2 Vector functions of a variable

Vectors were first studied in HELM booklet  9. A vector is a quantity that has magnitude and direction and combines together with other vectors according to the triangle law. Examples are (i) a velocity of 60 mph West and (ii) a force of 98.1 newtons vertically downwards.
It is often convenient to express vectors in terms of $\underset{̲}{i}$ , $\underset{̲}{j}$ and $\underset{̲}{k}$ , which are unit vectors in the $x$ , $y$ and $z$ directions respectively. Examples are $\underset{̲}{a}=3\underset{̲}{i}+4\underset{̲}{j}$ and $\underset{̲}{b}=2\underset{̲}{i}-2\underset{̲}{j}+\underset{̲}{k}$
The magnitudes of these vectors are $\left|\underset{̲}{a}\right|=\sqrt{3^2+4^2}=5$ and $\left|\underset{̲}{b}\right|=\sqrt{2^2+{\left(-2\right)}^2+1^2}=3$ respectively. In this case $\underset{̲}{a}$ and $\underset{̲}{b}$ are constant vectors, but a vector could be a function of an independent variable such as $t$ (which may represent time in certain applications).

Example 1

A particle is at the point A(3,0). At time $t=0$ it starts moving at a constant speed of $2\text{m}{\text{s}}^{-1}$ in a direction parallel to the positive $y$ -axis. Find expressions for the position vector, $\underset{̲}{r}$ , of the particle at time $t$ , together with its velocity $\underset{̲}{v}=\frac{d\underset{̲}{r}}{dt}$ and acceleration $\underset{̲}{a}=\frac{d^2\underset{̲}{r}}{d{t}^2}$ .

Solution

In the first second of its motion the particle moves 2 metres to B and it moves a further 2 metres in each subsequent second, to C, D, $\dots$ . Because it moves parallel to the $y$ -axis its velocity is $\underset{̲}{v}=2\underset{̲}{j}$ . As its velocity is constant its acceleration is $\underset{̲}{a}=\underset{̲}{0}$ .
The position of the particle at $t=0,1,2,3$ is given in the table.

In general, after $t$ seconds, the position vector of the particle is $\underset{̲}{r}=3\underset{̲}{i}+2t\underset{̲}{j}$

Example 2

The position vector of a particle at time $t$ is given by $\underset{̲}{r}=2t\underset{̲}{i}+t^2\underset{̲}{j}$ . Find its equation in Cartesian form and sketch the path followed by the particle.

Tabulating $\underset{̲}{r}=x\underset{̲}{i}+y\underset{̲}{j}$ at different times $t$ :

Solution

To find the Cartesian equation of the curve we eliminate $t$ between $x=2t$ and $y=t^2$ . Re-arrange $x=2t$ as $t=\frac{1}{2}x$ . Then $y=t^2={\left(\frac{1}{2}x\right)}^2=\frac{1}{4}{x}^2$ , which is a parabola. This is the path followed by the particle. See Figure 1.

Figure 1:

In general, a three-dimensional vector function of one variable $t$ is of the form

$\underset{̲}{u}=x\left(t\right)\underset{̲}{i}+y\left(t\right)\underset{̲}{j}+z\left(t\right)\underset{̲}{k}$ .

Such functions may be differentiated one or more times and the rules of differentiation are derived from those for ordinary scalar functions. In particular, if $\underset{̲}{u}$ and $\underset{̲}{v}$ are vector functions of $t$ and if $c$ is a constant, then:

Also, if a particle moves so that its position vector at time $t$ is $\underset{̲}{r}\left(t\right)=x\left(t\right)\underset{̲}{i}+y\left(t\right)\underset{̲}{j}+z\left(t\right)\underset{̲}{k}$ then the velocity of the particle is

$\underset{̲}{v}=\frac{d\underset{̲}{r}}{dt}=\underset{̲}{ṙ}=\frac{dx\left(t\right)}{dt}\underset{̲}{i}+\frac{dy\left(t\right)}{dt}\underset{̲}{j}+\frac{dz\left(t\right)}{dt}\underset{̲}{k}=ẋ\underset{̲}{i}+ẏ\underset{̲}{j}+\mathrm{\ż}\underset{̲}{k}$

and its acceleration is

$\underset{̲}{a}=\frac{d\underset{̲}{v}}{dt}=\frac{d^2\underset{̲}{r}}{d{t}^2}=\underset{̲}{\ddot{r}}=\frac{d^{2}x\left(t\right)}{d{t}^2}\underset{̲}{i}+\frac{d^{2}y\left(t\right)}{d{t}^2}\underset{̲}{j}+\frac{d^{2}z\left(t\right)}{d{t}^2}\underset{̲}{k}=ẍ\underset{̲}{i}+ÿ\underset{̲}{j}+\ddot{z}\underset{̲}{k}$

Example 3

Find the derivative (with respect to $t$ ) of the position vector $\underset{̲}{r}=t^2\underset{̲}{i}+3t\underset{̲}{j}+4\underset{̲}{k}$ . Also find a unit vector tangential to the curve traced out by the position vector at the point where $t=2$ .

Solution

Differentiating $\underset{̲}{r}$ with respect to $t$ ,

$\underset{̲}{ṙ}=\frac{d\underset{̲}{r}}{dt}=2t\underset{̲}{i}+3\underset{̲}{j}$

so

$\underset{̲}{ṙ}\left(2\right)=4\underset{̲}{i}+3\underset{̲}{j}$

A unit vector in this direction, which is tangential to the curve, is

$\frac{\underset{̲}{ṙ}\left(2\right)}{\left|\underset{̲}{ṙ}\left(2\right)\right|}=\frac{4\underset{̲}{i}+3\underset{̲}{j}}{\sqrt{4^2+3^2}}=\frac{4}{5}\underset{̲}{i}+\frac{3}{5}\underset{̲}{j}$

Example 4

For the position vectors $\underset{̲}{r}=3\underset{̲}{i}+2t\underset{̲}{j}$ and $\underset{̲}{r}=2t\underset{̲}{i}+t^2\underset{̲}{j}$ use the general expressions for velocity and acceleration to confirm the values of $\underset{̲}{v}$ and $\underset{̲}{a}$ found earlier in Examples 1 and 2.

Solution

1. $\underset{̲}{r}=3\underset{̲}{i}+2t\underset{̲}{j}$ . Then

   $\underset{̲}{v}=\frac{d\underset{̲}{r}}{dt}=\underset{̲}{ṙ}=\frac{d}{dt}\left(3\underset{̲}{i}+2t\underset{̲}{j}\right)=\frac{d\left(3\right)}{dt}\underset{̲}{i}+\frac{d\left(2t\right)}{dt}\underset{̲}{j}=0\underset{̲}{i}+2\underset{̲}{j}=2\underset{̲}{j}$

   and

   $\underset{̲}{a}=\frac{d\underset{̲}{v}}{dt}=\underset{̲}{\ddot{r}}=\frac{d}{dt}\left(2\underset{̲}{j}\right)=\frac{d\left(2\right)}{dt}\underset{̲}{j}=0\underset{̲}{j}=\underset{̲}{0}$

   which agree with those found earlier.
   
   

2. $\underset{̲}{r}=2t\underset{̲}{i}+t^2\underset{̲}{j}$ . Then

   $\underset{̲}{v}=\frac{d\underset{̲}{r}}{dt}=\underset{̲}{ṙ}=\frac{d}{dt}\left(2t\underset{̲}{i}+t^2\underset{̲}{j}\right)=\frac{d\left(2t\right)}{dt}\underset{̲}{i}+\frac{d\left(t^2\right)}{dt}\underset{̲}{j}=2\underset{̲}{i}+2t\underset{̲}{j}$

   and

   $\underset{̲}{a}=\frac{d\underset{̲}{v}}{dt}=\underset{̲}{\ddot{r}}=\frac{d}{dt}\left(2\underset{̲}{i}+2t\underset{̲}{j}\right)=\frac{d\left(2\right)}{dt}\underset{̲}{i}+\frac{d\left(2t\right)}{dt}\underset{̲}{j}=0\underset{̲}{i}+2\underset{̲}{j}=2\underset{̲}{j}$

   which agree with those found earlier.

Example 5

A particle of mass $m=1$ kg has position vector $\underset{̲}{r}$ . The torque (moment of force) $\underset{̲}{H}$ relative to the origin acting on the particle as a result of a force $\underset{̲}{F}$ is defined as $\underset{̲}{H}=\underset{̲}{r}\times \underset{̲}{F}$ , where, by Newton’s second law, $\underset{̲}{F}=m\underset{̲}{\ddot{r}}$ . The angular momentum (moment of momentum) $\underset{̲}{L}$ of the particle is defined as $\underset{̲}{L}=\underset{̲}{r}\times m\underset{̲}{ṙ}$ . Find $\underset{̲}{L}$ and $\underset{̲}{H}$ for the particle where 1. $\underset{̲}{r}=3\underset{̲}{i}+2t\underset{̲}{j}$ and 2. $\underset{̲}{r}=2t\underset{̲}{i}+t^2\underset{̲}{j}$ , and show that in each case the torque law $\underset{̲}{H}=\underset{̲}{\dot{L}}$ is satisfied.

Solution

1. Here $\underset{̲}{r}=3\underset{̲}{i}+2t\underset{̲}{j}$ so $\underset{̲}{ṙ}=2\underset{̲}{j}$ and $\underset{̲}{a}=\underset{̲}{0}$ . Then

   $\underset{̲}{L}=\underset{̲}{r}\times m\underset{̲}{ṙ}=\left(3\underset{̲}{i}+2t\underset{̲}{j}\right)\times 2\underset{̲}{j}=6\underset{̲}{k}$ so $\underset{̲}{\dot{L}}=\frac{d}{dt}\left(6\right)\underset{̲}{k}=\underset{̲}{0}$

   and

   $\underset{̲}{H}=\underset{̲}{r}\times \underset{̲}{F}=\underset{̲}{r}\times m\underset{̲}{\ddot{r}}=\left(3\underset{̲}{i}+2t\underset{̲}{j}\right)\times \underset{̲}{0}=\underset{̲}{0}$ giving $\underset{̲}{H}=\underset{̲}{\dot{L}}$ as required.

2. Here $\underset{̲}{r}=2t\underset{̲}{i}+t^2\underset{̲}{j}$ so $\underset{̲}{ṙ}=2\underset{̲}{i}+2t\underset{̲}{j}$ and $\underset{̲}{a}=2\underset{̲}{j}$ . Then
   

   $\underset{̲}{L}=\underset{̲}{r}\times m\underset{̲}{ṙ}=\left(2t\underset{̲}{i}+t^2\underset{̲}{j}\right)\times \left(2\underset{̲}{i}+2t\underset{̲}{j}\right)=\left(4t^2-2t^2\right)\underset{̲}{k}=2t^2\underset{̲}{k}$ so $\underset{̲}{\dot{L}}=4t\underset{̲}{k}$

   and

   $\underset{̲}{H}=\underset{̲}{r}\times \underset{̲}{F}=\underset{̲}{r}\times m\underset{̲}{\ddot{r}}=\left(2t\underset{̲}{i}+t^2\underset{̲}{j}\right)\times 2\underset{̲}{j}=4t\underset{̲}{k}$ giving $\underset{̲}{H}=\underset{̲}{\dot{L}}$ as required.

Task!

A particle moves so that its position vector is $\underset{̲}{r}=12t\underset{̲}{i}+\left(19t-5t^2\right)\underset{̲}{j}$ .

1. Find $\frac{d\underset{̲}{r}}{dt}$ and $\frac{d^2\underset{̲}{r}}{d{t}^2}$ .
2. When is the $\underset{̲}{j}$ -component of $\frac{d\underset{̲}{r}}{dt}$ equal to zero?
3. Find a unit vector normal to its trajectory when $t=1$ .

Answer

Task!

A particle moving at a constant speed around a circle moves so that

$\underset{̲}{r}=cos\left(\pi t\right)\underset{̲}{i}+sin\left(\pi t\right)\underset{̲}{j}$

1. Find $\frac{d\underset{̲}{r}}{dt}$ and $\frac{d^2\underset{̲}{r}}{d{t}^2}$ .

    

2. Find $\underset{̲}{r}\cdot \frac{d\underset{̲}{r}}{dt}$ and $\underset{̲}{r}\times \frac{d^2\underset{̲}{r}}{d{t}^2}$ .

Answer

Task!

1. If $\underset{̲}{r}=sin\left(2t\right)\underset{̲}{i}+cos\left(2t\right)\underset{̲}{j}+t^2\underset{̲}{k}$ and $\left(1+t^2\right){\left|\ddot{\underset{̲}{r}}\right|}^2=c{\left|\dot{\underset{̲}{r}}\right|}^2$ , find the value of $c$ .

Answer