result of a single-variable parametric equation. They are an extension to simple vector lines.

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Definition §

Parametric Curve ^definition

A parametric curve, $C$ is a differentiable map that arises from the single-parameter parametric equation $\vec{c}(t)$ describe as:

$$\vec{c}:X\subseteq \mathbb{R}\to {\mathbb{R}}^n$$

That is, $C$ is traced by $\vec{c}(t)$ as $t$ varies.

• $\vec{c}(t)$ is not unique. Even more importantly, the curve $C$ is independent of its parameterisation, i.e. any $\vec{c}(t)$ that maps to all the points of $C$ is allowed.
• However, different parameterisations can lead to different orientations of $C$.

Tangents & Normals §

Because a parametric curve are dependent on a single variable $t$, we can easily define directional vectors. Recall that

#tosee only works in ${\mathbb{R}}^3$??

Let $\vec{c}(t)$ be the parametric equation describing a curve $C$.

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Tangent Vector §

The tangent vector $\vec{T}(t)$ is given by the derivative of $\vec{c}(t)$:

$$\vec{T}(t)=\frac{\text{d}\vec{c}}{\text{d}t}$$

The unit tangent vector $\hat{T}(t)$ is a unit vector that is always tangential to the curve, and can be thought of as the direction the curve is going at the instantaneous time $t$:

$$\hat{T}(t)=\frac{\frac{\text{d}\vec{c}}{\text{d}t}}{\Vert \frac{\text{d}\vec{c}}{\text{d}t}\Vert }=\frac{{\vec{c}}^{{}^{\prime }}(t)}{\Vert {\vec{c}}^{{}^{\prime }}(t)\Vert }=\widehat{(c^{\prime }(t)})$$

Normal Vector §

The normal vector $\vec{N}(t)$ is given by the derivative of the unit tangent vector:

$$\vec{N}(t)=\frac{\text{d}\hat{T}}{\text{d}t}$$

Note: This vector is not necessarily a unit vector!

The unit tangent vector $\hat{N}(t)$ (usually denoted $\hat{n}$) is a unit vector that is always normal to the curve, and can be thought of as the direction perpendicular to the curve at the instantaneous time $t$:

$$\hat{n}(t)=\frac{\frac{\text{d}\hat{T}}{\text{d}t}}{\Vert \frac{\text{d}\hat{T}}{\text{d}t}\Vert }=\frac{{\hat{T}}^{{}^{\prime }}(t)}{\Vert {\hat{T}}^{{}^{\prime }}(t)\Vert }=\widehat{(T^{\prime }(t))}$$

Binormal Vector §

If $\vec{c}(t)$ is in ${\mathbb{R}}^3$ only , we can have a unit vector that is perpendicular to both the unit tangent vector and the unit normal vector. This is known as the unit binormal vector and is given by the cross product of both the other vectors:

$$\hat{B}(t)=\hat{T}(t)\times \hat{N}(t)$$

Order matters!

The unit binormal vector is defined exactly as $\hat{T}(t)\times \hat{N}(t)$, which means we cannot swap the order around! Doing $\hat{N}(t)\times \hat{T}(t)$ results in $-\hat{B}(t)$, which is not our ‘standard’ binormal vector.

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Orientation §

The orientation of a parametric curve is simply a choice of unit tangent vectors for each point of the curve:

• If the parametric equation observes this orientation as $t$ varies, it is *orientation-preserving.
• Note that every parametric curve has an implicit orientation, defined by the unit tangent vector at each point (and hence defined by how $t$ is parameterised):

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The choice of orientation also affects the normal vector, $\hat{n}$.

Since $\hat{n}$ is defined by taking the derivative of the tangent vector and normalising it, if we choose a different tangent vector, our normal vector changes as well!

For example, for the unit circle, we have two choices of parameterisations:

$$\begin{array}{r}x=\cos (t)y=\sin (t)t\in [0,2\pi )\\ x=\cos (t)y=\sin (t)t\in [0,-2\pi )\end{array}$$

The first one is oriented clockwise, (it is orientation-preserving for the clockwise orientation) while the second one is oriented anti-clockwise.

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Topology §

See Topology

Tangent Line §

We can always construct a tangent line to any point in a parametric equation using the vector line:

Let $\vec{c}(t)$ be a parametric equation. The tangent line to $C$, the curve parameterised by $\vec{c}$, at $t=t_0$ is given by:

$$\ell (s)=\vec{c}(t_0)+(s-t_0)\cdot \frac{\text{d}\vec{c}}{\text{d}t}(t_0)$$

I use $s$, but we could just as easily parameterise $\vec{c}$ with $t_1$ and $\ell$ with $t_2$. That is, we are using different parameters.

Differentiation Rules §

Let $\vec{b}$ and $\vec{c}$ be parametric curves with respect to the parameter $t$. Then:

\frac{\text{d}}{\text{d}t}(\vec{b} + \vec{c}) = \frac{\text{d}\vec{b}}{\text{d}t} + \frac{\text{d}\vec{c}}{\text{d}t}

* $\cdot$ is the [dot product](Dot%20Product.md)!$$ \frac{\text{d}}{\text{d}t} (\vec{b} \cdot \vec{c}) = \frac{\text{d}\vec{b}}{\text{d}t} \cdot \vec{c} + \frac{ \text{d}\vec{c} }{ \text{d}t } \cdot \vec{b}

• $\times$ is the cross product!$$ \frac{\text{d}}{\text{d}t} (\vec{b} \times \vec{c}) = \frac{ \text{d}\vec{b} }{ \text{d}t } \times \vec{c} + \vec{b} \times \frac{ \text{d}\vec{c} }{ \text{d}t }

Be very careful about the ordering, as the cross product is [non-commutative!](Cross%20Product.md#Properties) ### Examples >[!example] 1: Helix Parametric Curve > > **Identify the parametric curve parameterised by:** > $$ > \vec{c}(t) = (\cos (t), \sin(t), t) > $$ >>[!info]- Solution >> We can use the [trigonometric identity](Trigonometric%20&%20Hyperbolic%20Identities.md): $\cos(t)^2 + \sin(t)^2 = 1$. If we let $x = \cos (t)$ and $y = \sin(t)$, then we have: >> $$ >> \cos (t)^2 + \sin(t)^2 = 1 \implies x^2 + y^2 = 1 >> $$ >> This defines a circle, but in $\mathbb{R}^3$, it becomes a cylinder. However, our $z=t$ means that the circle moves higher and higher, resulting in a helical structure: >> > >>[!info]- Visuals >>,%20sin(t),%20t)$) >[!example] 2: Orientation-preserving Parametric Semicircle > > **Find an orientation-preserving parameterisation of the curve below:** > $$ > y = \sqrt{ a^2 - x^2 } > $$ > 1. **Oriented clockwise** > 2. **Oriented anti-clockwise** > >>[!info]- Solution >> The curve above describes the top semi-circle of radius $a$. We can convert the equation to [polar coordinates](Polar%20Coordinates.md): >> >> $$ >> y = \sqrt{ a^2 - x^2 } \implies y^2 = a^2 - x^2 \tag{$y>0$} \implies \dfrac{y^2}{a^2} + \dfrac{x^2}{a^2} = 1 >> $$ >> $$ >> \sin^2(t) + \cos^2(t) = 1 \implies y = a \sin(t),\ \ x = a \cos(t) >> $$ >> To ensure we plot a semi-circle instead of a full circle, we restrict the domain of $t$: >> >> 1. Clockwise orientation:$$ >> \vec{c}(t) = (a \cos (t), a \sin(t)) \quad\quad t \in [0,\pi] >> $$ >> 2. Anti-clockwise orientation: >> $$ >> \vec{c}(t) = (a \cos (t), - a \sin(t)) \quad\quad t \in [\pi, 2\pi] >> $$ >[!example] 3: Simplification via Differentiation Rules > > **Simplify the following derivative, assuming $u: \mathbb{R} \to\mathbb{R}^3$ and $u'''$exists:** > $$ > \frac{\text{d}}{\text{d}t} [(u'' \times u') \cdot u] > $$ >>[!info]- Solution >> $$ >> \begin{align} >> \frac{\text{d}}{\text{d}t} [(u'' \times u') \cdot u] &= \frac{\text{d}}{\text{d}t} (u'' \times u')\cdot u + \frac{\text{d}u}{\text{d}t}\cdot (u'' \times u') \tag{Dot product rule}\\ >> &= \quad\quad\quad\dots\quad\quad \ + u'\cdot (u'' \times u') \tag{$\frac{ \text{d}u }{ \text{d}t } =u'$}\\ >> &= \frac{\text{d}}{\text{d}t} (u'' \times u')\cdot u + 0 \tag{$u' \times u''$ is orthogonal to $u'$, hence dot product is 0} \\ >> &= \left(\frac{ \text{d}u'' }{ \text{d}t } \times u' + u'' \times \frac{ \text{d}u' }{ \text{d}t }\right) \cdot u \tag{Cross product rule}\\ >> &=(u''' \times u' + u'' \times u'')\cdot u \\ >> &=(u''' \times u') \cdot u \tag{Cross product of a vector with itself is 0} >> \end{align} >> $$ > ### Resources * https://en.wikipedia.org/wiki/Frenet%E2%80%93Serret_formulas