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Math Help - parametrized curve

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    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?
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    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}.
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    Quote Originally Posted by dori1123 View Post
    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}.
    Last edited by bkarpuz; September 10th 2008 at 11:14 PM. Reason: Note added.
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    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.
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