# Representing A Plane Curve By A Vector Valued Function

I am given the function $x^2+y^2=25$, and I am suppose to write this as a vector valued function.

I have always been awful at these sort of problems, even with parametric equations, which requires the same process. I just don't understand the concept of "just let x=t;" in this particular case, that just doesn't seem to work.

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Do parametrization $$x = 5\cos t \\ y = 5\sin t$$ so your vector valued function is $$\mathbf r(t) = 5(\cos t\ \hat{\mathbf i}+\sin t \hat{\mathbf j})$$
How did you know what to set $x$ and $y$ to? Through practice? –  Mack Feb 25 '13 at 23:56
Almost certainly. You can always set one of the variables to $t$, and then solve for the other one -- but you will get a more complicated, multi-part expression. –  user7530 Feb 26 '13 at 0:01
@EliMackenzie If I see something like $x^2+y^2$ I immediately switch to polar coordinates, i.e. $\sin$ and $\cos$ since $\sin^2t+\cos^2t = 1$. If it's $x^2-y^2=1$ then I do $\cosh t$ and $\sinh t$ substitution, since, again $\cosh^2t - \sinh^2t = 1$. As for which is which, it doesn't really matter. I could do $x = 5\sin t, y = 5\cos t$ and it'd be the same. –  Kaster Feb 26 '13 at 0:06