Question

When a particle is located at a distance x meters from the origin, a force of cos(pi x/3) newtons acts on it. How much work is done in moving the particle from x=1 to x=2?

Answer 1

use u-substitution

Answer 2

i thought that it was just W=FxD so it would be \[\int\limits_{1}^{2} x \cos (\pi x/3)\]

Answer 3

Alright, so do u understand that work is force X distance, and that u need to add to the integrals from x=1 to x=1.5 and from x=1.5 to x=2?

Answer 4

I'm confusing u rite?

Answer 5

no, that makes sense because now that i graph the function i see that half of it is in the positive region and half is in the negative, could i also just do the absolute value of the function from 1 to 2?

Answer 6

so in this case instead of the .0849 i come out with .3838 instead.

Answer 7

well actually that still does not give me the correct answer, possibly am i setting it up wrong?

Answer 8

Is there any multiple choice?

Answer 9

there is not, this is for my calc II class, and we are just talking about physics and engineering applications. so she doesn't need us to evaluate the integrals by hand, more just understand how to set up the integral and learn how this is used to help in real world applications.

Answer 10

now one thing that i look at and notice on some of the problems we did in class, i notice that the distance it most of the time absorbed by dx because it is a changing distance and then your endpoints of where it is being measured it taken up by the endpoints of the integral, so maybe its just the \[\int\limits_{1}^{2} \cos(\pi x/3) dx\]

Answer 11

Use u-substitution. I'll work the indefinite integral.

∫ {cos (πx/3)} dx

u = πx/3
du = π/3 dx
dx = 3/π dx

∫ {cos (πx/3)} dx
= (3/π) ∫ {cos u } du
= (3/π) {sin u} + C
= (3/π) sin (πx/3) + C

Answer 12

then evaluating (3/π) sin (πx/3) from 1 to 2 would give me .25587 when using absolute value

Answer 13

Did tht make sense? ^^

Answer 14

that is even if i added the definite integrals from 1 to 1.5 + 1.5 to 2

Answer 15

absolutely it made sense, but the only issue that I'm having is this is not giving me the correct answer. therefore; i think i am setting this up wrong

Answer 16

Ok start all over (scrape paper and all) and do it again and let me see wat it come's out to be.

Answer 17

alright so we are starting out with the original problem:

When a particle is located at a distance x meters from the origin, a force of cos(pi x/3) newtons acts on it. How much work is done in moving the particle from x=1 to x=2?

Answer 18

First let's find the integral of cos( piX/3). Let INT[] be the integral function. Let W = work
so we want
W = INT[cos(piX/3)dx]

Answer 19

U got this so far?

Answer 20

yes

Answer 21

k then...

Answer 22

let y = piX/3
dy/dx=pi/3
dx=(3/pi) dy
plug into your integral

INT[cos(piX/3)dx]=INT[cos(y)(3/pi)dy]
=(3/pi)sin(y)
plug back
=(3/pi)sin(piX/3)

Answer 23

right o

Answer 24

Give me an honest option rite now. Do u get it so far?

Answer 25

absolutely, this is something that i have worked on doing over and over for the past 6 months
i completely understand what you did when evaluating the integral of cos(pix/3)

Answer 26

K good!

Answer 27

now that we have the integral of the function i can plug in points to x from the integrated function

Answer 28

therefore the actual answer is 0

Answer 29

\[\int\limits_{1}^{1.5}\cos(\pi x/3)dx+\int\limits_{1.5}^{2}\cos(\pi x/3)dx =0\]

Answer 30

or this looks like this on my paper

\[\frac{ 3\pi }{ 3 } \sin(\frac{ \pi x }{ 3 })\left(\begin{matrix}2 \\ 1\end{matrix}\right)\]

Answer 31

i dont know how to show the line showing the endpoints but that is close enough

Answer 32

U did it! Does it feel good to no how to do it?? U said u were doing this for 6 months? Thts a long time!!!

Answer 33

wellllll not this problem, but evaluating integrals in calc I and calc II that is why i understood what you were doing when working on the problem. the physics applications such as this problem are brand new to me
Either way, thank you very much @e.cociuba

Answer 34

I didn't help u tht much but watev :)

Answer 35

Still... Thank you. i may have more questions on these physics and engineering application problems, but this will help me understand the other questions, hopefully.

Answer 36

Haha ok. :) ask away when ever I'm on :) Just name it and I will do my best to answer. np:)