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Next: Line Integrals Up: Vectors Previous: The Vector Product

Vector Calculus

Suppose that vector ${\bf a}$ varies with time, so that ${\bf a} = {\bf a} (t)$. 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} (38)

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} (39)

Suppose that ${\bf a}$ is, in fact, the product of a scalar $\phi(t)$ and another vector ${\bf b}(t)$. What now is the time derivative of ${\bf a}$? 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} (40)

which implies that
\begin{displaymath}
\frac{d {\bf a}}{dt} = \frac{d\phi}{dt}\, {\bf b} + \phi\, \frac{d {\bf b}}{dt}.
\end{displaymath} (41)

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} (42)

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} (43)

It can be seen that the laws of vector differentiation are fairly analogous to those in conventional calculus.


next up previous
Next: Line Integrals Up: Vectors Previous: The Vector Product
Richard Fitzpatrick 2007-07-14