Parametric equations intuition

Question

If I paramaterise a function like x= f(t) and y=g(t) like x=t and y=t^2 why why does eliminating the parameter give you the function that the parametric equations? It may be clear for y=x^2 but i'm not why it is the case for functions like x=cost and y=sint.

Answer 1

You have two mistakes in your premise that "eliminating the parameter give you the function [for] the parametric equations."

First, eliminating the parameter may not result in a function, meaning that $y$ may not be a function of $x$. In your example $x=\cos t$ and $y=\sin t$, eliminating the parameter ends up with the equation

$$x^2+y^2=1$$

But that is not the graph of a function, since for any $y$ where $-1<y<1$ there are two corresponding values of $x$, not just one. The graph is a circle, which does not pass the vertical line test. The same is true if we try to make $x$ a function of $y$. Therefore we end up with a relation between $x$ and $y$ but not a function.

Your second mistake is that eliminating the parameter from the two equations resulting in one equation in just $x$ and $y$ may end up with an Cartesian equation with a graph different from the original parametric one. For example, consider the parametric equations

$$x=t^2, \quad y=t^2+1$$

Eliminating $t$ from those equations, in any of several ways, results in the Cartesian equation

$$y=x+1$$

The graph of that is a line that has parts in the first, second, and third quadrants. However, the original parametric equations define a ray that lies only within the first quadrant (and one point on the positive $y$ axis, $(0,1)$). The Cartesian equation contains the parametric equations but is not equivalent to them.

So a correct statement is to say that eliminating the parameter variable (when possible) results in a Cartesian equation that whose graph contains the graph of the original parametric equations. I.e. all points in the parametric graph satisfy the Cartesian equation, but additional points may also satisfy it.

Is that clear? And do you understand why my final statement is true?