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?


2 Answers 2


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.

  • 3
    $\begingroup$ Downvote? What did I say wrong? $\endgroup$
    – GPerez
    Feb 3, 2014 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, 2018 at 0:27
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    $\begingroup$ @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, 2018 at 11:55
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    $\begingroup$ 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. $\endgroup$ Aug 13, 2019 at 11:52
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    $\begingroup$ True, but answers don’t always have to be aimed at the OP. Answers are written for everyone who might see this thread in the future. If someone has an insightful way of looking at something, it can be helpful to share. I got something out of this answer. $\endgroup$
    – littleO
    Sep 29, 2021 at 7:01

They are basically the same (different ways of expressing the same idea).

In the equation below, the LHS is a vector-valued function, and the RHS is a parametric form of the same function.

$$\vec{v} = \pmatrix{\sin{(t)} \\ \cos{(t)} \\ t} \equiv \matrix{x(t) = \sin{(t)} \\ y(t) = \cos{(t)} \\ z(t) = t}$$


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