Let be a parametricsmooth curve. The tangent vector is given by provided it exists and provided, where we have used a prime instead of the usual dot to indicate differentiation with respect to parametert.[1] The unit tangent vector is given by
Given the curvein, the unit tangent vector at is given byWhere the components of the tangent vector are found by taking the derivative of each corresponding component of the curve with respect to.
If is given parametrically in then-dimensional coordinate systemxi (here we have used superscripts as an index instead of the usual subscript) by orthen the tangent vector field is given byUnder a change of coordinatesthe tangent vector in theui-coordinate system is given bywhere we have used theEinstein summation convention. Therefore, a tangent vector of a smooth curve will transform as acontravariant tensor of order one under a change of coordinates.[2]
Let be a differentiable function and let be a vector in. We define thedirectional derivative in the direction at a point byThe tangent vector at the point may then be defined[3] as
Let be a differentiable manifold and let be the algebra of real-valued differentiable functions on. Then the tangent vector to at a point in the manifold is given by thederivation which shall be linear — i.e., for any and we have
Note that the derivation will by definition have the Leibniz property