Question

Hi! This is Calculus 3. I'm trying to understand linear approximation (I think). I have to find \[ \Delta z\], and write it in the form \[dz+\epsilon _1\Delta x+\epsilon _2 \Delta y\]. The example problem states that the two epsilons have limits of 0 as \[(\Delta x , \Delta y) \rightarrow (0,0)\]. I would appreciate knowing the significance of this! I have an equation \[z=4x^2y+2x^2\], but any z is fine to work with to demonstrate how stuff works, especially if it makes it simpler!

Any help, even conceptual help not related to any specific problem, would be appreciated!

Answer 1

Well, I'd imagine linear approximation in \(\mathbb{R}^3\) (3D graphh) would be a plane tangent to some point in the surface....

Answer 2

I'd also think you'd use the partial derivatives to find out the graph of that plane.

Answer 3

Alright! I can take a closer look at planes if that will help... Unless you mean linear approximation will help with planes.. I think I've memorized the formula for planes.

\[z=f_x(x,y)(x-x_0)+f_y(x,y)(y-y_0)\]

Answer 4

I think you mean \[
\Large L(x,y)= f(x_0,y_0) + f_x(x,y)(x-x_0)+f_y(x,y)(y-y_0)
\]

Answer 5

So we're working in \[R^3\] and so the delta in x and y will be from some ordered pair in the domain of f(x,y) to this plane.
I'll take your word for it :) So that equation is for linear approximation?

Answer 6

Since \(z=f(x,y)\)
\[ \large
f_x=\frac{\delta z}{\delta x}
\]

Answer 7

That formula gives you the equation of the plane tangent to \((x_0, y_0)\)

Answer 8

Alright, thanks. And are the terms added to f(x,y) are to adjust the z-value to the point on the tangent plane to f(x,y) at a certain \[(x_0,y_0)\]?

Answer 9

By adjust, I mean get to L(x,y) from f(x,y)...

Answer 10

terms added to f(x,y)?

Answer 11

Where L(x,y) = f(x0, y0) +.... Them :P
So are x0 and y0 the points at which we are creating a tangent plane?
I am a little lost in this process. Thank you for your help so far!

Answer 12

Consider that \[ \Large
f_x=\frac{\delta z}{\delta x} = \lim_{\Delta x \rightarrow 0}\frac{\Delta z}{\Delta x}
\]Might be why they are giving you certain info...

Answer 13

The \(f(x_0, y_0) \) is added just because it is needed. It's sort of similar to how you need the y intercept in a line, but a bit more complicated that at.

Answer 14

Hmm... I don't understand what the epsilons are exactly, unless they are \[f_x(x,y)\] and \[f_y(x,y)\].
I just reunderstood \[f_x\] now, thank you. ..Actually I might still be confused. With me, I feel like the conceptual stuff is what I need to understand, but it is difficult.

Answer 15

I think I'm going to continue on that which I have begun to understand thanks to you! Take care, and thanks for all your help!