What is the relationship between the gradient of a curve and its normal vector?

[mikewinifred]

grad of curve is its normal ?

i read in a book grad(f)/|grad(f)| is the normal vector of the curve... but actually it should be the cross product of the binormal rite ?? .. how come the binormal is missing here as this expression will give only the tangent ! .. Help pleasezzz

mike

 

[tiny-tim]

i read in a book grad(f)/|grad(f)| is the normal vector of the curve... but actually it should be the cross product of the binormal rite ?? .. how come the binormal is missing here as this expression will give only the tangent !

Hi mike!

(have a grad: ∇ and a curly d: ∂ and a squared: ² )

I think you're thinking of the gradient , ∇f, (of a 2D curve) as being dy/dx.

∇ operates on a scalar function of the vector (x,y).

So, for example, the curve y = x² can be written f = 0, where f = y - x².

Then ∇f = (∂f/∂y,∂f/∂x) = (1,-2x), which is perpendicular to the tangent, (2x,1).

 

[HallsofIvy]

First, since the text said "the" normal, you should have recognised that this is about two dimensions not 3 (there exist an entire circle of unit normals around a 3 dimensional curve). There is no "binormal" since that is perpendicular to both the tangent line and "principle" normal so you must have 3 dimensions to define it.

If a curve is given by f(x,y)= constant, then the derivative of f in the direction with angle $\theta$ to the x-axis is
$$\nabla f\cdot <cos(\theta), sin(\theta)>= \frac{\partial f}{\partial x}cos(\theta)+ \frac{\partial f}{\partial y}sin(\theta)$$

ON the curve itself, f is a constant so the derivative in that direction is 0. That is:
$$\nabla f\cdot <cos(\theta), sin(\theta)>= 0$$
which says that $grad f= \nabla f$ is normal to a vector pointing tangent to the curve.

In 3 dimensions, f(x,y,z)= constant is the equation of a surface and $\nabal f$ is normal to the surface.

 

[mikewinifred]

thx guys fog cleared !