I have 2 problems in the theme double point that I didn't understand.
1)How can I show that a curve has no double point at all?
2)How can I show that a curve has only 1 double point( or more if needed)?
If its ok can you explain in this example:
$$x=(t^2+π^2) \cos(t)$$
$$y=(t^2+π^2) \sin(t)$$
$$t\in \left[-\dfrac {4\pi}3,π \right]$$
Show that the curve have only 1 double point.
You need to show that there is only one pair $t_1\neq t_2$ in your domain such that $$ x(t_1)=x(t_2) $$ and $$ y(t_1)=y(t_2) $$ These two conditions lead to $\cos(t_2)/\cos(t_1)=\sin(t_2)/\sin(t_1)$ or $\tan(t_2)=\tan(t_1)$. This implies $t_2=t_1+k\pi$ for some integer $k$. $k=\pm 1$ does not work because $\sin$ and $\cos$ take opposite values so if you had $x(t_1)=x(t_2)$ and $y(t_1)=y(t_2)$ you would need both $\cos t_1=\sin t_1=0$ which is impossible. $k=\pm 2$ requires $t_1=-\pi$ hence $t_2=\pi$, which is a double point. Larger $k$'s are not allowed because the length of your $t$-domain is less than $3\pi$.
The only double point corresponds to $t_1=-\pi$