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So as it is, I'm now starting to cover vector-valued functions in my Calculus III class. While studying the topic, I noticed that it seemed to be the exact same thing as parametric equations. I know that I am probably missing an important difference between the two topics, but I can't seem to figure it out.

So the question is:

What is the difference between a set of parametric equations and a vector-valued function?

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The difference is that a parametrization has some extra properties. A vector valued function is a map $$f:U\subset\mathbb R^m\to V\subset\mathbb R^n$$

And parametric equations for a [portion of a] submanifold $M$ in Euclidean space (it's rare to parametrize things other than manifolds) is a map $$\varphi:U\subset\mathbb R^m\to M\subset\mathbb R^n$$ Where:

  • $U$ is open
  • $\varphi$ is a homeomorphism onto its image
  • $\operatorname{rank}D\varphi = m$ everywhere

What we could say then, is that a parametrization is always in the form of a vector valued function, but conversely, we use vector valued functions with nice properties to parametrize varieties.

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    $\begingroup$ Downvote? What did I say wrong? $\endgroup$ – GPerez Feb 3 '14 at 21:41
  • $\begingroup$ Why is a parametric derivative different than the derivative of a vector-valued function? There doesn't seem to be anything in your definition of parametric that breaks the symmetry between the components of the vector-valued output, yet they are treated differently in the parametric derivative. For example, y is treated as a function of t and x, while x is treated as a function of t. $\endgroup$ – user10478 Oct 21 '18 at 0:27
  • $\begingroup$ @user10478 I have never heard of the term "parametric derivative" and to be honest I can't make sense of the rest of your comment either, sorry. $\endgroup$ – GPerez Oct 22 '18 at 11:55
  • $\begingroup$ What did you say wrong? Someone learning calculus doesn't know about what is an homepomorphism or an algebraic structure. This is what is wrong. $\endgroup$ – İbrahim İpek Aug 13 at 11:52

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