- #1

AAMAIK

- 47

- 0

- TL;DR Summary
- Why the assumptions, that the tangent plane must pass through the point we wish to approximate the function nearby and the plane has the same slopes in the i and j directions as the surface does will make the tangent plane a good approximation to the surface?

In a manner analogues to the linearization of functions of a single variable to approximate the value of a function of two variables in the neighbourhood of a given point (x0,y0,z0) where z=f(x,y) using a tangent plane. The tangent plane must pass through the point we wish to approximate z nearby and the second condition which I don't quite understand why it makes the tangent plane a good approximate is that it must have the same slopes as the surface in the i and j directions. Therefore Z≈ z0+fx(x-x0)+fy(y-y0)

I can understand that the equation of the tangent plane will serve as a good approximation if we are changing x and keeping y=y0 or vice versa. Namely, if we are varying y and keeping x=x0. Then the lines z=z0+fx(x-x0) and z=z0+fy(y-y0) of the line will serve as a good approximation to the curve traced in the vertical plane y=0 and x=x0. But since x and y are free to take any values what if we decided to change both x and y how can I prove that the tangent plane with these set of conditions will be a good approximation to the curve.

I can understand that the equation of the tangent plane will serve as a good approximation if we are changing x and keeping y=y0 or vice versa. Namely, if we are varying y and keeping x=x0. Then the lines z=z0+fx(x-x0) and z=z0+fy(y-y0) of the line will serve as a good approximation to the curve traced in the vertical plane y=0 and x=x0. But since x and y are free to take any values what if we decided to change both x and y how can I prove that the tangent plane with these set of conditions will be a good approximation to the curve.