# Vector functions: solving for curves of intersection

 HW Helper P: 2,616 Vector functions: solving for curves of intersection x = t works because we usually assume $$t \in R$$ without having to write that our explictly (though you should). The "parameter function" f(t) has to be able to take on the all possible values of x. For example, consider the unit circle at the origin. x^2+ y^2 = 1. A parametric representation of this is: x = cos t y = sin t. We know that this works because cos t can take on any value betwee 1 and -1, just like sin t. But more importantly we know that his parametric representation satisfies the equation of the circle. Of course no one says you can't use a parametric representation like: $$x = \sin (2t)$$ $$y = \cos (2t)$$, just that in this case t isn't the angle between the x-axis and the line from the point (x,y) and the origin. Both these parametric representations satisfy the equation of the circle as well.