Proving formula for approximation of a plane tangent to Z

[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 $z_0$

 

[jvicens]

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Thank you for your post. I am glad that you have a better understanding of forum notation now. I will try my best to answer your questions and provide some insights on how to prove the formula for n-dimensions.

Firstly, to prove the formula for n-dimensions, we need to use the concept of partial derivatives. In simple terms, partial derivatives measure the rate of change of a function with respect to one of its variables while holding other variables constant. In your formula, a and b represent the partial derivatives of the function f with respect to x and y, respectively. Therefore, in n-dimensions, we will have n partial derivatives, each with respect to a different variable.

Secondly, the difference between proving it for any function f(a) and \vec{a}, or as typically shown for standard parametric cycloids in terms of \vec{r}(t) is the type of function that we are dealing with. In the first case, we are dealing with a general function f, while in the second case, we are dealing with a specific function that can be expressed parametrically. The method of proving may differ depending on the type of function, but the general concept of using partial derivatives to measure the rate of change remains the same.

Thirdly, in step #4, you have correctly shown that z_0=k(z_0), where k=2 in this case. This is because z_0 is a constant, and any scalar coefficient times z_0 will simply be z_0. This holds true for any scalar in n-dimensions as well. In order to show this formally, you can use the definition of a scalar as a quantity that has magnitude but no direction, and therefore, it will not affect the position of the point in n-dimensions.

Lastly, for linear approximations in n-dimensions, the coefficients of x, y, and z may vary, but they will still be related to the partial derivatives of the function f. This is because the linear approximation is based on the tangent plane to the function at a specific point, and the partial derivatives give us the slope of the tangent plane in each direction.

I hope this helps in your understanding. If you have any further questions or concerns, please do not hesitate to ask. Keep up the good work in exploring forum notation!