Title: Default for 12.5 Differentiation of Vectors
Course: HELM Workbooks 2023-2024

Differentiation of vectors

Consider the figure below.

[Image: See the text below.]

If \({\underline r}\) 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 \({\underline r}={\underline r}(t)\) to show the dependence upon time. Suppose that the object is at the point \(P\), with position vector \(\underline{r}(t)\) at time \(t\) and at the point \(Q\), with position vector \({\underline r}(t+\delta t)\), at the later time \(t+\delta t\), as shown in the figure below.

[Image: See the text above and below.]

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,

\[\mathrm{average} \; \mathrm{velocity} \; =\dfrac{\overrightarrow{PQ}}{\delta t}=\dfrac{{\underline r} (t+\delta t)-{\underline r}(t)}{\delta t}\]

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 \(\underline r\) with respect to \(t\). Not surprisingly we refer to this derivative as the instantaneous velocity, \({\underline v}\). By its very construction we see that the velocity vector is always tangential to the curve as the object moves along it. We have:

\[ \underline{v} = \lim_{\delta t \to 0} \frac{{\underline r}(t+\delta t)-{\underline r}(t)}{\delta t}=\frac{\mathrm{d}{\underline r}}{\mathrm{d}t}\]

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

\(\qquad {\underline r}(t)=x(t){\underline i}+y(t){\underline j}\)

Therefore,

\(\qquad {\underline r}(t+\delta t)=x(t+\delta t){\underline i}+y(t+\delta t){\underline j}\)

so that,

\[\begin{aligned} {\underline v}(t)&=\lim_{\delta t\to 0}\dfrac{x(t+\delta t){\underline i}+y(t+\delta t){\underline j}-x(t){\underline i}-y(t){\underline j}}{\delta t}\\ &= \lim_{\delta t\to 0}\left\{\dfrac{x(t+\delta t)-x(t)}{\delta t}{\underline i}+\dfrac{y(t+\delta t)-y(t)}{\delta t}{\underline j}\right\}\\ &=\dfrac{\mathrm{d}x}{\mathrm{d}t}{\underline i}+\dfrac{\mathrm{d}y}{\mathrm{d}t}{\underline j}\end{aligned}\]

This is often abbreviated to \({\underline v}=\dot{\underline r}=\dot x{\underline i}+\dot y{\underline j}\), 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.

Key Point

Given \(\qquad {\underline r}(t)=x(t){\underline i}+y(t){\underline j}+z(t){\underline k}\)

then the velocity vector is \[\qquad {\underline v} =\dot{\underline r}(t)=\dot x(t){\underline i}+\dot y(t){\underline j}+\dot z(t){\underline k}\] The magnitude of the velocity vector gives the speed of the object.

We can define the acceleration vector in a similar way, as the rate of change (i.e. the derivative) of the velocity with respect to the time: \[\qquad {\underline a}=\dfrac{\mathrm{d}{\underline v}}{\mathrm{d}t}=\dfrac{\mathrm{d}^2{\underline r}}{\mathrm{d}t^2}=\ddot {\underline r}=\ddot x{\underline i}+\ddot y{\underline j}+\ddot z{\underline k}\]

Example

If \({\underline w}=3t^2{\underline i}+\cos 2t{\underline j},\) find

1. \({\dfrac{\mathrm{d}{\underline w}}{\mathrm{d}t}}\)
2. \({\left|\dfrac{\mathrm{d}{\underline w}}{\mathrm{d}t}\right|}\)
3. \({\dfrac{\mathrm{d}^2{\underline w}}{\mathrm{d}t^2}}\)

Solution

1. If \({\underline w}=3t^2{\underline i}+\cos 2t{\underline j},\) then differentiation with respect to \(t\) yields: \({\dfrac{\mathrm{d}{\underline w}}{\mathrm{d}t}}=6t{\underline i}-2\sin 2t{\underline j}\)
2. \({\left|\dfrac{\mathrm{d}{\underline w}}{\mathrm{d}t}\right|}=\sqrt{(6t)^2+(-2\sin 2t)^2}=\sqrt{36t^2+4\sin^2 2t}\)
3. \({\dfrac{\mathrm{d}^2{\underline w}}{\mathrm{d}t^2}=6{\underline i}-4\cos 2t{\underline j}}\)

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

Key Point

If \({\underline w}\) and \({\underline z}\) are vectors and \(c\) is a scalar, all these being functions of time \(t\), then: \[\begin{aligned} \dfrac{\mathrm{d}}{\mathrm{d}t}({\underline w}+{\underline z})&=\dfrac{\mathrm{d}{\underline w}}{\mathrm{d}t}+\dfrac{\mathrm{d}{\underline z}}{\mathrm{d}t}\\ \dfrac{\mathrm{d}}{\mathrm{d}t}(c{\underline w})&=c\dfrac{\mathrm{d}{\underline w}}{\mathrm{d}t}+\dfrac{\mathrm{d}c}{\mathrm{d}t}{\underline w}\\ \dfrac{\mathrm{d}}{\mathrm{d}t}({\underline w}\cdot {\underline z})&={\underline w}\cdot \dfrac{\mathrm{d}{\underline z}}{\mathrm{d}t}+\dfrac{\mathrm{d}{\underline w}}{\mathrm{d}t}\cdot {\underline z}\\ \dfrac{\mathrm{d}}{\mathrm{d}t}({\underline w}\times {\underline z})&={\underline w}\times \dfrac{\mathrm{d}{\underline z}}{\mathrm{d}t}+\dfrac{\mathrm{d}{\underline w}}{\mathrm{d}t}\times {\underline z}\end{aligned}\]

Example

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

\({\dfrac{\mathrm{d}}{\mathrm{d}t}({\underline w}\cdot {\underline z})={\underline w}\cdot \dfrac{\mathrm{d}{\underline z}}{\mathrm{d}t}+\dfrac{\mathrm{d}{\underline w}}{\mathrm{d}t}\cdot {\underline z}}\)

Solution

\({\underline w}\cdot {\underline z}=(3t{\underline i}-t^2{\underline j})\cdot (2t^2{\underline i}+3{\underline j})=6t^3-3t^2.\)

Therefore \(\qquad \dfrac{\mathrm{d}}{\mathrm{d}t}({\underline w}\cdot {\underline z})=18t^2-6t \qquad (1)\)

Also \(\qquad \dfrac{\mathrm{d}{\underline w}}{\mathrm{d}t}=3{\underline i}-2t{\underline j}\qquad\text{and}\qquad \dfrac{\mathrm{d}{\underline z}}{\mathrm{d}t}=4t{\underline i}\)

\(\begin{aligned} \text{so}\qquad{\underline w}\cdot \dfrac{\mathrm{d}{\underline z}}{\mathrm{d}t}+{\underline z}\cdot \dfrac{\mathrm{d}{\underline w}}{\mathrm{d}t}&=(3t{\underline i}-t^2{\underline j})\cdot (4t{\underline i})+(2t^2{\underline i}+3{\underline j})\cdot (3{\underline i}-2t{\underline j})\\ &=12t^2+6t^2-6t\\ &=18t^2-6t \qquad\qquad (2)\end{aligned}\)

We have verified \({\dfrac{\mathrm{d}}{\mathrm{d}t}({\underline w}\cdot {\underline z})={\underline w}\cdot \dfrac{\mathrm{d}{\underline z}}{\mathrm{d}t}+\dfrac{\mathrm{d}{\underline w}}{\mathrm{d}t}\cdot {\underline z}}\) since (1) is the same as (2).

Example

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

\({\dfrac{\mathrm{d}}{\mathrm{d}t}({\underline w}\times {\underline z})={\underline w}\times \dfrac{\mathrm{d}{\underline z}}{\mathrm{d}t}+\dfrac{\mathrm{d}{\underline w}}{\mathrm{d}t}\times {\underline z}}\)

Solution

\({\underline w}\times {\underline z}=\left|\begin{array}{ccc}{\underline i}&{\underline j}&{\underline k}\\ 3t&-t^2&0\\ 2t^2&3&0 \end{array}\right| =(9t+2t^4){\underline k}\) \(\qquad \text{implying} \qquad\) \({\dfrac{\mathrm{d}}{\mathrm{d}t}({\underline w}\times {\underline z})=(9+8t^3){\underline k}} \qquad (1)\)

\(\begin{aligned} {\underline w}\times \dfrac{\mathrm{d}{\underline z}}{\mathrm{d}t}&=&\left|\begin{array}{ccc}{\underline i}&{\underline j}&{\underline k}\\ 3t&-t^2&0\\ 4t&0&0 \end{array}\right|=4t^3{\underline k}\qquad (2)\end{aligned}\)

\(\begin{aligned} \dfrac{\mathrm{d}{\underline w}}{\mathrm{d}t}\times {\underline z}&=&\left|\begin{array}{ccc}{\underline i}&{\underline j}&{\underline k}\\ 3&-2t&0\\ 2t^2&3&0 \end{array}\right|=(9+4t^3){\underline k}\qquad (3)\end{aligned}\)

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

Exercise

If \(\underline{r} = [value]\underline{i} + [value]\underline{j} + [value]\underline{k}\), find

1. \(\dfrac{d\underline{r}}{dt} =\) \(\underline{i} \; +\) \(\underline{j} \; +\) \(\underline{k}\).
2. \(\dfrac{d^2\underline{r}}{dt^2} =\) \(\underline{i} \; +\) \(\underline{j} \; +\) \(\underline{k}\).
Feedback: Solution We are given the vector: \(\underline{r} = [value]\underline{i} + [value]\underline{j} + [value]\underline{k}.\) 1. \(\dfrac{d\underline{r}}{dt} = [value]\underline{i} + [value]\underline{j} + [value]\underline{k}.\) 2. \(\dfrac{d^2\underline{r}}{dt^2} = \dfrac{d}{dt}\left(\dfrac{d\underline{r}}{dt}\right) = [value]\underline{i} + [value]\underline{j} + [value]\underline{k}.\)

Exercise

If \(\underline{B} = [value]\underline{i} + [value]\underline{j}\), find

1. \(\dfrac{d\underline{B}}{dt} =\) \(\underline{i} \; +\) \(\underline{j}.\)
2. \(\dfrac{d^2\underline{B}}{dt^2} =\) \(\underline{i} \; +\) \(\underline{j}.\)
Feedback: Solution We are given: \(\underline{B} = [value]\underline{i} + [value]\underline{j}.\) NB: The term \( [value]\) requires the product rule to be used when differentiating. Note that \( \dfrac{d}{dt}\left(e^{at}\right) = ae^{at}\), \( \dfrac{d}{dt}(\displaystyle\cos(t)) = -\displaystyle\sin(t)\) and that \( \dfrac{d}{dt}(\displaystyle\sin(t)) = \displaystyle\cos(t)\). 1. \(\dfrac{d\underline{B}}{dt} = ([value])\underline{i} + [value]\underline{j}\). 2. \(\dfrac{d^2\underline{B}}{dt^2} = \dfrac{d}{dt}\left(\dfrac{d\underline{B}}{dt}\right) = ([value])\underline{i} + [value]\underline{j}\).

Exercise

If \(\underline{r} = [value]\underline{i} + [value]\underline{j} - [value]\underline{k}\), evaluate \(\underline{r}\) and \(\underline{r'} = \dfrac{d\underline{r}}{dt}\) when \(t = 1.\)

1. \(\underline{r}(1) = \) \(\underline{i} \; +\) \(\underline{j} \; -\) \(\underline{k}.\)
2. \(\underline{r'}(1) =\) \(\underline{i} \; +\) \(\underline{j} \; +\) \(\underline{k}.\)
Feedback: Solution We are given \(\underline{r} = [value]\underline{i} + [value]\underline{j} - [value]\underline{k}\), and asked to evaluate \(\underline{r}\) and \(\underline{\dot{r}}\) at \(t = 1.\) 1. \( \underline{r}(1) = [value](1)^2\underline{i} \; + \; [value](1)\underline{j} \; - [value]\underline{k} = [value]\underline{i} \; + \; [value]\underline{j} \; - [value]\underline{k}\). 2. \( \underline{\dot{r}}(t)=[value]t\underline{i} \; + \; [value]\underline{j} ,\) so: \( \underline{\dot{r}}(1)= [value](1)\underline{i} \; + \; [value]\underline{j} = [value]\underline{i} \; + \; [value]\underline{j}\).

Exercise

Let \(\underline{w} = [value]\underline{i} \; [value]\underline{k}\) and \(\underline{z} = ([value])\underline{i} \; + \; [value]\underline{j} \; [value]\underline{k}\).

1. Find \(\underline{w}\cdot\underline{z} = \)
2. Find \(\dfrac{d\underline{w}}{dt} = \) \(\underline{i} \; +\) \(\underline{j} \; +\) \(\underline{k}\).
3. Find \(\dfrac{d\underline{z}}{dt} = \) \(\underline{i} \; +\) \(\underline{j} \; + \;\) \(\underline{k}\).
4. True or False: \(\dfrac{d}{dt}(\underline{w}\cdot\underline{z}) = \underline{w}\cdot \dfrac{d\underline{z}}{dt} + \dfrac{d\underline{w}}{dt}\cdot\underline{z}\).
Feedback: Solution We are given the vectors \(\underline{w} = [value]\underline{i} \; [value]\underline{k}\) and \(\underline{z} = ([value])\underline{i} \; + \; [value]\underline{j} \; [value]\underline{k}\). 1. \( \begin{align} \underline{w}\cdot\underline{z} &= \left( [value]\underline{i} \; + \; 0\underline{j} [value]\underline{k} \right) \cdot \left( ([value])\underline{i} \; + \; [value]\underline{j} \; [value]\underline{k} \right) \\ &= [value]\cdot([value]) + [value]\cdot[value] + ([value])\cdot([value]) \\ & = [value] \end{align}\) 2. \( \dfrac{d}{dt}([value]) = [value] \) and \( \dfrac{d}{dt}([value]) = [value] \). So we have \( \dfrac{d\underline{w}}{dt} = [value]\underline{i} \; + \; 0\underline{j} \; [value]\underline{k} \). 3. \( \dfrac{d}{dt}([value]) = [value] \) and \( \dfrac{d}{dt}[value] = [value] \) and \( \dfrac{d}{dt}([value]) = [value] \). So we have \( \dfrac{d\underline{z}}{dt} = [value]\underline{i} \; + \; [value]\underline{j} \; + \; [value]\underline{k}\). 4. \(\dfrac{d}{dt}(\underline{w}\cdot\underline{z}) = [value]\). \( \underline{w}\cdot \dfrac{d\underline{z}}{dt} + \dfrac{d\underline{w}}{dt}\cdot\underline{z} = ([value]) + ([value]) = [value] \). Therefore the statement is true. Alternatively, you could spot that the statement is essentially the product rule (which is true).

Exercise

Given \(\underline{r} = [value]\underline{i} \; + \; [value]\underline{j}\):

1. Find \(\underline{\dot{r}} =\) \(\underline{i} \; + \; \) \(\underline{j}\).
2. Find \(\underline{\ddot{r}} =\) \(\underline{i} \; + \; \) \(\underline{j}\).
3. Find \( |\underline{r}| = \)
4. Find \(\underline{r}\cdot\underline{\dot{r}} = \)
From your answer deduce whether the following statement is True or False:
The position vector \(\underline{r}\) and the velocity vector \(\underline{\dot{r}}\) are perpendicular.
Feedback: Solution We are given the vector: \(\underline{r} = [value]\underline{i} \; + \; [value]\underline{j}.\) 1. \( \dfrac{d}{dt}([value]) = [value] \) and \( \dfrac{d}{dt}([value]) = [value] \). So we have \( \underline{\dot{r}} = \dfrac{d}{dt}( \underline{r}) = [value]\underline{i} [value]\underline{j} \). 2. \( \dfrac{d}{dt}([value]) = [value] \) and \( \dfrac{d}{dt}([value]) = [value] \). So we have \( \underline{\ddot{r}} = [value]\underline{i} [value]\underline{j} \). 3. Using the trigonometric identity \( \displaystyle\sin^2(x) + \displaystyle\cos^2(x) = 1\) for all \(x\), we have \( \sqrt{[value]^2 + [value]^2} = 1\). 4. \( \left( [value]\underline{i} \; + \; [value]\underline{j} \right) \cdot \left( [value]\underline{i} [value]\underline{j} \right) = [value]\cdot[value] + [value]\cdot([value]) = [value]\).