Normal of Implicit Surface

Question

I am trying to understand the reasoning behind finding what the normal of an implicit surface is. I found this article on the web:

http://web.mit.edu/hyperbook/Patrikalakis-Maekawa-Cho/node27.html

that contains the information I need. I copied below the section of interest:

[QUOTE]

The unit normal vector for an implicit surface can be derived by considering two parametric curves

${\bf r}_1 = (x_1(t_1), y_1(t_1),z_1(t_1))^T$, ${\bf r}_2 = (x_2(t_2),z_2(t_2))^T$

lying on an implicit surface $f(x,y,z)=0$, and intersecting at point $P$ on the surface with different tangent directions. Thus we have the relations: $$ \begin{eqnarray} f(x_1(t_1), y_1(t_1), z(t_1))=0,\;\;\;f(x_2(t_2), y_2(t_2), z(t_2))=0\;. \end{eqnarray} $$

(3.5)

Total differentiation of (3.5) with respect to $t_1$ and $t_2$, respectively, yields:

$$f_x\frac{dx_1}{dt_1} + f_y\frac{dy_1}{dt_1} + f_z\frac{dz_1}{dt_1} = 0\;$$

(3.6)

$$f_x\frac{dx_2}{dt_2} + f_y\frac{dy_2}{dt_2} + f_z\frac{dz_2}{dt_2} = 0\;$$

(3.7)

Now if we multiply (3.6) by $\frac{dx_2}{dt_2}$ and subtract (3.7) multiplied by $\frac{dx_1}{dt_1}$, and if we multiply (3.6) by $\frac{dy_2}{dt_2}$ and subtract (3.7) by $\frac{dy_1}{dt_1}$ we can deduce the following relation

$$ \begin{eqnarray} f_x:f_y:f_z = \frac{dz_2}{dt_2}\frac{dy_1}{dt_1}- \frac{dz_1}{dt_1}\frac{dy_2}{dt_2}: \frac{dz_1}{dt_1}\frac{dx_2}{dt_2}- \frac{dz_2}{dt_2}\frac{dx_1}{dt_1}:\frac{dx_1}{dt_1}\frac{dy_2}{dt_2}- \frac{dx_2}{dt_2}\frac{dy_1}{dt_1}\;, \end{eqnarray} $$

(3.8)

which indicates that vector $\nabla f =(f_x, f_y, f_z)^T$ (also known as gradient of $f$) is in the direction of the cross product of the two tangent vectors at $P$, i.e. in the normal direction. Thus the unit normal vector of the implicit surface is given by:

$$ \begin{eqnarray} {\bf N} = \frac{(f_x, f_y, f_z)^T}{\sqrt{f_x^2 + f_y^2 + f_z^2}} = \frac{\nabla f}{|\nabla f|}\;, \end{eqnarray} $$

provided that

$|\nabla f|\neq 0$

[/QUOTE]

The problem is that I don't understand the demonstration at all. I understand the idea of the parametric curves passing through $P$ and that if you give the parameter $t_1$ and $t_2$ to these curves you get coordinates $x_1, y_1, z_1$ etc. which plugged in the equation of the implicit surface give 0. That's clear.

What I don't understand is what follows. Why do we sum up the results in (3.6) and (3.7)? Why do these multiplications and subtractions to get to 3.8 and what does the notation with the semi-colon (:) means in 3.8?

Answer 1

I think this is unnecessarily complicated. Here's how I might do it.

The claim is that the gradient $\nabla f(P)$ is a normal vector to your surface $f(x,y,z) = 0$ at a point $P = (x_0, y_0, z_0)$. An important property of the gradient is that it determines directional derivatives: for any vector $\bf u$, the directional derivative of $f$ at $P$ along $\bf u$ is ${\bf u} \cdot \nabla f(P)$. (Some authors restrict the term "directional derivative" to unit vectors: it doesn't matter here) In particular, when $\bf u$ is a tangent vector to the surface, the function $f$ is constant on the surface so the directional derivative is $0$. Thus the gradient $\nabla f(P)$ is orthogonal to all tangent vectors to the surface, which says exactly that it is a normal to the surface.

Note also that a differentiable surface will have two linearly independent tangent vectors at any point, spanning a $2$-dimensional subspace of $\mathbb R^3$. Since $\mathbb R^3$ is $3$-dimensional, the vectors orthogonal to this subspace form a $1$-dimensional space. Thus, if $\nabla f(P) \ne 0$, every normal vector to the surface at $P$ is a scalar multiple of $\nabla f(P)$. If you scale appropriately to get a unit vector, there are only two: $\nabla f/|\nabla f|$ and $-\nabla f/|\nabla f|$.

Answer 2

There are easier proofs than the one in the textbook, but if you want to understand that one...

First of all, keep in mind that the notation with a semi-colon means "is to ... as ... is to". That is to say, "a:b:c=d:e:f" means "a is to b is to c as d is to e is to f" or, in more words, "the proportion between a and b, and that between b and c, are respectively equal to the proportion between d and e and that between e and f".

I.e. $\frac{a}{b} = \frac{d}{e}$ and $\frac{b}{c}=\frac{e}{f}$, or more succintly $\frac{a}{d}=\frac{b}{e}=\frac{c}{f}$.

Now what is the textbook doing? What you basically have are two polynomials in $f_x,f_y,f_z$, and they are both set to $0$:

$\alpha_x f_x + \alpha_y f_y + \alpha_z f_z=0~~~~~~~~~\beta_x f_x + \beta_y f_y + \beta_z f_z=0$.

These are the two formulas in $3.6$. Now, if you multiply the first polynomial by the first coefficient of the second polynomial $\beta_x$, and the second polynomial by the first coefficient of the first polynomial $\alpha_x$, and you subtract, you cancel out the $f_x$ term, and collecting the $f_y$ and $f_z$ terms you can write:

$(\beta_x\alpha_y -\alpha_x\beta_y) f_y + (\beta_x\alpha_z - \alpha_x\beta_z) fz = 0$,

which allows you to write $f_y:f_z=(\beta_x\alpha_z - \alpha_x\beta_z): (\beta_x\alpha_y -\alpha_x\beta_y)$ (remember that $a:b$ means $\frac{a}{b}$). Now repeat the process cancelling out $f_y$...

If you are familiar with scalar products and cross products of vectors ($v \cdot u$ and $v \times u$), you can see what the textbook is saying immediately after $3.8$ from the two formulas in $3.6$ avoiding all the number crunching. 3.6 is telling you that:

1.$(f_x,f_y,f_z) \cdot (\frac{dx_1}{dt}\frac{dy_1}{dt}\frac{dz_1}{dt}) =0$

2.$(f_x,f_y,f_z) \cdot (\frac{dx_2}{dt}\frac{dy_2}{dt}\frac{dz_2}{dt}) =0$

i.e. $(f_x,f_y,f_z)$ is orthogonal to both $(\frac{dx_1}{dt}\frac{dy_1}{dt}\frac{dz_1}{dt})$ and $(\frac{dx_1}{dt}\frac{dy_1}{dt}\frac{dz_1}{dt})$ which in $3$ dimensions means that

3. $(f_x,f_y,f_z)$ is parallel to $(\frac{dx_1}{dt}\frac{dy_1}{dt}\frac{dz_1}{dt})\times(\frac{dx_1}{dt}\frac{dy_1}{dt}\frac{dz_1}{dt})$