Why Is the Normal Vector of a Tangent Plane Equal to the Gradient?

[chemicalsss]

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

 

[Petr Mugver]

You have to be a bit more precise: if a surface is defined by

f(\mathbf{x})=0

then

\nabla f\big|_{\mathbf{x}_0}

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:

0=df=\nabla f\cdot\mathbf{v}

 

[StalkerM]

Hi, there is a quick proof of this.
Suppose a surface:
F(x,y,z)=Costant

and a point:
P(x₀,y₀,z₀) \in 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
\Rightarrow The vector r'(t) (tangent to the surface) is perpendicular to the levele surface.