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Suppose that vector
varies with time, so that
. The time
derivative of the vector is defined
![\begin{displaymath}
\frac{d {\bf a}}{dt} = \lim_{\delta t\rightarrow 0} \left[\frac{{\bf a}(t+\delta t) - {\bf a}(t)}
{\delta t}\right].
\end{displaymath}](img202.png) |
(54) |
When written out in component form this becomes
![\begin{displaymath}
\frac{d {\bf a}}{dt} = \left(\frac{d a_x}{dt}, \frac{d a_y}{d t}, \frac{d a_z}{ d t}\right).
\end{displaymath}](img203.png) |
(55) |
Suppose that
is, in fact, the product of a scalar
and another vector
. What now is the time derivative of
? We have
![\begin{displaymath}
\frac{d a_x}{dt} = \frac{d}{dt}\!\left(\phi b_x\right) = \frac{d\phi}{dt} b_x + \phi
\frac{d b_x}{dt},
\end{displaymath}](img206.png) |
(56) |
which implies that
![\begin{displaymath}
\frac{d {\bf a}}{dt} = \frac{d\phi}{dt} {\bf b} + \phi \frac{d {\bf b}}{dt}.
\end{displaymath}](img207.png) |
(57) |
It is easily demonstrated that
![\begin{displaymath}
\frac{d}{dt}\left({\bf a}\cdot{\bf b}\right) = \frac{d{\bf a}}{dt}\cdot {\bf b} +{\bf a}\cdot\frac{d{\bf b}}{dt}.
\end{displaymath}](img208.png) |
(58) |
Likewise,
![\begin{displaymath}
\frac{d}{dt}\left({\bf a}\times{\bf b}\right) = \frac{d{\bf a}}{dt}\times{\bf b} + {\bf a}\times
\frac{d{\bf b}}{dt}.
\end{displaymath}](img209.png) |
(59) |
It can be seen that the laws of vector differentiation are analogous to those in
conventional calculus.
Next: Line integrals
Up: Vectors
Previous: The vector triple product
Richard Fitzpatrick
2006-02-02