# Given any parametric curve, finding its general form?

I'll illustrate the problem I'm trying to solve with an example.

Let's consider the equations

$$x = \cos (t)$$ $$y = \sin (t)$$

We know that these are a parametric form of the unit circle.

In this case, we can actually prove that without plotting because if we square and add the two equations, we get

$$x^2 + y^2 = \cos^2 (t) + \sin^2 (t) = 1$$

which is an equation in $x, y$ and describes the circle of radius $1$. But we didn't do it by some formal method, just by recognition.

I know that parametric equations that describe straight lines can easily be solved by substitution. For example,

$$x = 2t$$ $$y = 4t + 2$$ $$\implies y = 2x + 2$$

But what other methods exist of generally finding an algebraic equation if we are given parametric form of a curve?

The example that I'm particularly struck at is

$$x = t\cos(t)$$ $$y = t\sin(t)$$

If I try to square and add as I did for the example of the unit circle, I get a surface for a cone, whereas the answer actually is a spiral. So that's a wrong method.

But in general, what is the strategy to solve any given parametric curve's general/cartesian form?

• In general parametrised curves don't have any other useable representation. – Henrik supports the community May 18 '16 at 19:38
• ""In this case, we can actually prove that without plotting because if we square and add the two..." Actually, this doesn't prove the parametric equations give you the whole circle. In some cases functions satisfying $x(t)^2+y(t)^2 = 1$ give you only a part of the circle. – Michael Hardy May 18 '16 at 20:06
• @Henrik but what if we can verify that the parametric equations are resulting in a curve? Then the curve would definitely have an Cartesian form, correct? – Peeyush Kushwaha May 19 '16 at 7:28