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

Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
Berger describes sociologists as concerned with
Got questions? Join the online conversation and get instant answers!