I have to solve the following system:
$$ x'(t) \times y'(s) = 0 $$ $$ (y(s) - x(t)) \times y'(s) = 0 $$ Where: $$ x(t) = (cos(t), sin(t)) $$ $$ y(s) = (s, s^2) $$
using Newton's method for non-linear systems in GNU Octave. I deduce that what I have to do is a cross product, because the $$sin(x)$$ is 0 when the angle between the two vectors is 0: $$ x '(t) \times y' (s) = -2s \cdot sin (t) -cos (t) $$ $$ (y(s) -x (t)) \times y'(s) = 2 \cdot s ^ 2-2 \cdot s \cdot cos (t) + s ^ 2-sin (t) $$ However, when I take the initial guess with $$ (t_0, s_0) = (-3, -2) $$ iterating 10 times, I do not get the answer. Am I right to think that way?
If so, is the cross product between two curves calculated in this way?
To analyze the problem, consider the two equations $$f=2s \, \sin (t) +\cos (t)=0\tag1$$ $$g=3 s ^ 2-2 s \cos (t)-\sin (t)=0\tag2$$ From $(1)$, we have $s=-\frac{1}{2}\cot (t)$. Replacing in $(2)$ and simplifying, we end with $$g=\frac{1}{8} \csc ^2(t) (4 \sin (3 t)+3 \cos (2 t)-4 \sin (t)+3)\tag 3$$ and we end with the search of the zeros of $$h=4 \sin (3 t)+3 \cos (2 t)-4 \sin (t)+3\tag 4$$ If we graph functions $g$ or $h$,we can notice the problem. For $t <0$, there is no root between $-\pi$ and $0$. The first one seems to be close to $-3.9$ and the next is close to $-5.7$.
If we face such a problem in $1D$, I prefer not to think about what it could be in $2D$. If you prefer to stay in $2D$, use $$(t_0, s_0) = (-4.0, +0.5)$$ and calculations should converge.
For the $1D$ problem, you would have a very good approximation of the roots builting the Taylor expansion $$h=h(t_*)+\frac 12 h''(t_*) (t-t_*)^2+O((t-t_*)^2)$$ with $t_*=-\frac {3\pi}2$ where $h'(t)=0$. This would give, as estimates, $$t_\pm=-\frac{3 \pi }{2}\pm \frac{2}{\sqrt{13}}$$