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Tangent Vector

For a curve with Position Vector ${\bf r}(t)$, the unit tangent vector $\hat {\bf T}(t)$ is defined by

$\displaystyle \hat {\bf T}(t)$ $\textstyle \equiv$ $\displaystyle {{\bf r}'(t)\over \vert{\bf r}'(t)\vert} = {{d{\bf r}\over dt}\over \left\vert{d{\bf r}\over dt}\right\vert}$ (1)
  $\textstyle =$ $\displaystyle {{d{\bf r}\over dt}\over {ds\over dt}}$ (2)
  $\textstyle =$ $\displaystyle {d{\bf r}\over ds},$ (3)

where $t$ is a parameterization variable and $s$ is the Arc Length. For a function given parametrically by $(f(t), g(t))$, the tangent vector relative to the point $(f(t), g(t))$ is therefore given by
$\displaystyle x(t)$ $\textstyle =$ $\displaystyle {f'\over\sqrt{f'^2+g'^2}}$ (4)
$\displaystyle y(t)$ $\textstyle =$ $\displaystyle {g'\over\sqrt{f'^2+g'^2}}.$ (5)

To actually place the vector tangent to the curve, it must be displaced by $(f(t), g(t))$. It is also true that
$\displaystyle {d\hat {\bf T}\over ds}$ $\textstyle =$ $\displaystyle \kappa\hat {\bf N}$ (6)
$\displaystyle {d\hat {\bf T}\over dt}$ $\textstyle =$ $\displaystyle \kappa{ds\over dt}\hat {\bf N}$ (7)
$\displaystyle {[}\dot{\bf T},\ddot{\bf T},\raise7.5pt\hbox{.}\mkern0mu\raise7.5pt\hbox{.}\mkern0mu\raise7.5pt\hbox{.}\mkern-15mu{\bf T}]$ $\textstyle =$ $\displaystyle \kappa^5 {d\over ds}\left({\tau\over\kappa}\right),$ (8)

where ${\bf N}$ is the Normal Vector, $\kappa$ is the Curvature, and $\tau$ is the Torsion.

See also Curvature, Normal Vector, Tangent, Tangent Bundle, Tangent Plane, Tangent Space, Torsion (Differential Geometry)


References

Gray, A. ``Tangent and Normal Lines to Plane Curves.'' §5.5 in Modern Differential Geometry of Curves and Surfaces. Boca Raton, FL: CRC Press, pp. 85-90, 1993.




© 1996-9 Eric W. Weisstein
1999-05-26