# Proving formula for approximation of a plane tangent to Z

1. Jun 12, 2013

### CubicFlunky77

I think I've got the basics of forum notation now, thanks to Fredrick from my other thread. Here goes:

Show: $Z = z_0 + a(x-x_0) + b(y-y_0)$ where $a = f_x = \frac{\partial f}{\partial x}$ and $b = f_y = \frac{\partial f}{\partial y}$

I'm attempting this using the coordinate method, but how would I prove it generally (for n-dimensions)?

What is the difference between proving it for any function $f(a)$ and $\vec{a}$, or as is typically shown for standard parametric cycloids in terms of $\vec{r}(t)$?

So the coordinate proof is fairly straightforward, but there is an assumption at step #4 that I believe I'm showing incorrectly;

What I've done:

#1) If $\frac{\partial f}{\partial z} \approx \frac{\partial f}{\partial x}(x-x_0) + \frac{\partial f}{\partial y}(y-y_0)$, then

$f_z \approx f_x\Delta x + f_y\Delta y$

#2) If we let $L_1 + L_2$ be the sum of the linear systems that constitute a plane approximately tangent to $f(x,y)=Z$

then $L_1 = \frac{\partial f}{\partial x}(x_0,y_0)$$$~~\Rightarrow~~$$$$\left\{\begin{array}{1}z_0 + \frac{\partial f}{\partial x}(x-x_0)\\y=y_0\end{array}\right.$$ and $L_2 = \frac{\partial f}{\partial y}(y,y_0)$ $$~~\Rightarrow~~$$$$\left\{\begin{array}{1}z_0 + \frac{\partial f}{\partial y}(y-y_0)\\x=x_0\end{array}\right.$$

#3) $Z = L_1 + L_2 = 2z_0 + [\frac{\partial f}{\partial x}(x-x_0) + \underbrace{\frac{\partial f}{\partial x}(x_0-x_0)}_{= 0}] + [\frac{\partial f}{\partial y}(y-y_0) + \underbrace{\frac{\partial f}{\partial y}(y_0-y_0)}_{=0}]$
$= 2z_0 + \frac{\partial f}{\partial x}(x-x_0) + \frac{\partial f}{\partial y}(y-y_0)$

#4) Now I've got to show that $z_0=k(z_0)$, where, in this case, $k=2$ but in general it can be anything. This might seem like a dumb question, but is it alright for me to assume that any scalar coefficient times $z_0$ will simply be $z_0$. In three dimensions it seems to make perfect sense since a scalar (or any magnitude value) in space represents a point that will be the exact same point regardless of its coefficient. But how could I show this formally?

#5 Assuming $z_0=k(z_0)$, the formula is valid.

I'm speculating that for linear approximations in n-dimenions $x$ and $y$ will have varied, yet interrelated scalar subscripts as will the coefficient of $z_0$