# 3.2 Calculus of vector-valued functions  (Page 6/11)

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A particle moves on a circular path of radius b according to the function $\text{r}\left(t\right)=b\phantom{\rule{0.1em}{0ex}}\text{cos}\left(\omega t\right)\phantom{\rule{0.1em}{0ex}}\text{i}+b\phantom{\rule{0.1em}{0ex}}\text{sin}\left(\omega t\right)\phantom{\rule{0.1em}{0ex}}\text{j},$ where $\omega$ is the angular velocity, $d\theta \text{/}dt.$

Find the velocity function and show that $\text{v}\left(t\right)$ is always orthogonal to $\text{r}\left(t\right).$

$\text{r}\text{'}\left(t\right)=-b\omega \phantom{\rule{0.1em}{0ex}}\text{sin}\left(\omega t\right)\phantom{\rule{0.1em}{0ex}}\text{i}+b\omega \phantom{\rule{0.1em}{0ex}}\text{cos}\left(\omega t\right)\phantom{\rule{0.1em}{0ex}}\text{j}.$ To show orthogonality, note that $\text{r}\text{'}\left(t\right)·\phantom{\rule{0.1em}{0ex}}\text{r}\left(t\right)=0.$

Show that the speed of the particle is proportional to the angular velocity.

Evaluate $\frac{d}{dt}\left[\phantom{\rule{0.1em}{0ex}}\text{u}\left(t\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{u}^{\prime }\left(t\right)\right]$ given $\text{u}\left(t\right)={t}^{2}\phantom{\rule{0.1em}{0ex}}\text{i}-2t\phantom{\rule{0.1em}{0ex}}\text{j}+\phantom{\rule{0.1em}{0ex}}\text{k}.$

$0\phantom{\rule{0.1em}{0ex}}\text{i}+2\phantom{\rule{0.1em}{0ex}}\text{j}+4t\phantom{\rule{0.1em}{0ex}}\text{j}$

Find the antiderivative of $\text{r}\text{'}\left(t\right)=\text{cos}\left(2t\right)\phantom{\rule{0.1em}{0ex}}\text{i}-2\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{j}+\frac{1}{1+{t}^{2}}\phantom{\rule{0.1em}{0ex}}\text{k}$ that satisfies the initial condition $\text{r}\left(0\right)=3\phantom{\rule{0.1em}{0ex}}\text{i}-2\phantom{\rule{0.1em}{0ex}}\text{j}+\phantom{\rule{0.1em}{0ex}}\text{k}.$

Evaluate ${\int }_{0}^{3}\text{‖}t\phantom{\rule{0.1em}{0ex}}\text{i}+{t}^{2}\phantom{\rule{0.1em}{0ex}}\text{j}\text{‖}dt.$

$\frac{1}{3}\left({10}^{3\text{/}2}-1\right)$

An object starts from rest at point $P\left(1,2,0\right)$ and moves with an acceleration of $\text{a}\left(t\right)=\text{j}+2\phantom{\rule{0.1em}{0ex}}\text{k},$ where $\text{‖}\phantom{\rule{0.1em}{0ex}}\text{a}\left(t\right)\text{‖}$ is measured in feet per second per second. Find the location of the object after $t=2$ sec.

Show that if the speed of a particle traveling along a curve represented by a vector-valued function is constant, then the velocity function is always perpendicular to the acceleration function.

$\begin{array}{ccc}\hfill \text{‖}\phantom{\rule{0.1em}{0ex}}\text{v}\left(t\right)\text{‖}& =\hfill & k\hfill \\ \hfill \text{v}\left(t\right)·\phantom{\rule{0.1em}{0ex}}\text{v}\left(t\right)& =\hfill & k\hfill \\ \hfill \frac{d}{dt}\left(\phantom{\rule{0.1em}{0ex}}\text{v}\left(t\right)·\phantom{\rule{0.1em}{0ex}}\text{v}\left(t\right)\right)& =\hfill & \frac{d}{dt}k=0\hfill \\ \hfill \text{v}\left(t\right)·\phantom{\rule{0.1em}{0ex}}\text{v}\text{'}\left(t\right)+\phantom{\rule{0.1em}{0ex}}\text{v}\text{'}\left(t\right)·\phantom{\rule{0.1em}{0ex}}\text{v}\left(t\right)& =\hfill & 0\hfill \\ \hfill 2\phantom{\rule{0.1em}{0ex}}\text{v}\left(t\right)·\phantom{\rule{0.1em}{0ex}}\text{v}\text{'}\left(t\right)& =\hfill & 0\hfill \\ \hfill \text{v}\left(t\right)·\phantom{\rule{0.1em}{0ex}}\text{v}\text{'}\left(t\right)& =\hfill & 0.\hfill \end{array}$
The last statement implies that the velocity and acceleration are perpendicular or orthogonal.

Given $\text{r}\left(t\right)=t\phantom{\rule{0.1em}{0ex}}\text{i}+3t\phantom{\rule{0.1em}{0ex}}\text{j}+{t}^{2}\phantom{\rule{0.1em}{0ex}}\text{k}$ and $\text{u}\left(t\right)=4t\phantom{\rule{0.1em}{0ex}}\text{i}+{t}^{2}\phantom{\rule{0.1em}{0ex}}\text{j}+{t}^{3}\phantom{\rule{0.1em}{0ex}}\text{k},$ find $\frac{d}{dt}\left(\phantom{\rule{0.1em}{0ex}}\text{r}\left(t\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}\left(t\right)\right).$

Given $\text{r}\left(t\right)=⟨t+\text{cos}\phantom{\rule{0.1em}{0ex}}t,t-\text{sin}\phantom{\rule{0.1em}{0ex}}t⟩,$ find the velocity and the speed at any time.

$\text{v}\left(t\right)=⟨1-\text{sin}\phantom{\rule{0.1em}{0ex}}t,1-\text{cos}\phantom{\rule{0.1em}{0ex}}t⟩,$ $\text{speed}\phantom{\rule{0.2em}{0ex}}=-\phantom{\rule{0.1em}{0ex}}\text{v}\left(t\right)\text{‖}=\sqrt{4-2\left(\text{sin}\phantom{\rule{0.1em}{0ex}}t+\text{cos}\phantom{\rule{0.1em}{0ex}}t\right)}$

Find the velocity vector for the function $\text{r}\left(t\right)=⟨{e}^{t},{e}^{\text{−}t},0⟩.$

Find the equation of the tangent line to the curve $\text{r}\left(t\right)=⟨{e}^{t},{e}^{\text{−}t},0⟩$ at $t=0.$

$x-1=t,y-1=-t,z-0=0$

Describe and sketch the curve represented by the vector-valued function $\text{r}\left(t\right)=⟨6t,6t-{t}^{2}⟩.$

Locate the highest point on the curve $\text{r}\left(t\right)=⟨6t,6t-{t}^{2}⟩$ and give the value of the function at this point.

$\text{r}\left(t\right)=⟨18,9⟩$ at $t=3$

The position vector for a particle is $\text{r}\left(t\right)=t\phantom{\rule{0.1em}{0ex}}\text{i}+{t}^{2}\phantom{\rule{0.1em}{0ex}}\text{j}+{t}^{3}\phantom{\rule{0.1em}{0ex}}\text{k}.$ The graph is shown here:

Find the velocity vector at any time.

Find the speed of the particle at time $t=2$ sec.

$\sqrt{593}$

Find the acceleration at time $t=2$ sec.

A particle travels along the path of a helix with the equation $\text{r}\left(t\right)=\text{cos}\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{i}+\text{sin}\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{j}+t\phantom{\rule{0.1em}{0ex}}\text{k}.$ See the graph presented here:

Find the following:

Velocity of the particle at any time

$\text{v}\left(t\right)=⟨\text{−}\text{sin}\phantom{\rule{0.1em}{0ex}}t,\text{cos}\phantom{\rule{0.1em}{0ex}}t,1⟩$

Speed of the particle at any time

Acceleration of the particle at any time

$\text{a}\left(t\right)=-\text{cos}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{i}-\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{j}+0\phantom{\rule{0.1em}{0ex}}\text{j}$

Find the unit tangent vector for the helix.

A particle travels along the path of an ellipse with the equation $\text{r}\left(t\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{i}+2\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{j}+0\phantom{\rule{0.1em}{0ex}}\text{k}.$ Find the following:

Velocity of the particle

$\text{v}\left(t\right)=⟨\text{−}\text{sin}\phantom{\rule{0.1em}{0ex}}t,2\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}t,0⟩$

Speed of the particle at $t=\frac{\pi }{4}$

Acceleration of the particle at $t=\frac{\pi }{4}$

$\text{a}\left(t\right)=⟨-\frac{\sqrt{2}}{2},-\sqrt{2},0⟩$

Given the vector-valued function $\text{r}\left(t\right)=⟨\text{tan}\phantom{\rule{0.1em}{0ex}}t,\text{sec}\phantom{\rule{0.1em}{0ex}}t,0⟩$ (graph is shown here), find the following:

Velocity

Speed

$\text{‖}\phantom{\rule{0.1em}{0ex}}\text{v}\left(t\right)\text{‖}=\sqrt{{\text{sec}}^{4}t+{\text{sec}}^{2}t\phantom{\rule{0.1em}{0ex}}{\text{tan}}^{2}t}=\sqrt{{\text{sec}}^{2}t\left({\text{sec}}^{2}t+{\text{tan}}^{2}t\right)}$

Acceleration

Find the minimum speed of a particle traveling along the curve $\text{r}\left(t\right)=⟨t+\text{cos}\phantom{\rule{0.1em}{0ex}}t,t-\text{sin}\phantom{\rule{0.1em}{0ex}}t⟩$ $t\in \left[0,2\pi \right).$

2

Given $\text{r}\left(t\right)=t\phantom{\rule{0.1em}{0ex}}\text{i}+2\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{j}+2\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{k}$ and $\text{u}\left(t\right)=\frac{1}{t}\phantom{\rule{0.1em}{0ex}}\text{i}+2\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{j}+2\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{k},$ find the following:

$\text{r}\left(t\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}\left(t\right)$

$\frac{d}{dt}\left(\phantom{\rule{0.1em}{0ex}}\text{r}\left(t\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}\left(t\right)\right)$

$⟨0,2\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t\left(t-\frac{1}{t}\right)-2\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}t\left(1+\frac{1}{{t}^{2}}\right),2\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t\left(1+\frac{1}{{t}^{2}}\right)+2\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}t\left(t-\frac{2}{t}\right)⟩$

Now, use the product rule for the derivative of the cross product of two vectors and show this result is the same as the answer for the preceding problem.

Find the unit tangent vector T (t) for the following vector-valued functions.

$\text{r}\left(t\right)=⟨t,\frac{1}{t}⟩.$ The graph is shown here:

$\text{T}\left(t\right)=⟨\frac{{t}^{2}}{\sqrt{{t}^{4}+1}},\frac{-1}{\sqrt{{t}^{4}+1}}⟩$

$\text{r}\left(t\right)=⟨t\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}t,t\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t⟩$

$\text{r}\left(t\right)=⟨t+1,2t+1,2t+2⟩$

$\text{T}\left(t\right)=\frac{1}{3}⟨1,2,2⟩$

Evaluate the following integrals:

$\int \left({e}^{t}\phantom{\rule{0.1em}{0ex}}\text{i}+\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{j}+\frac{1}{2t-1}\phantom{\rule{0.1em}{0ex}}\text{k}\right)dt$

${\int }_{0}^{1}\phantom{\rule{0.1em}{0ex}}\text{r}\left(t\right)dt,$ where $\text{r}\left(t\right)=⟨\sqrt[3]{t},\frac{1}{t+1},{e}^{\text{−}t}⟩$

$\frac{3}{4}\phantom{\rule{0.1em}{0ex}}\text{i}+\text{ln}\left(2\right)\phantom{\rule{0.1em}{0ex}}\text{j}+\left(1-\frac{1}{e}\right)\phantom{\rule{0.1em}{0ex}}\text{j}$

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
Do somebody tell me a best nano engineering book for beginners?
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
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
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
how did you get the value of 2000N.What calculations are needed to arrive at it
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