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I have a vector $\tilde{p} = (x,y,z)$ (homogenous coordinates). The corresponding non-homogenous vector is $p = (x/z, y/z)$.

Now the $\tilde{p}$ is a result of some linear transform $R(\theta)$ of another vector $\tilde{u}$:

$$\tilde{p} = R\tilde{u}$$

I need to compute this derivative:

$$\frac{\partial p}{\partial \theta}$$

According to chain rule:

$$\frac{\partial p}{\partial \theta} = \frac{\partial p}{\partial \tilde{p}} \frac{\partial \tilde{p}}{\partial \theta}$$

The second partial derivative from the product on the right side is pretty obvious to me:

$$\frac{\partial \tilde{p}}{\partial \theta} = \frac{\partial}{\partial \theta}R\tilde{u}$$

But I don't understand the first one which should be (according to textbook):

$$\frac{\partial p}{\partial \tilde{p}} = \frac{\partial [x/z\quad y/z]}{\partial [x\quad y\quad z]} = \begin{bmatrix} 1/z & 0 & -x/z^{2} \\ 0 & 1/z & -y/z^{2} \end{bmatrix}$$

and what are the $x,y,z$ here? Does it come from $\tilde{u}$ or from $\frac{\partial}{\partial \theta}R\tilde{u}$ ? How is the matrix built?

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up vote 2 down vote accepted

You have two maps, namely $$f:\quad {\mathbb R}\to{\mathbb R}^3,\qquad \theta\mapsto \tilde p(\theta):=R(\theta)\tilde u\, $$ and $$g:\quad {\mathbb R}^3\to{\mathbb R}^2,\qquad \tilde p\mapsto p\ ,$$ and you want to know the derivative of the composition $$h:=g\circ f:\quad{\mathbb R}\to{\mathbb R}^2,\qquad \theta \mapsto g\bigl(f(\theta)\bigr) \ .$$ When we express everything in coordinates then we don't even need the chain rule: The map $f$ appears in the form $$\theta\mapsto \tilde p(\theta)=\bigl(x(\theta),y(\theta),z(\theta)\bigr)\ ,$$ and $g$ is given by $$(x,y,z)\mapsto\cases{\xi:=x/z \cr \eta:=y/z \cr}\ .$$ Therefore $h$ appears in coordinates as $$\theta\mapsto\cases{\xi(\theta):=x(\theta)/z(\theta) \cr \eta(\theta):=y(\theta)/z(\theta) \cr}\ ,$$ and we can compute $h'(\theta)=\bigl(\xi'(\theta), \eta'(\theta)\bigr)$ just using the quotient rule.

Using the chain rule it would work the following way: The chain rule says $$dh(\theta)=dg\bigl(f(\theta)\bigr)\circ df(\theta)\ .$$ Therefore $$h'(\theta)= dh(\theta).1=dg\bigl(f(\theta)\bigr)\circ df(\theta).1 =dg\bigl(f(\theta)\bigr).f'(\theta)\ .$$ The matrix $\bigl[dg\bigr]_{(x,y,z)}$ computes to $$\bigl[dg\bigr]_{x,y,z)}=\left[\matrix{{\partial\xi\over\partial x}&{\partial\xi\over\partial y}&{\partial\xi\over\partial z} \cr {\partial\eta\over\partial x}&{\partial\eta\over\partial y}&{\partial\eta\over\partial z}\cr}\right]_{(x,y,z)} =\left[\matrix{1/z&0&-x/z^2 \cr 0 &1/z&-y/z^2\cr}\right]\ .$$ When you put everything together you will see that both ways lead to the same end result.

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I did not realize that derivative of the $\mathbb{R}^{3}\rightarrow \mathbb{R}^{2}$ yelds that $2\times 3$ jacobian matrix. I like the first solution more, but thanks for providing the clear explanation of the latter, too! – Libor Feb 28 '12 at 14:35

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