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As an example consider $x(t)=t+\sin(t)$ and $y(t)=\cos(t)$ and the graph plotted between $t=0$ and $t=2\pi$. The parametric curve looks like 2 half hills. enter image description here

But if I work out $x^2 +y^2$ I get $t^2 + 2 t \sin(t) + \sin^2(t)$. This is always greater than one and so represents a circle of varying radius dependant on $t$. For a start why is the parametric equations not in one-to-one correspondence with the implicit function but also if I plot any of the implicit functions (for any $t$) then I do not get the same graph as the parametric graph.

I'm very confused. Would anyone be able to explain this?

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  • $\begingroup$ I got $x^2+y^2=t^2+2t\sin t+1$. But that does not represent a circle. Why should it? There is no such implicit function. To that end you would have to eliminate $t$, and that seems somewhat non-trivial. I guess you could try $t=\arccos y$ and then $x=\arccos y\pm\sqrt{1-y^2}$. In addition you would have to deal with the branches of inverse cosine. To summarize: I would deal with this curve only in the parametric form and forget explicit/implicit functions. $\endgroup$ – Jyrki Lahtonen Jan 8 '20 at 21:52
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    $\begingroup$ No you don't. When you substitute different values of $t$ you get different points on that curve (it is actually a cycloid). To get the impllicit function you need an equation on $x$ and $y$ alone, where $t$ disappears. It didn't disappear when you calculated $x^2+y^2$ so you can deduce that the curve is not a circle centered at the origin. $\endgroup$ – Jyrki Lahtonen Jan 8 '20 at 21:59
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    $\begingroup$ If, instead, your curve were parametrized as $x(t)=\cos t$, $y(t)=\sin t$, then you could see that $x^2+y^2=1$. The parameter $t$ disappeared, and now you can use that equation to do implicit differentiation and whatnot. Similarly, with parametrization $x=t^2-1$, $y=t^3-t$ you can show that $y^2=x^3+x^2$. Again $t$ disappeared from this equation and you can use it to differentiate implicitly. $\endgroup$ – Jyrki Lahtonen Jan 8 '20 at 22:02
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    $\begingroup$ For the purposes of implicit differentiation and such ALL the points on the curve must satisfy the SAME equation on $x$ and $y$ alone. No $t$. $\endgroup$ – Jyrki Lahtonen Jan 8 '20 at 22:03
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    $\begingroup$ Well, you do, but only because you’re doing something nonsensical. The $x$ and $y$ on the left-hand side started out as functions of $t$, so when you fix a specific value of $t$ you also have to set $x$ and $y$ accordingly. When you do that you end up with a tautology: the square of the distance of a point on the curve to the origin is equal to itself. Since the minimal distance from the origin is $1$, it should come as no surprise that the expression on the right-hand side is also never less than $1$. $\endgroup$ – amd Jan 8 '20 at 23:45
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Functions can be of either implicit or parametric representation. An implicit function is basically obtained by eliminating parameter t between them as in the cited example

$$x(t)= \cos^{-1} y(t)+ \sqrt{1-y(t)^2}$$

leaving no trace with any one to one mapped correspondences.

The parametrization of an implicit relation is by no means unique.

For example circle $x^2+y^2=1 $ has no unique parametrization. We have $ (x=\cos t, y=\sin t), (x= sech\, t, y=tanh\, t) $

But inter-mappings can be among parametric representations belonging to same curve.

But in any case mapping is not by finding distance to origin where parameter t is still available.

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