# Direction of the second derivative of an arclength parametrized curve

I have a question, it's so simple and stupid ._. If I have a planar curve parametrized by arc length, it's easy to show that the second derivate is orthogonal to the first derivate vector (tangent vector), thus if I rotate the tangent vector by an angle of $\pi/2$, I have a new vector (call it N), clearly N and the second derivate are parallel, the question is, is it true that the second derivate vector always lies "inside" the curve?

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If the curve isn't closed (e.g. a parabola), what's "inside"? – J. M. Aug 7 '11 at 19:04
I assume the curve is sufficiently smooth and lives in $\mathbb{R}^2$. Whether or not the normal vector always points toward some regions (defined by how the curve separates the plane's points) and not others depends on if it's closed or unbounded, convex or concave, and self-intersecting (as well as perhaps the order of the contacts). In full generality this becomes something of a classification problem. – anon Aug 7 '11 at 19:32
Im not saying "inside" in the sense of a closed curve, only in the sense that the side toward which the curve bends – Daniel Aug 7 '11 at 19:48

Of the two possibilities in 2D, the second derivative vector points in the direction the curve is turning. Basically, this is because 1) the second derivative vector (even with the arc length parametrization) can be thought of as an acceleration vector, and 2) the direction of the acceleration vector describes how the direction of the curve is changing.

To elaborate a bit:

1) Parametrizing a curve by arc length is equivalent to parametrizing the curve by time in which an object traveling at constant unit speed traces out the curve. So the first derivative vector ${\bf r}'(t)$ gives velocity at "time" $t$, and the second derivative vector ${\bf r}''(t)$ gives acceleration at $t$. (And ${\bf r}''(t)$ is orthogonal to ${\bf r}'(t)$ because the latter has constant length.)

2) The direction of the acceleration vector describes how the direction of the velocity vector is changing. Since the velocity vector points in the direction of the curve, this means that the direction of the acceleration vector describes how the direction of the curve itself is changing. So if the curve is turning in a particular direction, the direction of the acceleration vector has to track that change. For example, see the following (borrowed from the Wikipedia entry on acceleration).

The change in direction of the curve over the time interval between the green and blue dots is indicated by the difference between the green and blue arrows; i.e., the black vector ${\bf \Delta v}$. Since the acceleration vector ${\bf a}$ is just a scaled version of ${\bf \Delta v}$, its direction also indicates how the direction of the curve is changing and so points toward the side to which the curve turns. (Well, you have to let $\Delta t$ go to $0$ to get ${\bf a}$, but that doesn't change the fundamental behavior.)

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nice answer Thanks !!!! – Daniel Aug 9 '11 at 13:44

The second derivative points "in" on convex parts of a closed curve, "out" on concave parts.

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