1 Integration of vectors

If a vector depends upon time $t$ , it is often necessary to integrate it with respect to time. Recall that $\underset{̲}{i},\underset{̲}{j}$ and $\underset{̲}{k}$ are constant vectors and must be treated thus in any integration. Hence the integral,

$\underset{̲}{I}=\int \; <!--nolimits\; -->\left(f\left(t\right)\underset{̲}{i}+g\left(t\right)\underset{̲}{j}+h\left(t\right)\underset{̲}{k}\right)dt$

is evaluated as three scalar integrals i.e. $\underset{̲}{I}=\left(\int \; <!--nolimits\; -->f\left(t\right)dt\right)\underset{̲}{i}+\left(\int \; <!--nolimits\; -->g\left(t\right)dt\right)\underset{̲}{j}+\left(\int \; <!--nolimits\; -->h\left(t\right)dt\right)\underset{̲}{k}$

Example 1

If $\underset{̲}{r}=3t\underset{̲}{i}+t^2\underset{̲}{j}+\left(1+2t\right)\underset{̲}{k},$ evaluate ${\int \; <!--nolimits\; -->}_0^1\underset{̲}{r}dt.$

Solution

${\int \; <!--nolimits\; -->}_0^1\underset{̲}{r}dt=\left({\int \; <!--nolimits\; -->}_0^{1}3tdt\right)\underset{̲}{i}+\left({\int \; <!--nolimits\; -->}_0^1{t}^{2}dt\right)\underset{̲}{j}+\left({\int \; <!--nolimits\; -->}_0^1\left(1+2t\right)dt\right)\underset{̲}{k}$

$={\left[\frac{3t^2}{2}\right]}_0^1\underset{̲}{i}+{\left[\frac{t^3}{3}\right]}_0^1\underset{̲}{j}+{\left[t+t^2\right]}_0^1\underset{̲}{k}=\frac{3}{2}\underset{̲}{i}+\frac{1}{3}\underset{̲}{j}+2\underset{̲}{k}$

1.1 Trajectories

To simplify the modelling of the path of a body projected from a fixed point we usually ignore any air resistance and effects due to the wind. Once this initial model is understood other variables and effects can be introduced into the model.

A particle is projected from a point $O$ with velocity $\underset{̲}{u}$ and an angle $\theta$ above the horizontal as shown in Figure 1.

Figure 1

The only force acting on the particle in flight is gravity acting downwards, so if $m$ is the mass of the projectile and taking axes as shown, the force due to gravity is $-mg\underset{̲}{j}$ . Now using Newton’s second law (rate of change of momentum is equal to the applied force) we have

$\frac{d\left(m\underset{̲}{v}\right)}{dt}=-mg\underset{̲}{j}$

Cancelling the common factor $m$ and integrating we have

$\underset{̲}{v}\left(t\right)=-gt\underset{̲}{j}+\underset{̲}{c}$ where $\underset{̲}{c}$ is a constant vector .

However, velocity is the rate of change of position: $\underset{̲}{v}\left(t\right)=\frac{d\underset{̲}{r}}{dt}$ so

$\frac{d\underset{̲}{r}}{dt}=-gt\underset{̲}{j}+\underset{̲}{c}$

Integrating once more:

$\underset{̲}{r}\left(t\right)=-\frac{1}{2}g{t}^2\underset{̲}{j}+\underset{̲}{c}t+\underset{̲}{d}$ where $\underset{̲}{d}$ is another constant vector.

The values of these constant vectors may be determined by using the initial conditions in this problem: when $t=0$ then $\underset{̲}{r}=\underset{̲}{0}$ and $\underset{̲}{v}=\underset{̲}{u}$ . Imposing these initial conditions gives

$\underset{̲}{d}=\underset{̲}{0}$ and $\underset{̲}{c}=ucos\theta \underset{̲}{i}+usin\theta \underset{̲}{j}$ where $u$ is the magnitude of $\underset{̲}{u}$ . This gives

$\underset{̲}{r}\left(t\right)=utcos\theta \underset{̲}{i}+\left(utsin\theta -\frac{1}{2}g{t}^2\right)\underset{̲}{j}$ .

The interested reader might try to show why the path of the particle is a parabola.

Exercises

1. Given $\underset{̲}{r}=3sint\underset{̲}{i}-cost\underset{̲}{j}+\left(2-t\right)\underset{̲}{k},$ evaluate ${\int \; <!--nolimits\; -->}_0^{\pi }\underset{̲}{r}dt.$
2. Given $\underset{̲}{v}=\underset{̲}{i}-3\underset{̲}{j}+\underset{̲}{k}$ , evaluate:
   1. ${\int \; <!--nolimits\; -->}_0^1\underset{̲}{v}dt$ ,
   2. ${\int \; <!--nolimits\; -->}_0^2\underset{̲}{v}dt$
3. The vector, $\underset{̲}{a}$ , is defined by $\underset{̲}{a}=t^2\underset{̲}{i}+e^{-t}\underset{̲}{j}+t\underset{̲}{k}$ . Evaluate
   1. ${\int \; <!--nolimits\; -->}_0^1\underset{̲}{a}dt$ ,
   2. ${\int \; <!--nolimits\; -->}_2^3\underset{̲}{a}dt$ ,
   3. ${\int \; <!--nolimits\; -->}_1^4\underset{̲}{a}dt$
4. Let $\underset{̲}{a}$ and $\underset{̲}{b}$ be two three-dimensional vectors. Is the following result true?

   ${\int \; <!--nolimits\; -->}_{t_1}^{t_2}\underset{̲}{a}dt\times {\int \; <!--nolimits\; -->}_{t_1}^{t_2}\underset{̲}{b}dt={\int \; <!--nolimits\; -->}_{t_1}^{t_2}\underset{̲}{a}\times \underset{̲}{b}dt$

   where $\times$ denotes the vector product.

Answer