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Proving formula for approximation of a plane tangent to Z

  1. Jun 12, 2013 #1
    I think I've got the basics of forum notation now, thanks to Fredrick from my other thread. Here goes:

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

    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 [itex]f(a)[/itex] and [itex]\vec{a}[/itex], or as is typically shown for standard parametric cycloids in terms of [itex]\vec{r}(t)[/itex]?

    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 [itex]\frac{\partial f}{\partial z} \approx \frac{\partial f}{\partial x}(x-x_0) + \frac{\partial f}{\partial y}(y-y_0)[/itex], then

    [itex]f_z \approx f_x\Delta x + f_y\Delta y[/itex]

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

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

    #3) [itex]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}][/itex]
    [itex]= 2z_0 + \frac{\partial f}{\partial x}(x-x_0) + \frac{\partial f}{\partial y}(y-y_0)[/itex]

    #4) Now I've got to show that [itex]z_0=k(z_0)[/itex], where, in this case, [itex]k=2[/itex] 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 [itex]z_0[/itex] will simply be [itex]z_0[/itex]. 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 [itex]z_0=k(z_0)[/itex], the formula is valid.

    I'm speculating that for linear approximations in n-dimenions [itex]x[/itex] and [itex]y[/itex] will have varied, yet interrelated scalar subscripts as will the coefficient of [itex]z_0[/itex]
  2. jcsd
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