Gradient and tangent planes

  • #1

Main Question or Discussion Point

For a tangent plane to a surface, why is the normal vector for this plane equal to the gradient vector? Or is it not?
 

Answers and Replies

  • #2
279
0
You have to be a bit more precise: if a surface is defined by

[tex]f(\mathbf{x})=0[/tex]

then

[tex]\nabla f\big|_{\mathbf{x}_0}[/tex]

is a vector tangent to the surface at point x0. This is because it is orthogonal to the velocities of all possible curves that pass through x0:

[tex]0=df=\nabla f\cdot\mathbf{v}[/tex]
 
  • #3
6
0
Hi, there is a quick proof of this.
Suppose a surface:
F(x,y,z)=Costant

and a point:
P(x0,y0,z0) [tex]\in[/tex] surface.

Let C be a curve on the surface passing through P. This curve can be described by a vector function:
r(t)=(x(t),y(t),z(t))

let:
r(t0)=(x0,y0,z0)

C lies on the surface this implies that:
F(r(t))=Costant

differentiating (if F and r are differentiable) we have:
(∂F/∂x)(dx/dt)+(∂F/∂y)(dy/dt)+(∂F/∂z)(dz/dt)=0

∇F·r'(t)=0
[tex]\Rightarrow[/tex] The vector r'(t) (tangent to the surface) is perpendicular to the levele surface.
 
Last edited:

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