$A = \{ {(x^{2}-y^{2})((x-1)^{2}+y^{2}-1)=0}\}$
I know there are many definitions behind tangential vectors, but I don't know how to use them, especially when I can't make x and y independent.
$A = \{ {(x^{2}-y^{2})((x-1)^{2}+y^{2}-1)=0}\}$
I know there are many definitions behind tangential vectors, but I don't know how to use them, especially when I can't make x and y independent.
Let f by the function you have, so that A is the set of $(x,y)$ where f=0.
Note that the normal to the curve at $(x,y)$ must be proportional to $\nabla f$. So the tangent at that point must satisfy $$t \cdot \nabla f = 0$$ If $t_1$ and $t_2$ are the x and y components of the tangent vector, then the equation above yields their ratio. $$\frac{t_1}{t_2}=\frac{-f_y}{f_x}$$ where $f_x$ and $f_y$ are partial derivatives.
So you have the expression for the tangent at any $(x,y)$ on the curve. I think I can leave the computation of the two partials to you. It is simple in the case of this given function.