How do we show inverse relationship while converting parametric equation? Good Day
Suppose we have to convert parametric equation $(\cos t, \sin t)$ into cartesian equation.
We realize that $\cos^2 t + \sin^2 t = 1$, so $x^2 + y^2 = 1$.
Sorry, if I ask, but I don't understand that when doing this, every parametric point lies on the graph of the equation, but how do we show that every point on the graph of the equation can be represented as a parametric point (i.e. there are no "extra" points on the graph)?
This is a relatively simpler example. For example, how do we show that every point on $(x + y - 7) ^ 2 + (x - y + 1)^ 2 = 100$ can be represented as a parametric form $(5 \cos t + 5 \sin t + 3, 5 \cos t - 5 \sin t + 4)$ for some $t$?
 A: Quite often, the parametric equations will not produce all the points of the graph. Or, saying this another way, the graph will contain “extra” points besides those generated by the parametric equations.
A very simple example: suppose the given parametric equations are $x=\sin(t)$ and $y= \sin(t)$. Then obviously every point generated by these equations lies on the line $y=x$. But it’s also obvious that $x(t) \le 1$ and $y(t) \le 1$ for all values of $t$, so there are many points on the line $y=x$ that are not “covered” by the parametric equations. The point $(x,y) = (7,7)$, for example.
In general, suppose we’re given parametric equations $x=u(t)$ and $y= v(t)$. We may be able to find a function $F:\mathbb R^2 \to \mathbb R$ such that $F(u(t), v(t)) = 0$ for all $t$. This shows that the curve generated by the parametric equations lies within the set $S = \{(x,y) \in \mathbb R^2 : F(x,y)= 0 \}$. But to show that the curve covers all of $S$ we have to show that for any given point $(a,b) \in S$ there exists a parameter value $t$ such that $u(t)=a$ and $v(t) = b$. This is often not easy to do.
