Parametric Surface

A parametric surface is a differentiable map that constructs a surface from as a result of a two-variable (bivariate) parametric equation.

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Definition §

Parametric Surface ^definition

A parametric surface, $S$ is a differentiable map that arises from the two-parameter parametric equation $\vec{\varphi }(s,t)$ describe as:

$$\vec{\varphi }:X\subseteq {\mathbb{R}}^2\to {\mathbb{R}}^n$$

That is, $S$ is constructed by $\vec{\varphi }(s,t)$ as $s$ and $t$ vary. If $\vec{\varphi }$ is a $C^1$ map, then $S$ is said to be a differentiable or $C_1$ surface.

In most cases, we take: $\vec{\varphi }:X\subseteq {\mathbb{R}}^2\to {\mathbb{R}}^3$

Tangents & Normals §

Tangents and normals over parametric surfaces are defined very similarly to that of tangent planes for bivariate functions since a bivariate function is a type of parametric surface!

Let’s assume the parametric equation $\vec{\varphi }$ can be represented as:

$$\vec{\varphi }(s,t)=(x(s,t),y(s,t),z(s,t))$$

where $x,y,z:{\mathbb{R}}^2\to \mathbb{R}$ are scalar functions.

We start by taking two lines $c_1$ and $c_2$, parallel to the $s$ and $t$-axis respectively. Under some parametric equation $\vec{\varphi }$, these get mapped onto a surface $S$.

The equations for the lines are:

$$c_1:\{(s,t)|t=t_0\}{c}_2:\{(s,t)|s=s_0\}$$

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We can then treat $c_1$ and $c_2$ as parametric curves, and use the same equation of tangents & normals for them (though we do now use partial derivatives).

So the tangent vector to the curve $c_1$ is also tangent with respect to $s$, because $c_1$ is parallel to the $s$-axis. Hence, for a given value of $s_0$ we have:

$$\overrightarrow{T_s}(s_0)=\frac{\partial \vec{\varphi }}{\partial s}{|}_{s=s_0}=\left(\frac{\partial x}{\partial s},\frac{\partial y}{\partial s},\frac{\partial z}{\partial s}\right){|}_{(s,t)=(s_0,t_0)}$$

Similarly for $c_2$ we have a tangent vector that is also parallel to $t$:

$$\overrightarrow{T_t}(t_0)=\frac{\partial \vec{\varphi }}{\partial t}{|}_{t=t_0}=\left(\frac{\partial x}{\partial t},\frac{\partial y}{\partial t},\frac{\partial z}{\partial t}\right){|}_{(s,t)=(s_0,t_0)}$$

Non-orthogonal Tangent Vectors!

Unlike with bivariate functions, our two tangent vectors are not parallel to the $x$ and $y$ axis! This means they don’t have to be orthogonal (perpendicular) to each other! That is, we don’t need to have them being at 90 degrees to each other or equivalently, we can have their dot product being non-zero: ${\vec{T}}_s(s_0)\cdot {\vec{T}}_t(t_0)\ne 0$

Unit Tangent Vectors §

Hence the two tangent unit vectors for a given $(s_0,t_0)$ are:

$${\hat{T}}_s(s_0,t_0)=\frac{\frac{\partial \vec{\varphi }}{\partial s}}{\Vert \frac{\partial \vec{\varphi }}{\partial s}\Vert }{|}_{(s,t)=(s_0,t_0)}$$

$${\hat{T}}_t(s_0,t_0)=\frac{\frac{\partial \vec{\varphi }}{\partial t}}{\Vert \frac{\partial \vec{\varphi }}{\partial t}\Vert }{|}_{(s,t)=(s_0,t_0)}$$

Tangent planes can be constructed using these two tangent vectors.

Unit Normal Vector §

The unit normal vector is constructed the same way as the unit binormal vector in parametric curves. That is, it is constructed from the cross product of the two unit tangent vectors. However, unlike the unit binormal vector, we have two choices of normal vectors:

$$\hat{N}(s_0,t_0)={\hat{T}}_s(s_0,t_0)\times {\hat{T}}_t(s_0,t_0)$$

$$\widehat{N^{\ast }}(s_0,t_0)={\hat{T}}_t(s_0,t_0)\times {\hat{T}}_s(s_0,t_0)$$

By the non-commutativity of the cross product, we know that:

$$\hat{N}=-\widehat{N^{\ast }}$$

So both $\hat{N}$ and ${\hat{N}}^{\ast }$ are both valid choices for normal vectors!

A parametric surface is considered smooth at a point $(s_0,t_0)$ if $\hat{N}(s_0,t_0)\ne \vec{0}$ (this also means ${\hat{N}}^{\ast }$ is non-zero).

Examples §

• A cone can be a parameterised surface via the equation:$$ x = s \cos (t) \quad\quad y = s \sin(t) \quad\quad z = s \tag{$t\in [0,2\pi ],s\ge 0$}

• A torus (donut) of radius $R$ and thickness $r$ can be a parameterised surface via the equation:$$ x = (R + r \cos(\varphi))\cos(\theta) \quad\quad y = (R+ r \cos(\varphi))\sin(\theta) \quad\quad z = r \sin(\varphi) \tag{$\theta ,\phi \in [0,2\pi ]$}

 >[!example] 1: Paraboloid as a Parametric Surface > > **Parameterise the surface of the paraboloid: $z = x^2 + y^2$** >>[!info]- Solution >> There are actually two common ways we can approach this problem: >> >> 1. Using [Cartesian coordinates](Cartesian%20Coordinates.md): >> $$ >> x = s \quad\quad y = t \quad\quad z = s^2 + t^2 \tag{$s,t \in \mathbb{R}$} >> $$ >> >> 2. Using [cylindrical coordinates](Cylindrical%20Coordinates.md): >> $$ >> x = \rho \cos(\phi)\quad\quad y = \rho \sin (\phi) \quad\quad z = x^2 + y^2 = \rho^2 \tag{$\rho \geq 0, \phi \in [0,2\pi)$} >> $$ > >>[!info]- Visuals >> >[!example] 2: Normal vectors to a Parametric Cone ^e2 > > **Find a normal vector to the cone parameterised by:** > $$ > \vec{\Phi}: x = s\cos (t) \quad\quad y = s \sin(t) \quad\quad z = s \tag{$s\geq 0, t \in [0,2\pi)$} > $$ >>[!info]- Solution >> The two tangent vectors (parallel to $s$ and $t$-axis) are given by: >> $$ >> \vec{T}_{s} = \frac{ \partial \vec{\Phi} }{ \partial s } =( \cos (t), \sin(t), 1) \quad\quad \vec{T}_{t} = \frac{ \partial \vec{\Phi} }{ \partial t } = (-s \sin(t), s \cos (t), 0) >> $$ >> (A) normal vector (well, technically a [vector field](Vector%20Field.md)) is given by: >> $$ >> \vec{N} = \vec{T}_{s} \times \vec{T}_{t} = \det\begin{bmatrix} >> \hat{i} & \hat{j} & \hat{k} \\ >> \cos(t) & \sin(t) & 1 \\ >> -s \sin(t) & s \cos(t) & 0 \\ >> \end{bmatrix} >> =(-s \cos(t), -s \sin(t), s (\cos^2(t) + \sin^2 (t)) >> $$ >> Hence: >> $$ >> \vec{N} (s,t) = (-s \cos (t), -s \sin(t), s) >> $$ >> >> Note: $\vec{N} = \vec{0}$ at $(s,t) = (0,0)$, which corresponds to the vertex of the cone. Here the normal vector being zero implies there is no [tangent plane](Tangent%20Plane.md), nor is the cone [differentiable](Bivariate%20Differentiability.md) at that point. Also, check the visuals to see that this choice of normal vector orients the cone 'inside'. > >>[!info]- Visuals >>