# Math Help - parametrized curve

1. ## parametrized curve

I am taking a Differential Geometry class, and we started talking about parametrized curve. There is something I don't understand. A parametrized curve alpha(t) has the property that its second derivative is identically zero, then what can be said about alpha(t)? Can someone help?

2. In mechanics, one would say: if the acceleration (second derivative) is zero, then the speed is constant, and therefore the point moves along a line, uniformly.

In other words, $\alpha(t)=\alpha(0)+t\vec{v}$ for some vector $\vec{v}$.

3. Originally Posted by dori1123
I am taking a Differential Geometry class, and we started talking about parametrized curve. There is something I don't understand. A parametrized curve alpha(t) has the property that its second derivative is identically zero, then what can be said about alpha(t)? Can someone help?
its a geodesic, this means, if you have two points $A$ and $B$ on the surface $M$, and $\alpha$ passes trought this points, then the shortest way on $M$ to travel from $A$ to $B$ is traveling trough the curve $\alpha$, provided that its a georesic, i.e. $\alpha^{\prime\prime}\equiv0$.

Note. Recall that on a plain, the shortest way to reach from $A$ to $B$ is the way combined combined by a line, which is obtained by solving the differential equation $\alpha^{\prime\prime}(t)=0\text{ for }t\in\mathbb{R}$.

4. It depends indeed on what is meant by $\alpha''(t)$ in your Geometry class, dori1123.

If it is just the usual derivative in $\mathbb{R}^3$, then my first answer suffices.

It may however also denote the projection of $\alpha''(t)$ (defined in $\mathbb{R}^3$ the usual way) on the tangent plane of $\mathcal{M}$ at $\alpha(t)$. In this case, this is an "intrinsic way" (independent of the embedding of $\mathcal{M}$ in $\mathbb{R}^3$) to define the second derivative of the parametrized curve. It can be defined in an more general context using affine connections and is then denoted $\nabla_{\alpha'(t)}\alpha'(t)$. And $\nabla_{\alpha'(t)}\alpha'(t)=0$ is the equation of the geodesics, as bkarpuz writes.

On a surface in $\mathbb{R}^3$, geodesics are characterized by the fact that their acceleration lies in the orthogonal of the tangent plane: $\alpha''(t)\in T_{\alpha(t)}^\perp$. In other words, for geodesics, the projection of $\alpha''(t)$ on the tangent plane at $\alpha(t)$ is $0$.