Suppose I have the following triangle in $\mathbb{R}^2$, $ A = \{ (x,y) \, | \, 0 \leq y \leq x \leq 1 \}$. I now perform the coordinate transformation $ z = \frac{y}{x}, u= x $. I now want to express the set $A$ as coordinates $(u,z)$, thus $ A = \{ (u,z) \, | \, \dots\}$. What will be the expression for this?
Asked
Active
Viewed 114 times
0
-
What happens with the origin? Where $x=y=0,z=\frac{0}{0}$? So, you mean: $$A=\left{(\cos\alpha,\sin\alpha), \alpha\in\left[0+2k\pi,\frac{\pi}{4}+2k\pi\right],k\in\mathbb Z\right},;\text{here};k=0?$$ – PinkyWay Mar 29 '20 at 14:33
-
We set z=0 if x=0 – PROB123 Mar 29 '20 at 14:36
-
Then my trigonometric notation doesn't make sense since $\sin\alpha=\cos\alpha$ for $\alpha=\frac{\pi}{4}+2k\pi$ – PinkyWay Mar 29 '20 at 14:43
1 Answers
0
Consider the map $[0,1]\times[0,1]\to A$ given by $$ (u,z)\to (z, uz)\ .$$
- It is well defined,
- maps $(0,0)$ to $(0,0)$, but also maps each $(u,0)$ to $(0,0)$, this is for the purposes of an integration irrelevant,
- it is injective on the part of all $(u,z)$ with $z\ne 0$, since from $(z,uz)=(z',u'z')$ we get immediately $z=z'$, then $u=u'$,
- it is surjective.
It is thus simpler to write the coordinate transformation on the part of $A$ of all $(x,y)$ with $0<y\le x\le 1$.
dan_fulea
- 32,856