What is the image of the set $$A=\{ (x,y) : 0\le x \le a\ , \ 1\le y\}$$
under the transformation $(x,y) \rightarrow (u,v)$ where $$u=x/y$$ $$v=x$$
The parameter $a$ is positive. I got a 'triangle' in the $(u,v)$ plane, that is the bit below the diagonal of the square with side $a$ with its lower left corner at the origin. But I'm having trouble quickly seeing this, is there a nice method for doing such exercises?
You should get a triangle in the $u$-$v$ plane that is above the diagonal of the square you describe, which has vertices at the origin, at $(a, a)$ and at $0, a$. It does not however, contain $(0, a)$: we get that for $a = 1$, $$(x, y) = (0.25, 100) \mapsto (u, v) = (0.0025, 0.25)$$