On covarient derivative
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On covarient derivative
Bearing in mind this exercise , prove the following assertion :
Let \( \displaystyle V \) and \( \displaystyle W \) be two differential vector fields along the parametrized curve \( \displaystyle \alpha : I \longrightarrow S \) such that \( \displaystyle \Big| V(t) \Big| = \Big| W(t) \Big| = 1 , \, \forall t \in I \), where \( \displaystyle I \subset \mathbb{R} \) is an interval and \( \displaystyle S \) is a regular surface. Then \[ \displaystyle \Big[ \frac{DW}{dt} \Big] - \Big[ \frac{DV}{dt} \Big] = \frac{d \phi}{dt} \]where \( \displaystyle \phi \) is one of the differentiable determinations of the angle from \( \displaystyle V \) to \( \displaystyle W \).
Let \( \displaystyle V \) and \( \displaystyle W \) be two differential vector fields along the parametrized curve \( \displaystyle \alpha : I \longrightarrow S \) such that \( \displaystyle \Big| V(t) \Big| = \Big| W(t) \Big| = 1 , \, \forall t \in I \), where \( \displaystyle I \subset \mathbb{R} \) is an interval and \( \displaystyle S \) is a regular surface. Then \[ \displaystyle \Big[ \frac{DW}{dt} \Big] - \Big[ \frac{DV}{dt} \Big] = \frac{d \phi}{dt} \]where \( \displaystyle \phi \) is one of the differentiable determinations of the angle from \( \displaystyle V \) to \( \displaystyle W \).
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