jaguar7 said:
is the normal just grad(f(x0,y0,z0))? If so, how exactly does this work out to be so? Explain? Thanks... :D
& is the calculus section the most appropriate place to put this question? thanks again. :)
This question is poorly phrased. The normal to
what? A normal is a vector perpendicular to some surface and just the function, f(x, y, z), does not determine any surface. The gradient vector, of a function, at a given point, is, as Office Shredder says, normal to the tangent plane of the graph of the surface defined by
f(x, y, z)= constant.
We can write the "directional derivative", the rate of change of the function f in the direction that makes angles \theta, \phi, and \psi with the positive x, y, and z axes, respectively, as
\frac{\partial f}{\partial x}cos(\theta)+ \frac{\partial f}{\partial y}cos(\phi)+ \frac{\partial f}{\partial z}cos(\psi)
which is exactly the same as the dot product
\nabla f\cdot cos(\theta)\vec{i}+ cos(\phi)\vec{j}+ cos(\psi)\vec{k}
and now cos(\theta)\vec{i}+ cos(\phi)\vec{j}+ cos(\psi)\vec{k} is the unit vector in the given direction.
If f(x,y,z) is a constant on a given surface, the derivative in any direction tangent to that surface must be 0. That is, \nabla f has 0 dot product with any vector tangent to the surface and so is normal to the surface.