Thread: Tangent plane to a surface

Is it correct to assume that taking the dot product of the gradient and the derivative of a surface function will give me the equation of the tangent plane at any point on the surface? My notes are not entirely clear about this. Thanks in advance.

Also, would a question of this nature be better suited in the calculus section?

To answer this question properly, I have be sure that you know how to write the equation of a plane given a point on the plane and its normal. Do you?

If so then the gradient to a surface at a point is the normal of the tangent plane at that point.

maybe i'm confusing the derivative of a vector function curve. I know that it represents the tangent at any point on a curve, and if I know the normal vector as well at that point (the gradient?), then the dot product of the two equal to zero should give me a plane tangent to the curve at that point

I understand that when looking at surfaces, the tangent plane at a point (x1, y1, z1) of the function F(x,y,z) is going to be:

∇F(x,y,z) . <(x - x1), (y - y1), (z - z1)> = 0

and my visualization of this is the following:

It should be for the tangent plane to the field at .

But remember that the gradient involves partial derivatives.

super, pretty much what the notes and readings say. Thanks for the help!