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theEric
 4 years ago
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!
theEric
 4 years ago
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!

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well, I'd imagine linear approximation in \(\mathbb{R}^3\) (3D graphh) would be a plane tangent to some point in the surface....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'd also think you'd use the partial derivatives to find out the graph of that plane.

theEric
 4 years ago
Best ResponseYou've already chosen the best response.0Alright! 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. \[z=f_x(x,y)(xx_0)+f_y(x,y)(yy_0)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think you mean \[ \Large L(x,y)= f(x_0,y_0) + f_x(x,y)(xx_0)+f_y(x,y)(yy_0) \]

theEric
 4 years ago
Best ResponseYou've already chosen the best response.0So 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Since \(z=f(x,y)\) \[ \large f_x=\frac{\delta z}{\delta x} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That formula gives you the equation of the plane tangent to \((x_0, y_0)\)

theEric
 4 years ago
Best ResponseYou've already chosen the best response.0Alright, thanks. And are the terms added to f(x,y) are to adjust the zvalue to the point on the tangent plane to f(x,y) at a certain \[(x_0,y_0)\]?

theEric
 4 years ago
Best ResponseYou've already chosen the best response.0By adjust, I mean get to L(x,y) from f(x,y)...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0terms added to f(x,y)?

theEric
 4 years ago
Best ResponseYou've already chosen the best response.0Where 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!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Consider 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...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The \(f(x_0, y_0) \) 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.

theEric
 4 years ago
Best ResponseYou've already chosen the best response.0Hmm... 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.

theEric
 4 years ago
Best ResponseYou've already chosen the best response.0I 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!
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