# 3.2 Calculus of vector-valued functions  (Page 3/11)

 Page 3 / 11

Now for some examples using these properties.

## Using the properties of derivatives of vector-valued functions

Given the vector-valued functions

$\text{r}\left(t\right)=\left(6t+8\right)\phantom{\rule{0.1em}{0ex}}\text{i}+\left(4{t}^{2}+2t-3\right)\phantom{\rule{0.1em}{0ex}}\text{j}+5t\phantom{\rule{0.1em}{0ex}}\text{k}$

and

$\text{u}\left(t\right)=\left({t}^{2}-3\right)\phantom{\rule{0.1em}{0ex}}\text{i}+\left(2t+4\right)\phantom{\rule{0.1em}{0ex}}\text{j}+\left({t}^{3}-3t\right)\phantom{\rule{0.1em}{0ex}}\text{k},$

calculate each of the following derivatives using the properties of the derivative of vector-valued functions.

1. $\frac{d}{dt}\left[\phantom{\rule{0.1em}{0ex}}\text{r}\left(t\right)·\phantom{\rule{0.1em}{0ex}}\text{u}\left(t\right)\right]$
2. $\frac{d}{dt}\left[\phantom{\rule{0.1em}{0ex}}\text{u}\left(t\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{u}^{\prime }\left(t\right)\right]$
1. We have ${r}^{\prime }\left(t\right)=6\phantom{\rule{0.1em}{0ex}}\text{i}+\left(8t+2\right)\phantom{\rule{0.1em}{0ex}}\text{j}+5\phantom{\rule{0.1em}{0ex}}\text{k}$ and ${u}^{\prime }\left(t\right)=2t\phantom{\rule{0.1em}{0ex}}\text{i}+2\phantom{\rule{0.1em}{0ex}}\text{j}+\left(3{t}^{2}-3\right)\phantom{\rule{0.1em}{0ex}}\text{k}.$ Therefore, according to property iv.:
$\begin{array}{cc}\hfill \frac{d}{dt}\left[\phantom{\rule{0.1em}{0ex}}\text{r}\left(t\right)·\phantom{\rule{0.1em}{0ex}}\text{u}\left(t\right)\right]& ={r}^{\prime }\left(t\right)·\phantom{\rule{0.1em}{0ex}}\text{u}\left(t\right)+\phantom{\rule{0.1em}{0ex}}\text{r}\left(t\right)·\phantom{\rule{0.1em}{0ex}}{u}^{\prime }\left(t\right)\hfill \\ & =\left(6\phantom{\rule{0.1em}{0ex}}\text{i}+\left(8t+2\right)\phantom{\rule{0.1em}{0ex}}\text{j}+5\phantom{\rule{0.1em}{0ex}}\text{k}\right)·\left(\left({t}^{2}-3\right)\phantom{\rule{0.1em}{0ex}}\text{i}+\left(2t+4\right)\phantom{\rule{0.1em}{0ex}}\text{j}+\left({t}^{3}-3t\right)\phantom{\rule{0.1em}{0ex}}\text{k}\right)\hfill \\ & \phantom{\rule{1em}{0ex}}+\left(\left(6t+8\right)\phantom{\rule{0.1em}{0ex}}\text{i}+\left(4{t}^{2}+2t-3\right)\phantom{\rule{0.1em}{0ex}}\text{j}+5t\phantom{\rule{0.1em}{0ex}}\text{k}\right)·\left(2t\phantom{\rule{0.1em}{0ex}}\text{i}+2\phantom{\rule{0.1em}{0ex}}\text{j}+\left(3{t}^{2}-3\right)\phantom{\rule{0.1em}{0ex}}\text{k}\right)\hfill \\ & =6\left({t}^{2}-3\right)+\left(8t+2\right)\left(2t+4\right)+5\left({t}^{3}-3t\right)\hfill \\ & \phantom{\rule{1em}{0ex}}+2t\left(6t+8\right)+2\left(4{t}^{2}+2t-3\right)+5t\left(3{t}^{2}-3\right)\hfill \\ & =20{t}^{3}+42{t}^{2}+26t-16.\hfill \end{array}$
2. First, we need to adapt property v. for this problem:
$\frac{d}{dt}\left[\phantom{\rule{0.1em}{0ex}}\text{u}\left(t\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{u}^{\prime }\left(t\right)\right]={u}^{\prime }\left(t\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{u}^{\prime }\left(t\right)+\phantom{\rule{0.1em}{0ex}}\text{u}\left(t\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u″}\left(t\right).$

Recall that the cross product of any vector with itself is zero. Furthermore, $\text{u″}\left(t\right)$ represents the second derivative of $\text{u}\left(t\right)\text{:}$

$\text{u″}\left(t\right)=\frac{d}{dt}\left[\phantom{\rule{0.1em}{0ex}}{u}^{\prime }\left(t\right)\right]=\frac{d}{dt}\left[2t\phantom{\rule{0.1em}{0ex}}\text{i}+2\phantom{\rule{0.1em}{0ex}}\text{j}+\left(3{t}^{2}-3\right)\phantom{\rule{0.1em}{0ex}}\text{k}\right]=2\phantom{\rule{0.1em}{0ex}}\text{i}+6t\phantom{\rule{0.1em}{0ex}}\text{k}.$

Therefore,

$\begin{array}{cc}\hfill \frac{d}{dt}\left[\phantom{\rule{0.1em}{0ex}}\text{u}\left(t\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{u}^{\prime }\left(t\right)\right]& =0+\left(\left({t}^{2}-3\right)\phantom{\rule{0.1em}{0ex}}\text{i}+\left(2t+4\right)\phantom{\rule{0.1em}{0ex}}\text{j}+\left({t}^{3}-3t\right)\phantom{\rule{0.1em}{0ex}}\text{k}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(2\phantom{\rule{0.1em}{0ex}}\text{i}+6t\phantom{\rule{0.1em}{0ex}}\text{k}\right)\hfill \\ & =|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ {t}^{2}-3& 2t+4& {t}^{3}-3t\\ 2& 0& 6t\end{array}|\hfill \\ & =6t\left(2t+4\right)\phantom{\rule{0.1em}{0ex}}\text{i}-\left(6t\left({t}^{2}-3\right)-2\left({t}^{3}-3t\right)\right)\phantom{\rule{0.1em}{0ex}}\text{j}-2\left(2t+4\right)\phantom{\rule{0.1em}{0ex}}\text{k}\hfill \\ & =\left(12{t}^{2}+24t\right)\phantom{\rule{0.1em}{0ex}}\text{i}+\left(12t-4{t}^{3}\right)\phantom{\rule{0.1em}{0ex}}\text{j}-\left(4t+8\right)\phantom{\rule{0.1em}{0ex}}\text{k}.\hfill \end{array}$

Given the vector-valued functions $\text{r}\left(t\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{i}+\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{j}-{e}^{2t}\phantom{\rule{0.1em}{0ex}}\text{k}$ and $\text{u}\left(t\right)=t\phantom{\rule{0.1em}{0ex}}\text{i}+\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{j}+\text{cos}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{k},$ calculate $\frac{d}{dt}\left[\phantom{\rule{0.1em}{0ex}}\text{r}\left(t\right)·\phantom{\rule{0.1em}{0ex}}{r}^{\prime }\left(t\right)\right]$ and $\frac{d}{dt}\left[\phantom{\rule{0.1em}{0ex}}\text{u}\left(t\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{r}\left(t\right)\right].$

$\frac{d}{dt}\left[\phantom{\rule{0.1em}{0ex}}\text{r}\left(t\right)·\phantom{\rule{0.1em}{0ex}}{r}^{\prime }\left(t\right)\right]=8{e}^{4t}$

$\begin{array}{l}\frac{d}{dt}\left[\phantom{\rule{0.1em}{0ex}}\text{u}\left(t\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{r}\left(t\right)\right]\\ =-\left({e}^{2t}\left(\text{cos}\phantom{\rule{0.1em}{0ex}}t+2\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t\right)+\text{cos}\phantom{\rule{0.1em}{0ex}}2t\right)\phantom{\rule{0.1em}{0ex}}\text{i}+\left({e}^{2t}\left(2t+1\right)-\text{sin}\phantom{\rule{0.1em}{0ex}}2t\right)\phantom{\rule{0.1em}{0ex}}\text{j}+\left(t\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}t+\text{sin}\phantom{\rule{0.1em}{0ex}}t-\text{cos}\phantom{\rule{0.1em}{0ex}}2t\right)\phantom{\rule{0.1em}{0ex}}\text{k}\end{array}$

## Tangent vectors and unit tangent vectors

Recall from the Introduction to Derivatives that the derivative at a point can be interpreted as the slope of the tangent line to the graph at that point. In the case of a vector-valued function, the derivative provides a tangent vector to the curve represented by the function. Consider the vector-valued function $\text{r}\left(t\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{i}+\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{j}.$ The derivative of this function is ${r}^{\prime }\left(t\right)=-\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{i}+\text{cos}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{j}.$ If we substitute the value $t=\pi \text{/}6$ into both functions we get

$\text{r}\left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}\phantom{\rule{0.1em}{0ex}}\text{i}+\frac{1}{2}\phantom{\rule{0.1em}{0ex}}\text{j}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{r}^{\prime }\left(\frac{\pi }{6}\right)=-\frac{1}{2}\phantom{\rule{0.1em}{0ex}}\text{i}+\frac{\sqrt{3}}{2}\phantom{\rule{0.1em}{0ex}}\text{j}.$

The graph of this function appears in [link] , along with the vectors $\text{r}\left(\frac{\pi }{6}\right)$ and ${r}^{\prime }\left(\frac{\pi }{6}\right).$

Notice that the vector ${r}^{\prime }\left(\frac{\pi }{6}\right)$ is tangent to the circle at the point corresponding to $t=\pi \text{/}6.$ This is an example of a tangent vector    to the plane curve defined by $\text{r}\left(t\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{i}+\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{j}.$

## Definition

Let C be a curve defined by a vector-valued function r, and assume that ${r}^{\prime }\left(t\right)$ exists when $t={t}_{0}.$ A tangent vector v at $t={t}_{0}$ is any vector such that, when the tail of the vector is placed at point $\text{r}\left({t}_{0}\right)$ on the graph, vector v is tangent to curve C. Vector ${r}^{\prime }\left({t}_{0}\right)$ is an example of a tangent vector at point $t={t}_{0}.$ Furthermore, assume that ${r}^{\prime }\left(t\right)\ne \phantom{\rule{0.1em}{0ex}}0.$ The principal unit tangent vector    at t is defined to be

$\text{T}\left(t\right)=\frac{{r}^{\prime }\left(t\right)}{\text{‖}\phantom{\rule{0.1em}{0ex}}{r}^{\prime }\left(t\right)\text{‖}},$

provided $\text{‖}\phantom{\rule{0.1em}{0ex}}{r}^{\prime }\left(t\right)\text{‖}\ne 0.$

The unit tangent vector is exactly what it sounds like: a unit vector that is tangent to the curve. To calculate a unit tangent vector, first find the derivative ${r}^{\prime }\left(t\right).$ Second, calculate the magnitude of the derivative. The third step is to divide the derivative by its magnitude.

## Finding a unit tangent vector

Find the unit tangent vector for each of the following vector-valued functions:

1. $\text{r}\left(t\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{i}+\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{j}$
2. $\text{u}\left(t\right)=\left(3{t}^{2}+2t\right)\phantom{\rule{0.1em}{0ex}}\text{i}+\left(2-4{t}^{3}\right)\phantom{\rule{0.1em}{0ex}}\text{j}+\left(6t+5\right)\phantom{\rule{0.1em}{0ex}}\text{k}$

1. $\begin{array}{cccc}\text{First step:}\hfill & \hfill {r}^{\prime }\left(t\right)& =\hfill & \mathrm{-sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{i}+\text{cos}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{j}\hfill \\ \text{Second step:}\hfill & \hfill \text{‖}\phantom{\rule{0.1em}{0ex}}{r}^{\prime }\left(t\right)\text{‖}& =\hfill & \sqrt{{\left(\text{−}\text{sin}\phantom{\rule{0.1em}{0ex}}t\right)}^{2}+{\left(\text{cos}\phantom{\rule{0.1em}{0ex}}t\right)}^{2}}=1\hfill \\ \text{Third step:}\hfill & \hfill \text{T}\left(t\right)& =\hfill & \frac{{r}^{\prime }\left(t\right)}{\text{‖}\phantom{\rule{0.1em}{0ex}}{r}^{\prime }\left(t\right)\text{‖}}=\frac{\text{−sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{i}+\text{cos}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{j}}{1}=\text{−sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{i}+\text{cos}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{j}\hfill \end{array}$

2. $\begin{array}{cccc}\text{First step:}\hfill & \hfill {u}^{\prime }\left(t\right)& =\hfill & \left(6t+2\right)\phantom{\rule{0.1em}{0ex}}\text{i}-12{t}^{2}\phantom{\rule{0.1em}{0ex}}\text{j}+6\phantom{\rule{0.1em}{0ex}}\text{k}\hfill \\ \text{Second step:}\hfill & \hfill \text{‖}\phantom{\rule{0.1em}{0ex}}{u}^{\prime }\left(t\right)\text{‖}& =\hfill & \sqrt{{\left(6t+2\right)}^{2}+{\left(-12{t}^{2}\right)}^{2}+{6}^{2}}\hfill \\ & & =& \sqrt{144{t}^{4}+36{t}^{2}+24t+40}\hfill \\ & & =\hfill & 2\sqrt{36{t}^{4}+9{t}^{2}+6t+10}\hfill \\ \text{Third step:}\hfill & \hfill \text{T}\left(t\right)& =\hfill & \frac{{u}^{\prime }\left(t\right)}{\text{‖}\phantom{\rule{0.1em}{0ex}}{u}^{\prime }\left(t\right)\text{‖}}=\frac{\left(6t+2\right)\phantom{\rule{0.1em}{0ex}}\text{i}-12{t}^{2}\phantom{\rule{0.1em}{0ex}}\text{j}+6\phantom{\rule{0.1em}{0ex}}\text{k}}{2\sqrt{36{t}^{4}+9{t}^{2}+6t+10}}\hfill \\ & & =\hfill & \frac{3t+1}{\sqrt{36{t}^{4}+9{t}^{2}+6t+10}}\phantom{\rule{0.1em}{0ex}}\text{i}-\frac{6{t}^{2}}{\sqrt{36{t}^{4}+9{t}^{2}+6t+10}}\phantom{\rule{0.1em}{0ex}}\text{j}+\frac{3}{\sqrt{36{t}^{4}+9{t}^{2}+6t+10}}\phantom{\rule{0.1em}{0ex}}\text{k}\hfill \end{array}$

where we get a research paper on Nano chemistry....?
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Ali
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Damian
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Professor
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Rafiq
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LITNING
scanning tunneling microscope
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Rafiq
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Mahi
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brayan
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Kyle
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Joe
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