Question

multivariable calc: What's a good way to think about the gradient? I know the gradient is the normal vector to a function and is also a derivative in the x, y, and z directions. In one dimension, though, the derivative gives you the tangent to the function. I've also seem remember that for the curvature of a function, the first derivative of a function gives you the tangent vector and the second derivative gives you the normal to the curve.

Answer 1

I also know that the equation of a tangent plane of \(f(x,y,z)=k\) at \( (x_0,y_0,z_0)\) is

\( f_x (x_0,y_0,z_0)(x-x_0)+f_y (x_0,y_0,z_0)(y-y_0)+f_z (x_0,y_0,z_0)(z-z_0) = 0\)

Answer 2

One thing that confuses me is that the gradient gives you a vector \(\vec{n}\) that is perpendicular to the normal curves of a 3D surface, but \(\vec{n}\) is not the same as the normal to the tangent plane at that point.

Answer 3

a simple way to start looking at it, from differentials

\(dF = F_x dx + F_y dy = <F_x, F_y>\bullet<dx, dy> = \nabla F \bullet<dx, dy>\)

if \(F(x,y) = c\), then \(dF = 0\)

and \(<dx, dy>\) lies on F, it has to

so \(\nabla (F) \bullet<dx, dy> = 0\)

so you have a normal vector....