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Differentiation of vectors

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(Last Updated On: May 25, 2020)

Differentiation of vectors. Hello friends, today it’s all about differentiation of vectors. Have a look!!



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Differentiation of vectors

Differentiation of vectors

‘Differentiation of vectors’ is a quite simple straightforward thing.

It happens when the vector has a parametric form like \textbf{AB} = r^2 \textbf{i}+ (2r - 11) \textbf{j}+3 \textbf{k}.

Also, the differentiation of vectors follows standard rules of differentiation.

Where to use ‘differentiation of vectors’

Differentiation of vectors is used to get the equation of unit tangent vector in vector analysis.

Just have a look at these two examples!!

Examples of differentiation of vectors

Disclaimer: None of these examples are mine. I have chosen these from some book or books. I have also given the due reference at the end of the post.

So here is the first example.

Example 1

According to Stroud and Booth (2011)* “If \textbf{F} = u \textbf{i}+ (1 - u) \textbf{j}+3u \textbf{k} and \textbf{G} = 2\textbf{i}- (1+u)\textbf{j}-u^2\textbf{k}, determine (a) \cfrac{\text{d}}{\text{d}u}(\textbf{F}.\textbf{G}); (b) \cfrac{\text{d}}{\text{d}u}(\textbf{F}\times\textbf{G}); (c) \cfrac{\text{d}}{\text{d}u}(\textbf{F}+\textbf{G}).”

Solution

Now here I have two vectors, \textbf{F} and \textbf{G}.

Also I know that the vector \textbf{F} is \textbf{F} = u \textbf{i}+ (1 - u) \textbf{j}+3u \textbf{k}. Also the vector \textbf{G} is \textbf{G} = 2\textbf{i}- (1+u)\textbf{j}-u^2\textbf{k}.

So I will start with part (a).

(a)

As I can see, I have to get the value of \cfrac{\text{d}}{\text{d}u}(\textbf{F}.\textbf{G}).

Now (\textbf{F}.\textbf{G}) means the scalar product of two vectors \textbf{F} and \textbf{G}.

As per the formula of scalar product of two vectors, the value of (\textbf{F}.\textbf{G}) is

    \[(\textbf{F}.\textbf{G}) = (u)(2)+(1-u)[-(1+u)]+(3u)(-u^2).\]

Now I’ll simplify it to get 

    \begin{eqnarray*} (\textbf{F}.\textbf{G}) &=& 2u -(1-u)(1+u)-3u^3\\ &=& 2u - (1 - u^2) - 3 u^3\\&=& 2u -1+ u^2 - 3 u^3.\end{eqnarray*}

Thus I get the value of (\textbf{F}.\textbf{G}) as

(1)   \begin{equation*} (\textbf{F}.\textbf{G}) = 2u -1+ u^2 - 3 u^3. \end{equation*}

Now I will differentiate both sides of equation (1) with respect to u. For that, I have to use the standard formulas for differentiation.

So I get

    \[\frac{\text{d}}{\text{d}u} (\textbf{F}.\textbf{G}) = \cfrac{\text{d}}{\text{d}u} \left[2u -1+ u^2 - 3 u^3\right].\]

And this means 

    \[\frac{\text{d}}{\text{d}u} (\textbf{F}.\textbf{G}) = 2(1) -0 + 2u - 3(3) u^2.\]

Next, I’ll simplify it to get

    \[\frac{\text{d}}{\text{d}u} (\textbf{F}.\textbf{G}) = 2+2u-9u^2.\]

Hence I can conclude that \cfrac{\text{d}}{\text{d}u} (\textbf{F}.\textbf{G}) = 2+2u-9u^2 is the answer to part (a).

Now I move to part (b).

(b)

Here I have to get the value of \cfrac{\text{d}}{\text{d}u}(\textbf{F}\times \textbf{G}).

Now (\textbf{F}\times \textbf{G}) means the vector product of two vectors \textbf{F} and \textbf{G}.

As per the formula of vector product of two vectors, the value of (\textbf{F}\times \textbf{G}) is 

    \[ (\textbf{A}\times\textbf{B}) = \left|\begin{array}{ccc} \textbf{i} & \textbf{j} & \textbf{k}  \\ u & 1 - u & 3u \\ 2 & - (1+u) & -u^2 \end{array} \right|. \]

So this gives

    \begin{eqnarray*} (\textbf{F}\times\textbf{G}) &=& \textbf{i}[(1-u)(-u^2)+(3u)(1+u)] - \textbf{j} [(u)(-u^2)-(2)(3u)] \\ &&+ \textbf{k}[-(u)(1+u)-(2)(1-u)]\\ &=& \textbf{i}[-u^2+u^3+3u+3u^2] - \textbf{j} [-u^3-6u] \\ &&+ \textbf{k}[-u-u^2-2+2u].\end{eqnarray*}

Next I will simplify it to get

    \begin{eqnarray*} (\textbf{F}\times\textbf{G}) &=& \textbf{i}[u^3+3u+2u^2] - \textbf{j} [-u^3-6u] + \textbf{k}[-u^2-2+u]\\ &=& \textbf{i}[u^3+2u^2+3u] + \textbf{j} [u^3+6u] - \textbf{k}[u^2-u+2]\\ &=& [u^3+2u^2+3u] \textbf{i} + [u^3+6u]\textbf{j} - [u^2-u+2]\textbf{k}. \end{eqnarray*}

Thus I get the value of (\textbf{F}\times\textbf{G}) as

(2)   \begin{equation*} (\textbf{F}\times\textbf{G}) = [u^3+2u^2+3u] \textbf{i} + [u^3+6u]\textbf{j} - [u^2-u+2]\textbf{k}. \end{equation*}

Again I will use standard formulas in differentiation to differentiate both sides of equation (2) with respect to u.

Therefore it will be

    \begin{eqnarray*}\frac{\text{d}}{\text{d}u}(\textbf{F}\times\textbf{G}) &=& \left[\frac{\text{d}}{\text{d}u}\left(u^3+2u^2+3u\right)\right] \textbf{i} + \left[\frac{\text{d}}{\text{d}u}\left(u^3+6u\right)\right]\textbf{j} \\ &&- \left[\frac{\text{d}}{\text{d}u}\left(u^2-u+2\right)\right]\textbf{k}.\end{eqnarray*}

So this gives

    \[\frac{\text{d}}{\text{d}u}(\textbf{F}\times\textbf{G})= (3u^2+4u+3)\textbf{i} +(3u^2+6)\textbf{j}-(2u-1)\textbf{k}.\]

Hence I can conclude that \cfrac{\text{d}}{\text{d}u} (\textbf{F}\times\textbf{G}) = (3u^2+4u+3)\textbf{i} +(3u^2+6)\textbf{j}-(2u-1)\textbf{k} is the answer to part (b).

Now I move to part (c).

(c)

Here I have to get the value of \cfrac{\text{d}}{\text{d}u}(\textbf{F}+\textbf{G}).

Now (\textbf{F} +\textbf{G}) means adding two vectors \textbf{F} and \textbf{G}.

Thus it will be

    \[(\textbf{F} +\textbf{G}) = [(u) + (2)]\textbf{i} + [(1-u)-(1+u)] \textbf{j} + [3u- u^2]\textbf{k}.\]

Now I’ll simplify it to get

    \begin{eqnarray*} (\textbf{F} +\textbf{G}) &=& [u + 2]\textbf{i} + [1-u-1-u] \textbf{j} + [3u- u^2]\textbf{k} \\ &=& [u + 2]\textbf{i} - [2u] \textbf{j} + [3u- u^2]\textbf{k}.\end{eqnarray*}

Next, I will differentiate both sides with respect to u to get

    \begin{eqnarray*} \frac{\text{d}}{\text{d}u}(\textbf{F}+\textbf{G}) &=& \left[\frac{\text{d}}{\text{d}u} (u + 2)\right]\textbf{i} - \left[\frac{\text{d}}{\text{d}u}(2u) \right]\textbf{j}+ \left[\frac{\text{d}}{\text{d}u}(3u- u^2)\right]\textbf{k}\\ &=& (1+0)\textbf{i}-(2)\textbf{j}+(3-2u)\textbf{k}\\ &=& \textbf{i} - 2 \textbf{j} + (3-2u)\textbf{k}. \end{eqnarray*}

Hence I can conclude that \cfrac{\text{d}}{\text{d}u} (\textbf{F}+\textbf{G}) = \textbf{i} - 2 \textbf{j} + (3-2u)\textbf{k} is the answer to part (c).

This is the end of example 1. Now I will work on the second example.



Example 2

According to Stroud and Booth (2011)* “If \textbf{r} = (t^2 + 3t) \textbf{i} - 2 \sin 3t \textbf{j}+3e^{2t} \textbf{k}, determine (a) \cfrac{\text{d}\textbf{r}}{\text{d}t}; (b) \cfrac{\text{d}^2\textbf{r}}{\text{d}t^2}; (c) the value of \left|\cfrac{\text{d}^2\textbf{r}}{\text{d}t^2}\right| at t = 0.”

Solution

Now here the given vector is \textbf{r} = (t^2 + 3t) \textbf{i} - 2 \sin 3t \textbf{j}+3e^{2t} \textbf{k}. So I will start with part (a) of the problem.

(a)

Now I have to get the value of \cfrac{\text{d}\textbf{r}}{\text{d}t}.

Like the first example, here also I will differentiate the vector \textbf{r} following standard rules of differentiation.

Thus it will be

    \[\frac{\text{d}\textbf{r}}{\text{d}t} =\left[ \frac{\text{d}}{\text{d}t}(t^2 + 3t) \right]\textbf{i} - \left[ \frac{\text{d}}{\text{d}t}( 2 \sin 3t) \right]\textbf{j}+\left[ \frac{\text{d}}{\text{d}t}(3e^{2t}) \right]\textbf{k}.\]

Next, I’ll simplify it to get

    \[\frac{\text{d}\textbf{r}}{\text{d}t} = (2t+3)\textbf{i}-(6 \cos 3t)\textbf{j}+(6e^{2t})\textbf{k}\\&=& (2t+3)\textbf{i}-6 \cos 3t\textbf{j}+6e^{2t}\textbf{k}.\]

This is the answer to part (a). Now I’ll move to part (b).

(b)

Next, I have to get the value of \cfrac{\text{d}^2\textbf{r}}{\text{d}t^2}.

Now, I have already got the value of \cfrac{\text{d}\textbf{r}}{\text{d}t} from part (a).

Thus I will differentiate both sides of \cfrac{\text{d}\textbf{r}}{\text{d}t} with respect to t to get the value of \cfrac{\text{d}^2\textbf{r}}{\text{d}t^2} as

    \[\frac{\text{d}^2\textbf{r}}{\text{d}t^2} = \left[ \frac{\text{d}}{\text{d}t}(2t+3) \right]\textbf{i} - \left[ \frac{\text{d}}{\text{d}t}(6 \cos 3t) \right]\textbf{j}+\left[\frac{\text{d}}{\text{d}t}(6e^{2t}) \right]\textbf{k}.\]

So it will be 

    \[\frac{\text{d}^2\textbf{r}}{\text{d}t^2} = \textbf{i} + 18 \sin 3t \textbf{j} + 12 e^{2t} \textbf{k}.\]

Hence I can conclude that \cfrac{\text{d}^2\textbf{r}}{\text{d}t^2} = 2 \textbf{i} + 18 \sin 3t \textbf{j} + 12 e^{2t} \textbf{k} is the answer to part (b) of this example.

Next, I will solve the next part (c).

(c)

Now I have to get the value of \left|\cfrac{\text{d}^2\textbf{r}}{\text{d}t^2}\right| at t = 0.

First of all, I will get the value of \left|\cfrac{\text{d}^2\textbf{r}}{\text{d}t^2}\right|.

Now, \left|\cfrac{\text{d}^2\textbf{r}}{\text{d}t^2}\right| means the absolute value of \cfrac{\text{d}^2\textbf{r}}{\text{d}t^2}.

Suppose \textbf{x} is a vector with \textbf{x} = a_1\textbf{i} +a_2\textbf{j} +a_3\textbf{k}.

Then the absolute value of the vector \textbf{x} will be

    \[|\textbf{x}|=\sqrt{a_1^{2}+a_2^{2}+a_3^{2}}.\]

So, in this example, the value of \left|\cfrac{\text{d}^2\textbf{r}}{\text{d}t^2}\right| will be

    \[\left|\frac{\text{d}^2\textbf{r}}{\text{d}t^2}\right| = \sqrt{(2)^2+(18 \sin 3t)^2+(12 e^{2t})^2}.\]

Hence it gives 

    \[\left|\frac{\text{d}^2\textbf{r}}{\text{d}t^2}\right| = \sqrt{4+(18)^2 \sin^2 3t + (12)^2 e^{4t}}.\]

Now at t = 0, it will be

    \begin{eqnarray*} \left|\frac{\text{d}^2\textbf{r}}{\text{d}t^2}\right| &=& \sqrt{4+(18)^2 \sin^2 3.0 + (12)^2 e^{4. 0}} \\ &=& \sqrt{4+(18)^2 \sin^2 0 + (12)^2 e^{0}}.\end{eqnarray*}

Also, I already know that \sin 0 = 0 and e^{0}=1.

Hence I can say that

    \[(18)^2 \sin^2 0 = (18)^2.0 = 0\]

and

    \[(12)^2 e^{0}= (12)^2.1 = 144.\]

Thus the value of \left|\cfrac{\text{d}^2\textbf{r}}{\text{d}t^2}\right| will be

    \begin{eqnarray*} \left|\frac{\text{d}^2\textbf{r}}{\text{d}t^2}\right| &=&\sqrt{4+144+0} \\&=&\sqrt{148}\\ &=&2\sqrt{37}. \end{eqnarray*}

Thus my conclusion is that the value of \left|\cfrac{\text{d}^2\textbf{r}}{\text{d}t^2}\right| at t = 0 is 2\sqrt{37}.

So here ends my second as well as the last example.

Dear friends, this is the end of my today’s post. Thank you very much for reading this. Please let me know how you feel about it. Soon I will be back again with a new post. Till then, bye, bye!!

*Reference: K. A. Stroud and Dexter J. Booth (2011): Advanced engineering mathematics, Industrial Press, Inc.; 5th edition (March 8, 2011), Chapter: Vector analysis 1, Further problems 22, p. 815, Q. No. 5 (Example 1);  Q. No. 18 (Example 2).

Filed Under: Engineering mathematics 2, Vector analysis, Vector differential calculus

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About Dr. Aspriha Peters

Trained mathematician & hobby academic. Curious about nature as well as an aspiring blogger.

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