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Given the curve $C$, $C = {(x,y):x^2+y^2=1}$, $n=\langle x,y\rangle$ is normal to $C$.

Consider the vector field $F$ defined by $F=\langle y,-x\rangle$.

Is the vector field $F$ tangent to $C$ or normal to $C$ at points on $C$?

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What happens at $(1,0)$? – Matt Nov 20 '12 at 3:54
what do you mean? – Jack F Nov 20 '12 at 3:56
the vectors are orthogonal but I don't get it since <y,-x> shows vectors that go in the circle style – Jack F Nov 20 '12 at 3:58
Draw the vector field $F$ and it becomes obvious. – Herp Derpington Nov 20 '12 at 5:33
If $n$ is normal to $C$, and also $F$ is orthogonal to $n$, then $F$ is "normal to the normal", making it actually tangent (in dimension 2). – coffeemath Nov 20 '12 at 7:15

It might be helpful to visualize what's happening here: for example, graphing the unit circle ($x^2 + y^2 = 1$) and the vector field $F$. Then consider what is happening at given points of the circle, e.g. at $(1, 0)$, and $(0, 1$.

In your comment above you observed that the vectors in $F$ go in "circle style". I think what you are observing each vector in $F$ is tangent to $C$, and tangent at some point $(x, y)$ of $C$, with each vector directed counter-clockwise.

We know that for each point $(x, y)$ that lies on $C$, the vector $n=\langle x, y\rangle$ is normal to $C$ (it's a given) at that point, and so at the point $(1, 0)$, $n$ lies along the $x$-axis, pointing in the positive $x$ direction. The vectors in $F:\;$ are each orthogonal to the corresponding vectors $n$, as you point out in your comment above. So the vector $f\in F$ is perpendicular to the x-axis at that point, and hence, must be tangent to $C$ at $(1, 0)$.

Likewise, consider what is happening at $(0, 1)$: At this point, $n$ lies along the y-axis, directed in positive $y$ direction (and normal to $C$). $f \in F$, at this point on $C$ is a "horizontal" vector, directed counter-clockwise, and hence tangent to $C$ at $(0, 1)$

You are correct that the vectors in $F = \langle -y, x \rangle$ are orthogonal to the corresponding vectors $n = \langle x, y \rangle$. We just need to know what's happening with vectors in $F$ at points of $C$, knowing $n$ is normal to $C$.

Included immediately below is a representation of vectors $n = \langle x, y\rangle$ for $(x,y) \in C$, normal to $C$:

$\quad\quad\quad\quad\quad$ radial normal vectors

See the image below for an representation of vectors tangent to the unit circle, representing vectors in $F$:

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$tangent vectors

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Jack: Does this give you a better understanding of the problem? – amWhy Nov 21 '12 at 0:59

You should memorize once and for all that, if ${\bf r}:=(x,y)$ is a vector in ${\mathbb R}^2$ attached to $(0,0)$, then ${\bf r}':=(-y,x)$ is the vector obtained by rotating ${\bf r}$ ninety degrees counterclockwise, in other words: ${\pi\over2}$ in the positive direction.

Given the circle $C:\ x^2+y^2=1$ we all know that at each point ${\bf r}:=(x,y)\in C$ the vector ${\bf r}$, attached to $(x,y)$, points in the outward radial direction vs. $C$, which is orthogonal to the tangential direction of $C$ at $(x,y)$. It follows that the vector ${\bf r}':=(-y,x)$ attached to $(x,y)$ points in the tangential direction of $C$ at $(x,y)$, to be exact: in the "positive" direction of $C$ when $C$ is described counterclockwise. Therefore your vector ${\bf F}=-{\bf r}'$ is tangential to $C$ as well at $(x,y)$.

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