I'm struggling to understand how to "demonstrate" that a transformation maps one region onto another. For example:
Consider the region $T$ given by $0≤p≤1$, $0≤q≤1$. Let $R$ be the region $0≤x≤1$, $0≤y≤1$.
Question: Demonstrate that the transformation $p(x) = 4x -4x^2$, $q=y$ maps $R$ onto $T$.
I wasn't exactly sure as what to do. I toyed around a bit and got:
$$dp = (4-8x)dx$$ and $$dq =dy$$
To be honest, I have no clue what I was doing as I haven't come across this style of question. If anybody could shed some light on how to tackle this type of question, it would be appreciated.
Thanks
There is no single fool proof method for problems of this kind. At any rate differentiation might by an auxiliary move in certain situations, but is not required a priori. We are given the map $$f:\quad{\mathbb R}^2\to{\mathbb R}^2,\qquad (x,y)\mapsto (p,q):=(4x-4x^2,y)\ ,$$ and are told to prove that $f(T)=R$.
Since $q(x,y)=y$ we immediately can infer that a horizontal segment $y=c\in[0,1]$ in $T$ is mapped to the segment $q=c$ in $R$.
The graph of the function $$x\mapsto p(x)\qquad (0\leq x\leq 1)\tag{1}$$ in the (auxiliary) $(x,p)$-plane is an arc of a parabola through the points $(0,0)$ and $(1,0)$, and culminating at $\bigl({1\over2},1\bigr)$. (In order to obtain this culmination point we had to differentiate!) It follows that the function $(1)$ takes all values in $[0,1[\ $ exactly twice, the value $1$ exactly once, and does not take any values outside the interval $[0,1]$.
Altogether we can conclude that $f$ maps each segment $y=c\in[0,1]$ in $T$ onto the corresponding segment $q=c$ in $R$. Therefore the claim $f(T)=R$ is true.