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Vector Calculus
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}](img3364.png) |
(1320) |
When written out in component form this becomes
![\begin{displaymath}
\frac{d {\bf a}}{dt} \equiv \left(\frac{d a_x}{dt}, \frac{d a_y}{d t}, \frac{d a_z}{ d t}\right).
\end{displaymath}](img3365.png) |
(1321) |
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}](img3368.png) |
(1322) |
which implies that
![\begin{displaymath}
\frac{d {\bf a}}{dt} = \frac{d\phi}{dt}\, {\bf b} + \phi\, \frac{d {\bf b}}{dt}.
\end{displaymath}](img3369.png) |
(1323) |
Moreover, 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}](img3370.png) |
(1324) |
and
![\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}](img3371.png) |
(1325) |
Hence, it can be seen that the laws of vector differentiation are analogous to those in
conventional calculus.
Next: Line Integrals
Up: Vector Algebra and Vector
Previous: Vector Triple Product
Richard Fitzpatrick
2011-03-31