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## 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?

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

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

then

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

is a vector tangent to the surface at point x

[tex]0=df=\nabla f\cdot\mathbf{v}[/tex]

- #3

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Hi, there is a quick proof of this.

Suppose a surface:

F(x,y,z)=Costant

and a point:

P(x_{0},y_{0},z_{0}) [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(t_{0})=(x_{0},y_{0},z_{0})

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.

Suppose a surface:

F(x,y,z)=Costant

and a point:

P(x

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(t

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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