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MATH
[1251]

Proofs from Derivatives and level-set geometry

All proofs from Derivatives and level-set geometry.

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Let γ(t)\boldsymbol{\gamma}(t) be a differentiable curve in SS with γ(t0)=x0\boldsymbol{\gamma}(t_0)=\mathbf{x}_0. Since FF has the constant value cc along the curve,

F(γ(t))=c.F(\boldsymbol{\gamma}(t))=c.

The chain rule gives

∇F(γ(t0))⋅γ′(t0)=0.\nabla F(\boldsymbol{\gamma}(t_0)) \cdot \boldsymbol{\gamma}'(t_0)=0.

Thus the gradient is perpendicular to every tangent vector supplied by a curve in the level set. It is therefore a normal vector at x0\mathbf{x}_0.

  1. [1]:Gradient normal to a level set

The graph is the level surface F=z−f(x,y)=0F=z-f(x,y)=0. Its gradient at the base point is

∇F(x0,y0,z0)=(−fx(x0,y0)−fy(x0,y0)1).\nabla F(x_0,y_0,z_0) = \begin{pmatrix} -f_x(x_0,y_0)\\ -f_y(x_0,y_0)\\ 1 \end{pmatrix}.

The point-normal equation is therefore

(−fx(x0,y0)−fy(x0,y0)1)⋅(x−x0y−y0z−z0)=0.\begin{pmatrix} -f_x(x_0,y_0)\\ -f_y(x_0,y_0)\\ 1 \end{pmatrix} \cdot \begin{pmatrix} x-x_0\\ y-y_0\\ z-z_0 \end{pmatrix} =0.

Solving for zz and using z0=f(x0,y0)z_0=f(x_0,y_0) gives the stated equation.

  1. [1]:Tangent plane to a graph