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I've seen two definitions of a normal vector to a curve in $\mathbb{R^2}$. Suppose we have a parametrisation of our curve: $r(t)=(x(t),y(t)),$ Then differentiating once gives us a tangent vector, differentiating again gives us a normal vector. Also $(y'(t),-x'(t))$ is normal, but I don't see how these two vectors always point in same (or opposite) direction (since we're only in $\mathbb{R^2}$).

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    $\begingroup$ $(y’,-x’)$ is normal in the sense that it is orthogonal to the tangent vector for all $t$. $\endgroup$ – TM Gallagher Apr 10 at 21:03
  • $\begingroup$ So it's normal to the curve for all $t$? I still don't understand how the other definition isn't the same as what you just said $\endgroup$ – Displayname Apr 10 at 21:29
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    $\begingroup$ In general $r’’$ is not orthogonal to $r’$ for all time. You need to consider the unit tangent vector $T=r’/||r’||$ and the unit normal $N=T’/||T’||$. The rescalings are maybe what is causing the confusion. You also need to be careful when reading multiple definitions: sometimes the authors will assume a unit speed parameterization (i.e. one for which $||r’||=1$ for all $t$) in which case $r’’$ is orthogonal to $r’$ for all time. $\endgroup$ – TM Gallagher Apr 10 at 21:52
  • $\begingroup$ Ah okay that makes sense now, I think I just ignored the scaling but obviously the magnitude of the tangent is a function of $t$ and so I was only ever considering a particular case as you mentioned above. Thank you. $\endgroup$ – Displayname Apr 10 at 22:00

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