The graph of a function z=f(x,y) can always be parametrized as (x,y,f(x,y)). Show that in this case the "old way" (using the derivative and gradient) and the "new way" (Parametrizing the surface and using T vectors) for finding the tangent plane at a point always agree.
I have no idea how to do this problem. I think he wants us to show that two different ways lead to the same tangent plane at a point. one is using the partial derivatives, the second is doing it via T vectors. Can someone run me through the steps? Thanks in advance!
The "old way", was to define a surface.
$S: F(x,y,z) = k$ and then find $dS = \nabla F = (\frac {\partial F}{\partial x},\frac {\partial F}{\partial y},\frac {\partial F}{\partial z})$
And the "new way" defines the surface parmetrically.
$x = f(s,t)\\ y = g(s,t)\\ z = h(s,t)$
$dS = (\frac {\partial f}{\partial s},\frac {\partial g}{\partial s},\frac {\partial h}{\partial s})\times(\frac {\partial f}{\partial t},\frac {\partial g}{\partial t},\frac {\partial h}{\partial t})$
But there is no reason $x,y$ can't be your parameters.
Show that:
$\nabla (z- f(x,y)) = (1,0,\frac {\partial z}{\partial x})\times(0,1,\frac {\partial z}{\partial y})$