finding the differentialF

Poirot

Banned
let f:R^3->R^2 be given by $f(x,y,z)=(x^2+y^2+z^2,xyz)$ I want to find it's differential at a point (x,y,z). I can find the represnting matrix w.r.t standard basis i.e. the matrix of partial derivatives but how do I use this to find ,say, the differential of f at (1,0,-1)?

Ackbach

Indicium Physicus
Staff member
Re: finding the differential

You can do
$$df= \frac{ \partial f}{ \partial x}dx+ \frac{ \partial f}{ \partial y}dy+ \frac{ \partial f}{ \partial y}dy,$$
where each sum is vector addition. That is, $df$ is a vector made up of
$$df= \begin{bmatrix} df_{1} \\ df_{2}\end{bmatrix},$$
and you follow the usual rules for finding the differential of each component.

Poirot

Banned
Re: finding the differential

Is it usual just to present it as the matrix? Also, how do you recover the vector in R^2 from the matrix?

Poirot

Banned
Re: finding the differential

So the differential at (1,0,-1) (in matrix form) is [2,0,-2]
.................................................................... [0,-1,0]

but what is this in R^2?

Ackbach

Indicium Physicus
Staff member
Re: finding the differential

I get
$$df(1,0,-1)=\begin{bmatrix}2\,dx-2 \, dy \\ -dy \end{bmatrix}.$$
As a slight correction to Jameson's post, I think I would write it as
$$df= \begin{bmatrix} f_{1x} & f_{1y} & f_{1z} \\ f_{2x} &f_{2y} &f_{2z}\end{bmatrix} \begin{bmatrix} dx \\ dy \\ dz \end{bmatrix}.$$

Jameson

Yep, that's exactly it. The dimensions of my matrix didn't work with the function mapping R^3 to R^2. You're also right that the (x, y, z) coordinate should be evaluated for each partial derivative in the matrix. Thanks for clearing this up. 