# Normal Vector in the place

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 $$\mathbb{R^2}$$).

• $(y’,-x’)$ is normal in the sense that it is orthogonal to the tangent vector for all $t$. – TM Gallagher Apr 10 at 21:03
• 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 – Displayname Apr 10 at 21:29
• 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. – TM Gallagher Apr 10 at 21:52
• 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. – Displayname Apr 10 at 22:00