# Thread: linear approximation of mapping

1. ## linear approximation of mapping

question says to find the linear approximation of the given mapping at the indicated point

$\displaystyle f(x,y) = (xe^y + cos(y), x, x+e^y); (1,0)$

is this related to the tangent plane to a graph

just take it and use it in this formula

$\displaystyle z = f(x_{0}, y_{0}) + \frac{\partial f}{\partial x}(x_{0},y_{0})(x-x_{0}) + \frac{\partial f}{\partial y}(x_{0},y_{0})(y-y_{0})$

2. ## Re: linear approximation of mapping

Yes, and then substitute the point in.

3. ## Re: linear approximation of mapping

actually leads me to another question

$\displaystyle f(x,y)$ has 3 outputs

how does that work with a point in 2-space?

$\displaystyle z= (2,1,2) + (1,1,1)(x-1)+(1,0,1)(y-0)$ ??

4. ## Re: linear approximation of mapping

A linear function, from $\displaystyle R^n$ to $\displaystyle R^m$ can be represented by an n by m matrix.

In particular the linear approximation to $\displaystyle f(x, y)= (xe^y+ cos(y), x, x+ e^y)$ at any point, (x, y), has as "slope"
$\displaystyle \begin{bmatrix}e^y & xe^y- sin(y) \\ 1 & 0 \\ 1 & e^y\end{bmatrix}$

At (1, 0) that is $\displaystyle \begin{bmatrix} 1 & 1 &\\ 1 & 0 \\ 1 & 1\end{bmatrix}$

Since f(1, 0)= (1+ 1, 1, 1+ 1)= (2, 1, 2), the approximation to the function at (1, 0) is
$\displaystyle \begin{bmatrix}2 \\ 1 \\ 2\end{bmatrix}+ \begin{bmatrix} 1 & 1 &\\ 1 & 0 \\ 1 & 1\end{bmatrix}\begin{bmatrix}x-1 \\ y \end{bmatrix}$
$\displaystyle = \begin{bmatrix}2+ (x- 1)+ y\\ 1+ (x- 1) \\ 2+ (x- 1)+ y\end{bmatrix}= \begin{bmatrix}x+ y+ 1 \\ x\\ x+ y+ 1\end{bmatrix}$.

That is exactly your $\displaystyle z= (2, 1, 2)+ (1, 1, 1)(x- 1)+ (1, 0, 1)(y- 0)$