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

$$\frac{d {\bf a}}{dt} = \lim_{\delta t\rightarrow 0} \left[\frac{{\bf a}(t+\delta t) - {\bf a}(t)} {\delta t}\right].$$ (54)

When written out in component form this becomes

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

$$\frac{d a_x}{dt} = \frac{d}{dt}\!\left(\phi b_x\right) = \frac{d\phi}{dt} b_x + \phi \frac{d b_x}{dt},$$ (56)

which implies that

$$\frac{d {\bf a}}{dt} = \frac{d\phi}{dt} {\bf b} + \phi \frac{d {\bf b}}{dt}.$$ (57)

It is easily demonstrated that

$$\frac{d}{dt}\left({\bf a}\cdot{\bf b}\right) = \frac{d{\bf a}}{dt}\cdot {\bf b} +{\bf a}\cdot\frac{d{\bf b}}{dt}.$$ (58)

Likewise,

$$\frac{d}{dt}\left({\bf a}\times{\bf b}\right) = \frac{d{\bf a}}{dt}\times{\bf b} + {\bf a}\times \frac{d{\bf b}}{dt}.$$ (59)

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