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The article for Inverse function theorem uses this function in an example:

${\displaystyle F(x,y)={\begin{bmatrix}{e^{x}\cos y}\\{e^{x}\sin y}\\\end{bmatrix}}}$

This current article (Vector-valued function) uses this notation:

${\displaystyle \mathbf {r} (t)=f(t)\mathbf {\hat {i}} +g(t)\mathbf {\hat {j}} }$

The Inverse function theorem and Jacobian articles seem to be using a kind of shorthand.

If they are accurate, it could be helpful in deciphering the column-vector notation to have something like the following in this current article:

${\displaystyle \mathbf {r} (t)=f(t)\mathbf {\hat {i}} +g(t)\mathbf {\hat {j}} ={\begin{bmatrix}{f(t)}\\{g(t)}\\\end{bmatrix}}}$

${\displaystyle \mathbf {r} (x,y)=f(x,y)\mathbf {\hat {i}} +g(x,y)\mathbf {\hat {j}} ={\begin{bmatrix}{f(x,y)}\\{g(x,y)}\\\end{bmatrix}}}$

Ac44ck 17:28, 18 May 2007 (UTC)[reply]

The column-vector notation seems to be shorthand for matrix multiplication an implied row-vector (which consists of unit spacial-vectors) by the column-vector:

${\displaystyle \mathbf {r} (x,y)=f(x,y)\mathbf {\hat {i}} +g(x,y)\mathbf {\hat {j}} ={\begin{bmatrix}\mathbf {\hat {i}} ~~\mathbf {\hat {j}} \end{bmatrix}}*{\begin{bmatrix}{f(x,y)}\\{g(x,y)}\\\end{bmatrix}}}$

Ac44ck 19:38, 18 May 2007 (UTC)[reply]

The graph shows the curve traced through space by the end of a vector rather than showing a vector. Interpreting the graph of the given vector-valued function could be helped by showing a vector for some value of 't'. Ac44ck 23:33, 19 May 2007 (UTC)[reply]

A vector-valued function does not necessarily take only scalar input arguments, as asserted by this article. This point was in part brought up by a previous comment on this page written in 2007, yet the problem persists. In general, vector valued function have both operands as vectors. Even wikipedia's own article on jacobians mentions this. See http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant. Further, this article states the input operand is a real number. Such a constraint is of course ridiculous.

Agreed, I have fixed the first paragraph, so it does not make such a ridiculous claim. However, I am afraid that the article still focuses too heavily on this specific scalar parameter case. Well at least its not claimed to be the only type of Vector valued function to exist anymore.MATThematical (talk) 18:11, 22 February 2010 (UTC)[reply]