# What is the defined direction of the curvature vector at a point?

I'm asking this only because I could not find it anywhere on the web (tried Wikipedia and Google searches).

The only hint I found was in this image on Wikipedia, which seems to indicate that the radius of curvature is directed towards the centre of the osculating circle, which would mean the curvature vector itself is directed in the opposite direction.

But there's no clear definition anywhere. So: how is the direction of the curvature vector defined?

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Naïvely, I would think the curvature vector should be the second derivative of the position vector with respect to arc length, which makes it point inwards. – Harald Hanche-Olsen Oct 10 '12 at 6:45
I have the same question. Curvature is a scalar value that defines the rate at which a curve is changing direction with respect to arc length, but what is the vector that tells us the direction that this curve is changing with respect to arc length? My theory is that it is: [dx/ds,dy/ds,dz/ds]; however, I don't know how you'd compute a vector such as this. Tangent, normal, and binormal vectors do not point in the direction of curvature. – user74712 Apr 27 '13 at 20:51
@Marcus -- the formula you gave will calculate the curve tangent. You need to differentiate again to get curvature. – bubba Apr 28 '13 at 8:04

Suppose we have a curve $\mathbf r$ that is parameterised by arclength, $s$, and let $\mathbf t$ be the unit tangent of $\mathbf r$. The curvature $\kappa$ is defined by $$\kappa =\left\Vert{\frac{d{\mathbf t}}{ds}}\right\Vert = \left\Vert{\frac{d^2{\mathbf r}}{ds^2}}\right\Vert.$$ Note that this implies that $\kappa \ge 0$.
The unit vector parallel to $\tfrac{d{\mathbf t}}{ds}$ is denoted by $\mathbf n$ and is called the principal normal of the curve.
So, we have $$\frac{d{\mathbf t}}{ds} = \kappa {\mathbf n}$$ Since $\kappa \ge 0$, this last equation says that the vector $\mathbf n$ is a non-negative multiple of $\tfrac{d{\mathbf t}}{ds}$, which means that $\mathbf n$ points roughly in the direction that $\mathbf t$ is turning towards. In other words, roughly speaking, $\mathbf n$ points towards the concave side of the curve. Or saying it another way, $\mathbf n$ points towards the center of the osculating circle.