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Explicit vs Implicit Surface

Let’s say that you have a surface given by $z=f(x,y)$.

You want to find the tangent plane to that surface at a given point $p_0 = (x_0, y_0, z_0)$. There are two tangent plane equations to choose from.

The explicit equation

$$ z = f(x_0, y_0) + f_x(x_0, y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0) $$

Or the implicit equation

$$ 0 = \nabla f \cdot (p - p_0 ) $$ where $p = (x, y, z)$

What’s the difference?

If you have a scalar function $g(x,y,z)$, you can define a couple things from its scalar field

  • A level set of a function $g(x,y,z)$ and a constant $C$ is the set of points in space that share the same value $C$. This is given by $g(x,y,z)=C$, where C is a constant. Different constants result in different level sets.
  • The equation $g(x,y,z)=C$ represents an isosurface.
    Every point on this surface has a value of C. An isosurface is always a level set, but a level set is not necessarily an isosurface. The set of points in a level set may may not form a surface, or the level set may not even be in 3D, ie: $f(x,y)=C$
  • The equation $g(x,y,z)=0$ represents an implicit surface, an isosurface with $C=0$

Gradient as normal

The gradient of a function $g(x,y,z)=C$ at a point is equal to its normal.

Quick proof

Let there be a parametric curve $r$ that lies on the surface of g

$$ r(t)=\begin{bmatrix} x(t) \\ y(t) \\ z(t) \end{bmatrix} $$

Apply multivariable chain rule to the implicit surface $\frac{d}{dt}g(r(t))=0$

$$ \frac{\partial g}{\partial x} \frac{dx}{dt} + \frac{\partial g}{\partial y} \frac{dy}{dt} + \frac{\partial g}{\partial z} \frac{dy}{dt} = 0 $$

$$ ( \frac{\partial g}{\partial x} \frac{\partial g}{\partial y} \frac{\partial g}{\partial z} ) \cdot ( \frac{\mathrm{d}x}{\mathrm{d}t} \frac{\mathrm{d}y}{\mathrm{d}t} \frac{\mathrm{d}z}{\mathrm{d}t} ) = 0 $$

$$ \nabla g \cdot r^{\prime}= 0 $$

This tells us that the gradient at a point $\nabla g(x_0, y_0,z_0)$ is orthogonal to the tangent vector $r^{\prime}(x_0,y_0,z_0)$ at that point on a curve that lies on the surface, so $n=\nabla g$

Using the gradient

The implicit surface $g(x,y,z) = 0$ represents the same surface as $z=f(x,y)$ by setting

$$ g(x,y,z)=f(x,y)-z $$

Then the normal $n$ is given by $\nabla g(x,y,z)$

$$ \begin{align*} n &= \nabla g \\ n &= \nabla ( f-z ) \\ n &= \begin{bmatrix} f_x \\ f_y \\ -1 \end{bmatrix} \end{align*} $$

The equation of a plane is $(\begin{bmatrix} x \\ y \\ z \end{bmatrix} -\begin{bmatrix} x_0 \\ y_0 \\ z_0 \end{bmatrix})\cdot n = 0$

Expanding it out, we have

$$ \begin{align*} \\ 0 &= (\begin{bmatrix} x \\ y \\ z \end{bmatrix} -\begin{bmatrix} x_0 \\ y_0 \\ z_0 \end{bmatrix})\cdot \begin{bmatrix} f_x \\ f_y \\ -1 \end{bmatrix} \\ 0 &= f_x(x-x_0)+ f_y(y-y_0) + (-1)(z-z_0) \\ z &= z_0 + f_x(x-x_0)+ f_y(y-y_0) \end{align*} $$
Which is the explicit tangent plane equation from the beginning.

Normal from Cross Product

An alternate method of computing the normal is to use the cross product of the partial derivative tangent vectors.

A small change in $x$ by $dx$ in produces a change in $z$ by $\frac{\partial z}{\partial x}dx$.

In this diagram, the tangent vector on the x-z plane is $\vec{a}=(dx,0,\frac{\partial z}{\partial x}dx)$
The tangent vector on the y-z plane is $\vec{b}=(0,dy,\frac{\partial z}{\partial y}dy)$

We can divide out $\mathrm{d}x$ and $\mathrm{d}y$ to get proportional tangent vectors.

$\vec{a} \propto (1,0,z_x)$ and $\vec{b} \propto (0,1,z_y)$. Crossing them gives the normal.

$$ \begin{align*} \\ n &= (1,0,z_x) \times (0,1,z_y) \\ n &= \begin{bmatrix} -z_x \\ -z_y \\ 1 \end{bmatrix} \end{align*} $$
Which gives the same normal as before, but flipped the other direction

Example

Find the equation of the tangent plane to the surface $z = f(x,y) = 4-x^2 - y^2$ at the point $(1,2)$

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The point $p_0$ in 3d is given by

$$ \begin{align*} \\ p_0 &= (1,2,f(1,2)) \\ p_0 &= (1,2,-1) \\ \end{align*} $$

The implicit surface $g$ is given by

$$ \begin{align*} \\ g(x,y,z) &= f(x,y) - z \\ g(x,y,z) &= 4 - x^2 - y^2 - z \\ \end{align*} $$

The normal $n$ as a function of position is given by

$$ \begin{align*} \\ n &=\nabla g \\ n &=(-2x, -2y, -1) \\ \end{align*} $$

The normal $n_0$ at point $p_0$ is given by

$$ \begin{align*} \\ n_0 &= n(p_0) \\ n_0 &= n(1,2,-1) \\ n_0 &= (-2(1), -2(2), -1) \\ n_0 &= (-2, -4, -1) \\ \end{align*} $$

Finally, the equation of the tangnet plane is given by

$$ \begin{align*} \\ 0 &= (p-p_0) \cdot n_0 \\ 0 &= (\begin{bmatrix} x \\ y \\ z \end{bmatrix} - \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}) \cdot \begin{bmatrix} -2 \\ -4 \\ -1 \end{bmatrix} \\ 0 &= -2(x-1) -4(y-2) + z \\ z &= -2(x-1) -4(y-2) \\ \end{align*} $$

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References