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Analiza matematyczna funkcji wielu zmiennych

Kurs: Analiza matematyczna funkcji wielu zmiennych > Rozdział 4

Lekcja 13: Strumień pola w 3D (artykuły)

Unit normal vector of a surface

Learn how to find the vector that is perpendicular, or "normal", to a surface. You will need this skill for computing flux in three dimensions.

Kontekst

Partial derivatives of parametric surfaces
- In particular, make sure you have a strong intuition for the partial derivatives of a function parameterizing a surface, and what they represent.
Cross product (video)

Do czego zmierzamy

If a surface is parameterized by a function start bold text, v, end bold text, with, vector, on top, left parenthesis, t, comma, s, right parenthesis, the unit normal vector to this surface is given by the expression
$\begin{aligned} \pm \dfrac{ \left( \dfrac{\partial \vec{\textbf{v}}}{\partial \blueE{t}}(t, s) \right) \times \left( \dfrac{\partial \vec{\textbf{v}}}{\partial \redE{s}}(t, s) \right) }{ \left| \left( \dfrac{\partial \vec{\textbf{v}}}{\partial \blueE{t}}(t, s) \right) \times \left( \dfrac{\partial \vec{\textbf{v}}}{\partial \redE{s}}(t, s) \right) \right| } \end{aligned}$
You always have two choices for a unit vector function. If a surface is closed, like a sphere or a torus, those choices can be interpreted as outward-facing and inward-facing vectors.
This is useful for the idea of flux in three-dimensions, covered in the next article.

Unit normal vector

Let's say you have some surface, S. If a vector at some point on S is perpendicular to S at that point, it is called a normal vector (of S at that point). More precisely, you might say it is perpendicular to the tangent plane of S at that point, or that it is perpendicular to all possible tangent vectors of S at that point.

When a normal vector has magnitude 1, it is called a unit normal vector. Notice, there will always be two unit normal vectors, each pointing in opposite directions:

Why do we care? To compute surface integrals in a vector field, also known as three-dimensional flux, you will need to find an expression for the unit normal vectors on a given surface. This will take the form of a multivariable, vector-valued function, whose inputs live in three dimensions (where the surface lives), and whose outputs are three-dimensional vectors.

Example: How to compute a unit normal vector

Consider the surface described by the following parametric function:

$\begin{aligned} \vec{\textbf{v}}(t, s) = \left[ \begin{array}{c} t + 1\\ s \\ s^2 - t^2 + 1 \end{array} \right] \end{aligned}$

In the range where minus, 2, is less than or equal to, t, is less than or equal to, 2 and minus, 2, is less than or equal to, s, is less than or equal to, 2, here's what that surface looks like:

For what follows, I am assuming you know that the two partial derivatives of a parametric surface give vectors which are each tangent to the surface, but in different directions.

Step 1: Find a (not necessarily unit) normal vector

Concept check: Which of the following will give a vector which is perpendicular to the surface parameterized by start bold text, v, end bold text, with, vector, on top at the point start bold text, v, end bold text, with, vector, on top, left parenthesis, 1, comma, minus, 2, right parenthesis?

Wybierz 1 odpowiedź:
(Choice A)
$\begin{aligned} \quad \left( \dfrac{\partial \vec{\textbf{v}}}{\partial \blueE{t}}(1, -2) \right) \times \left( \dfrac{\partial \vec{\textbf{v}}}{\partial \redE{s}}(1, -2) \right) \end{aligned}$
(Choice B)
$\begin{aligned} \quad \left( \dfrac{\partial \vec{\textbf{v}}}{\partial \blueE{t}}(1, -2) \right) \cdot \left( \dfrac{\partial \vec{\textbf{v}}}{\partial \redE{s}}(1, -2) \right) \end{aligned}$
(Choice C)
$\begin{aligned} \quad \left( \dfrac{\partial \vec{\textbf{v}}}{\partial \blueE{t}}(1, -2) \right) + \left( \dfrac{\partial \vec{\textbf{v}}}{\partial \redE{s}}(1, -2) \right) \end{aligned}$

[Odpowiedź]

This is a pretty complicated expression, with two vector-valued partial derivatives and a cross product. If you have computed some surface integrals before, you are all-too familiar with the expression and how ugly it can be to compute.

Once again, here's how start bold text, v, end bold text, with, vector, on top, left parenthesis, t, comma, s, right parenthesis is defined:

$\begin{aligned} \vec{\textbf{v}}(t, s) = \left[ \begin{array}{c} t + 1\\ s \\ s^2 - t^2 + 1 \end{array} \right] \end{aligned}$

Concept check: Now compute the cross product of the partial derivatives of start bold text, v, end bold text, with, vector, on top. Do this for a general point left parenthesis, t, comma, s, right parenthesis, meaning each component of your answer will be a function of t and s. As described in the previous problem, this will give you a function for normal vectors of the surface.

$\begin{aligned} \left( \dfrac{\partial \vec{\textbf{v}}}{\partial \blueE{t}}(t, s) \right) \times \left( \dfrac{\partial \vec{\textbf{v}}}{\partial \redE{s}}(t, s) \right) = \end{aligned}$
start bold text, i, end bold text, with, hat, on top, plus
start bold text, j, end bold text, with, hat, on top, plus
start bold text, k, end bold text, with, hat, on top

[Odpowiedź]

For example, if we plugged in left parenthesis, t, comma, s, right parenthesis, equals, left parenthesis, 1, comma, minus, 2, right parenthesis, here's what we'd get:

$\begin{aligned} \left[ \begin{array}{c} 2(1) \\ -2(-2) \\ 1 \end{array} \right] = \left[ \begin{array}{c} 2 \\ 4 \\ 1 \end{array} \right] \end{aligned}$

This is a vector which is perpendicular to the surface at the point start bold text, v, end bold text, with, vector, on top, left parenthesis, 1, comma, minus, 2, right parenthesis. However, it is not a unit vector, as you can see by computing its magnitude:

square root of, 2, squared, plus, 4, squared, plus, 1, squared, end square root, equals, square root of, 4, plus, 16, plus, 1, end square root, equals, square root of, 21, end square root

Step 2: Make that a unit normal vector

So we have this expression

\begin{aligned} \left[ \begin{array}{c} 2t \\ -2s \\ 1 \end{array} \right] \end{aligned}

that gives us a normal vector for each point start bold text, v, end bold text, with, vector, on top, left parenthesis, t, comma, s, right parenthesis. The next step is to massage this a bit to get an expression for a unit normal vector.

Concept check: What is the unit normal vector to our surface at the point start bold text, v, end bold text, with, vector, on top, left parenthesis, 1, comma, minus, 2, right parenthesis?

start bold text, i, end bold text, with, hat, on top, plus
start bold text, j, end bold text, with, hat, on top, plus
start bold text, k, end bold text, with, hat, on top

[Odpowiedź]

Concept check: More generally, what is the unit normal vector to our surface at an arbitrary point start bold text, v, end bold text, with, vector, on top, left parenthesis, t, comma, s, right parenthesis, as a function of t and s?

start bold text, i, end bold text, with, hat, on top, plus
start bold text, j, end bold text, with, hat, on top, plus
start bold text, k, end bold text, with, hat, on top

[Odpowiedź]

Bada boom bada bang, you've got yourself a unit normal vector.

If you plug in any value left parenthesis, t, start subscript, 0, end subscript, comma, s, start subscript, 0, end subscript, right parenthesis to this expression, you will get a vector which has magnitude 1, and is normal to the surface parameterized by the function start bold text, v, end bold text, with, vector, on top at the point start bold text, v, end bold text, with, vector, on top, left parenthesis, t, start subscript, 0, end subscript, comma, s, start subscript, 0, end subscript, right parenthesis.

Choosing orientation

Notice, if you multiply your function for a unit normal vector by minus, 1, it will still produce unit normal vectors. They will all just point in the opposite directions. The choice of direction for the unit normal vectors of your surface is what's called an orientation of that surface.

You will see the significance of this in the next article on three-dimensional flux. In short, orienting your surface is analogous to giving a one-dimensional curve a direction.

When your surface is closed, like a sphere or a torus, the two choices for unit normal vectors are often called outward-facing and inward-facing unit normal vectors.

Podsumowanie

Given a surface parameterized by a function start bold text, v, end bold text, with, vector, on top, left parenthesis, t, comma, s, right parenthesis, to find an expression for the unit normal vector to this surface, take the following steps:
Krok 1: Get a (non necessarily unit) normal vector by taking the cross product of both partial derivatives of start bold text, v, end bold text, with, vector, on top, left parenthesis, t, comma, s, right parenthesis:
$\begin{aligned} \left( \dfrac{\partial \vec{\textbf{v}}}{\partial \blueE{t}}(t, s) \right) \times \left( \dfrac{\partial \vec{\textbf{v}}}{\partial \redE{s}}(t, s) \right) \end{aligned}$
Krok 2: Turn this vector-expression into a unit vector by dividing it by its own magnitude:

$\begin{aligned} \dfrac{ \left( \dfrac{\partial \vec{\textbf{v}}}{\partial \blueE{t}}(t, s) \right) \times \left( \dfrac{\partial \vec{\textbf{v}}}{\partial \redE{s}}(t, s) \right) }{ \left| \left( \dfrac{\partial \vec{\textbf{v}}}{\partial \blueE{t}}(t, s) \right) \times \left( \dfrac{\partial \vec{\textbf{v}}}{\partial \redE{s}}(t, s) \right) \right| } \end{aligned}$

You can also multiply this expression by minus, 1, and it will still give unit normal vectors.
The main reason for learning this skill is to compute three-dimensional flux.

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