Vector calculus

Next: Line integrals Up: Vectors Previous: The vector triple 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

$\begin{displaymath} \frac{d {\bf a}}{dt} = \lim_{\delta t\rightarrow 0} \left[\frac{{\bf a}(t+\delta t) - {\bf a}(t)} {\delta t}\right]. \end{displaymath}$

(54)

When written out in component form this becomes

$\begin{displaymath} \frac{d {\bf a}}{dt} = \left(\frac{d a_x}{dt}, \frac{d a_y}{d t}, \frac{d a_z}{ d t}\right). \end{displaymath}$

(55)

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

$\begin{displaymath} \frac{d a_x}{dt} = \frac{d}{dt}\!\left(\phi b_x\right) = \frac{d\phi}{dt} b_x + \phi \frac{d b_x}{dt}, \end{displaymath}$

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which implies that

$\begin{displaymath} \frac{d {\bf a}}{dt} = \frac{d\phi}{dt} {\bf b} + \phi \frac{d {\bf b}}{dt}. \end{displaymath}$

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It is easily demonstrated that

$\begin{displaymath} \frac{d}{dt}\left({\bf a}\cdot{\bf b}\right) = \frac{d{\bf a}}{dt}\cdot {\bf b} +{\bf a}\cdot\frac{d{\bf b}}{dt}. \end{displaymath}$

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Likewise,

$\begin{displaymath} \frac{d}{dt}\left({\bf a}\times{\bf b}\right) = \frac{d{\bf a}}{dt}\times{\bf b} + {\bf a}\times \frac{d{\bf b}}{dt}. \end{displaymath}$

(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