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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

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

(A.63)

When written out in component form this becomes

$\displaystyle \frac{d {\bf a}}{dt} \equiv \left(\frac{d a_x}{dt}, \frac{d a_y}{d t}, \frac{d a_z}{ d t}\right).$

(A.64)

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

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

(A.65)

which implies that

$\displaystyle \frac{d {\bf a}}{dt} = \frac{d\phi}{dt}\, {\bf b} + \phi\, \frac{d {\bf b}}{dt}.$

(A.66)

Moreover, it is easily demonstrated that

$\displaystyle \frac{d}{dt}\left({\bf a}\cdot{\bf b}\right) = \frac{d{\bf a}}{dt}\cdot {\bf b} +{\bf a}\cdot\frac{d{\bf b}}{dt},$

(A.67)

and

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

(A.68)

Hence, it can be seen that the laws of vector differentiation are analogous to those in conventional calculus.

Next: Line Integrals Up: Vectors and Vector Fields Previous: Vector Triple Product

Richard Fitzpatrick 2016-01-22