What is the relationship between the gradient and the normal vector?

[jaguar7]

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. :)

 

[Office_Shredder]

The tangent plane is defined to be the plane of all points (x,y,z) the equation

$$\nabla f(x_0,y_0,z_0) \cdot (x,y,z)=\nabla f(x_0,y_0,z_0) \cdot (x_0, y_0, z_0)$$
(you may not have seen it written down this way, but look at what your definition of the tangent plane and make sure you understand that you can write it in this form).

From this equation, what can you say about a vector lying along the tangent plane (this is different from picking a point in the plane) when compared to the gradient?

 

[HallsofIvy]

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.