1 Differentiation of vectors

Consider Figure 31.

Figure 31

If $\underset{̲}{r}$ represents the position vector of an object which is moving along a curve $C$ , then the position vector will be dependent upon the time, $t$ . We write $\underset{̲}{r}=\underset{̲}{r}\left(t\right)$ to show the dependence upon time. Suppose that the object is at the point $P$ , with position vector r at time $t$ and at the point $Q$ , with position vector $\underset{̲}{r}\left(t+\delta t\right)$ , at the later time $t+\delta t$ , as shown in Figure 32.

Figure 32

Then $\overrightarrow{PQ}$ represents the displacement vector of the object during the interval of time $\delta t.$ The length of the displacement vector represents the distance travelled, and its direction gives the direction of motion. The average velocity during the time from $t$ to $t+\delta t$ is defined as the displacement vector divided by the time interval $\delta t,$ that is,

average velocity   $=\frac{\overrightarrow{PQ}}{\delta t}=\frac{\underset{̲}{r}\left(t+\delta t\right)-\underset{̲}{r}\left(t\right)}{\delta t}$

If we now take the limit as the interval of time $\delta t$ tends to zero then the expression on the right hand side is the derivative of $\underset{̲}{r}$ with respect to $t$ . Not surprisingly we refer to this derivative as the instantaneous velocity , $\underset{̲}{v}$ . By its very construction we see that the velocity vector is always tangential to the curve as the object moves along it. We have:

$\underset{̲}{v}=\underset{\delta t\to 0}{lim}\frac{\underset{̲}{r}\left(t+\delta t\right)-\underset{̲}{r}\left(t\right)}{\delta t}=\frac{d\underset{̲}{r}}{dt}$

Now, since the $x$ and $y$ coordinates of the object depend upon time, we can write the position vector $\underset{̲}{r}$ in Cartesian coordinates as:

$\underset{̲}{r}\left(t\right)=x\left(t\right)\underset{̲}{i}+y\left(t\right)\underset{̲}{j}$

Therefore,

$\underset{̲}{r}\left(t+\delta t\right)=x\left(t+\delta t\right)\underset{̲}{i}+y\left(t+\delta t\right)\underset{̲}{j}$

so that,

This is often abbreviated to $\underset{̲}{v}=\dot{\underset{̲}{r}}=ẋ\underset{̲}{i}+ẏ\underset{̲}{j}$ , using notation for derivatives with respect to time. So we see that the velocity vector is the derivative of the position vector with respect to time. This result generalizes in an obvious way to three dimensions as summarized in the following Key Point.

Example 6

If $\underset{̲}{w}=3t^2\underset{̲}{i}+cos2t\underset{̲}{j},$ find

1. $\frac{d\underset{̲}{w}}{dt}$
2. $\left|\frac{d\underset{̲}{w}}{dt}\right|$
3. $\frac{d^2\underset{̲}{w}}{d{t}^2}$

Solution

1. If $\underset{̲}{w}=3t^2\underset{̲}{i}+cos2t\underset{̲}{j},$ then differentiation with respect to $t$ yields: $\frac{d\underset{̲}{w}}{dt}=6t\underset{̲}{i}-2sin2t\underset{̲}{j}$
2. $\left|\frac{d\underset{̲}{w}}{dt}\right|=\sqrt{{\left(6t\right)}^2+{\left(-2sin2t\right)}^2}=\sqrt{36t^2+4{sin}^{2}2t}$
3. $\frac{d^2\underset{̲}{w}}{d{t}^2}=6\underset{̲}{i}-4cos2t\underset{̲}{j}$

It is possible to differentiate more complicated expressions involving vectors provided certain rules are adhered to as summarized in the following Key Point.

Example 7

If $\underset{̲}{w}=3t\underset{̲}{i}-t^2\underset{̲}{j}$ and $\underset{̲}{z}=2t^2\underset{̲}{i}+3\underset{̲}{j}$ , verify the result

$\frac{d}{dt}\left(\underset{̲}{w}\cdot \underset{̲}{z}\right)=\underset{̲}{w}\cdot \frac{d\underset{̲}{z}}{dt}+\frac{d\underset{̲}{w}}{dt}\cdot \underset{̲}{z}$

Solution

$\underset{̲}{w}\cdot \underset{̲}{z}=\left(3t\underset{̲}{i}-t^2\underset{̲}{j}\right)\cdot \left(2t^2\underset{̲}{i}+3\underset{̲}{j}\right)=6t^3-3t^2.$

Therefore $\frac{d}{dt}\left(\underset{̲}{w}\cdot \underset{̲}{z}\right)=18t^2-6t$ (1)

Also $\frac{d\underset{̲}{w}}{dt}=3\underset{̲}{i}-2t\underset{̲}{j}\text{and}\frac{d\underset{̲}{z}}{dt}=4t\underset{̲}{i}$

We have verified $\frac{d}{dt}\left(\underset{̲}{w}\cdot \underset{̲}{z}\right)=\underset{̲}{w}\cdot \frac{d\underset{̲}{z}}{dt}+\frac{d\underset{̲}{w}}{dt}\cdot \underset{̲}{z}$ since (1) is the same as (2).

Example 8

If $\underset{̲}{w}=3t\underset{̲}{i}-t^2\underset{̲}{j}$ and $\underset{̲}{z}=2t^2\underset{̲}{i}+3\underset{̲}{j}$ , verify the result

$\frac{d}{dt}\left(\underset{̲}{w}\times \underset{̲}{z}\right)=\underset{̲}{w}\times \frac{d\underset{̲}{z}}{dt}+\frac{d\underset{̲}{w}}{dt}\times \underset{̲}{z}$

Solution

$\underset{̲}{w}\times \underset{̲}{z}=\left|\begin{array}{ccc}\hfill \underset{̲}{i}\hfill & \hfill \underset{̲}{j}\hfill & \hfill \underset{̲}{k}\hfill \\ \hfill 3t\hfill & \hfill -t^2\hfill & \hfill 0\hfill \\ \hfill 2t^2\hfill & \hfill 3\hfill & \hfill 0\hfill \end{array}\right|=\left(9t+2t^4\right)\underset{̲}{k}$ implying $\frac{d}{dt}\left(\underset{̲}{w}\times \underset{̲}{z}\right)=\left(9+8t^3\right)\underset{̲}{k}$ (1)

We can see that (1) is the same as (2) $+$ (3) as required.

Exercises

1. If $r=3t\underset{̲}{i}+2t^2\underset{̲}{j}+t^3\underset{̲}{k},$ find
   1. $\frac{d\underset{̲}{r}}{dt}$
   2. $\frac{d^2\underset{̲}{r}}{d{t}^2}$
2. Given $\underset{̲}{B}=t{e}^{-t}\underset{̲}{i}+cost\underset{̲}{j}$ find
   1. $\frac{d\underset{̲}{B}}{dt}$
   2. $\frac{d^2\underset{̲}{B}}{d{t}^2}$
3. If $\underset{̲}{r}=4t^2\underset{̲}{i}+2t\underset{̲}{j}-7\underset{̲}{k}$ evaluate $\underset{̲}{r}$ and $\frac{d\underset{̲}{r}}{dt}$ when $t=1.$
4. If $\underset{̲}{w}=t^3\underset{̲}{i}-7t\underset{̲}{k}$ and $\underset{̲}{z}=\left(2+t\right)\underset{̲}{i}+t^2\underset{̲}{j}-2\underset{̲}{k}$
   1. find $\underset{̲}{w}\cdot \underset{̲}{z}$ ,
   2. find $\frac{d\underset{̲}{w}}{dt}$ ,
   3. find $\frac{d\underset{̲}{z}}{dt}$ ,
   4. show that $\frac{d}{dt}\left(\underset{̲}{w}\cdot \underset{̲}{z}\right)=\underset{̲}{w}\cdot \frac{d\underset{̲}{z}}{dt}+\frac{d\underset{̲}{w}}{dt}\cdot \underset{̲}{z}$
5. Given $\underset{̲}{r}=sint\underset{̲}{i}+cost\underset{̲}{j}$
   1. find  $\dot{\underset{̲}{r}}$ ,
   2. find  $\ddot{\underset{̲}{r}}$ ,
   3. find  $\left|\underset{̲}{r}\right|$
   4. Show that the position vector $\underset{̲}{r}$ and velocity vector $\underset{̲}{ṙ}$ are perpendicular.

Answer