Express the region $D=\{(x,y): x^2\leq y \leq x^2+x^3, x>0 \}$ as the union of cubic curves

Question

Let $D=\{(x,y): x^2\leq y \leq x^2+x^3, x>0 \}$

I know the family of curves $\gamma(t)=x^2+tx^3$ belong to D, for $t\in [0,1]$.

It is true that for every $(x,y)\in D$ there exist a unique $t\in[0,1]$ such that $y=x^2+tx^3$?

Note it would be equivalent to prove:

$t=\frac{y-x^2}{x^3}$ is always between $0$ and $1$ for $(x,y)\in D$.

I'm trying to prove the union of these curves is exactly D.

Answer 1

For every $(x,y)\in D$ it is not difficult to show that there is $\textit{no more than one}$ $t\in [0,1]$ such that $y=x^{2}+tx^{3}$. Indeed, suppose there are two values $t_{0},t_{1}\in [0,1]$ such that $$x^{2}+t_{0}x^{3}=x^{2}+t_{1}x^{3}$$ This implies $$t_{0}x^{3}=t_{1}x^{3}$$ Now, since no point in $D$ has a non-positive $x$-coordinate, we can cancel the $x^{3}$ from both sides of this equation, yielding $$t_{0}=t_{1}$$ That for a given $(x,y)\in D$ there is $\textit{at most one}$ value of $t\in [0,1]$ such that $y=x^{2}+tx^{3}$ follows from the definition of the set $D$: we know that $$x^{2}\leq y\leq x^{2}+x^{3}$$ and since the mapping $\varphi:[0,1]\rightarrow [x^{2},x^{2}+x^{3}]$ given by $$t\mapsto (1-t)x^{2}+t(x^{2}+x^{3})$$ is onto (this follows from the Intermediate Value Theorem and the fact that $\varphi$ is continuous), there must be some $t_{0}\in [0,1]$ such that $$y=\varphi(t_{0})=(1-t_{0})x^{2}+t_{0}(x^{2}+x^{3})=x^{2}+t_{0}x^{3}$$

Answer 2

Sketch: Let be $\alpha_2(x)=x^2+x^3$ and $\alpha_1(x)=x^2$, defined in $\Bbb{R_{>0}}.$

1. Note that, for every $x>0$, $\alpha_1(x)<\alpha_2(x)$, i.e., $\alpha_1(\Bbb{R_{>0}})\cap\alpha_2(\Bbb{R_{>0}})=\emptyset.$

2. If $(x_0,y_0)\in D$, then $y_0\in [x_0^2,x_0^2+x_0^3]$.

3. Let be $\lambda: [0,1]\to \Bbb{R}$ defined by $\lambda(t)=(1-t)x_0^2+t(x_0^2+x_0^3)$, i.e., the curve $\lambda$ connects $x_0^2$ to $x_0^2+x_0^3.$

4. Since $\lambda(0)=x_0^2$ and $\lambda(1)=x_0^2+x_0^3$ and $y_0\in [x_0^2,x_0^2+x_0^3]$, if $y_0\ne x_0^2$ and $y_0\ne x_0^2+x_0^3$, there is $t_0\in(0,1)$ such that $\lambda (t_0)=y_0$, by Intermediate value theorem.

So, for every $(x_0,y_0)\in D$,there is $t_0\in [0,1]$ such that $y_0=x_0^2+t_0x_0^3.$