Question

Vectors/line integral help please :)

Find the work done by the force field:
F=(x+y)i + (x-z)j +(z-y)k
in moving an object from (1,0,-1) to (0,-2,3) along a smooth path.

Have tried this over and over and cant seem to get the right answer.

Answer 1

how much does the object weigh?

Answer 2

Not sure, it wasnt given in the question. The context of the question is line and surface integrals where the answer is of form \[W=\int\limits_{}^{}f.dr\] where f and r are vectors.

Answer 3

use the following parameters
x = 1-t
y = -2*t
z = -1+4*t

they were gotten by
(1-t)*r0+t*r1

Gives you the parametrization of a line segment.

Answer 4

1/2 :)

Answer 5

\[dr = dxi +dyj +dzk\]
\[f=(x+y)i +(x-z)j +(z-y)k\]
\[f.dr=(x+y)dx +(x-z)dy +(z-y)dz\]

Answer 6

using the paremeters find the derivitives of each one. put them in x' i +y' j + z' k form

then plug in into your vector function like this

(1-t)*i+(-2*t)*j+(-1+4*t)*k

take dotproduct

and get
-7+15*t

and intigrate that from 0 to 1 wrt

Answer 7

I made a mistake plugging back in

Answer 8

@binary3i can u tell me ur final answer plzzz. I too did in this way. Wanna check the answer.

Answer 9

-9+37*t is when you plug back to F

Answer 10

19/2 is my new answer

Answer 11

I thnk it shud b x^2/2 and z^2/2. Isnt t?

Answer 12

It looks wrong to me

Answer 13

binary's

Answer 14

there has to be a derivitives somewhere.

Answer 15

yeah i made a mistake, their i fought to divided the X^2 and Z^2 bye 2

Answer 16

Nw what do u get @binary3i

Answer 17

19/2

I am 99.99999% sure of it :)

Answer 18

Can u explain ur method a bit in details if u dnt mind @timo86m.

Answer 19

plzzz

Answer 20

its 31/2

Answer 21

use the following parameters
x = 1-t
y = -2*t
z = -1+4*t

they were gotten by
(1-t)*r0+t*r1
r0 is intial r1 is end point

Gives you the parametrization of a line segment.
----------
now do this
r(t)= 1- t i + -2t j + -1+4t z
r'(t)=-i-2*j+4*k
F(x,y,z)= (x+y)i + (x-z)j +(z-y)k the original equation
Use your parameters to plug into z y and z

as in ((1-t) + ( -2*t)) i +.... and so on

take dot product of F(x,y,z) dot r'(t)

I got
-7+15*t


then just intigrat that from 0 to 1 with respect to x :)

Answer 22

x y and z not z y and z :P

Answer 23

I too am getting 19/2 in my method.:) Thanxxx a lot for the explanation.

Answer 24

OOH yay :D

Answer 25

similar problem from book

Answer 26

but then whats wrong with mine?

Answer 27

Ohh no am getting both of the answers.:(

Answer 28

several problems I can tell right away

because the results have to be a number

you taking integrals of with 2 variables will result in a variable left over.

as in take integral of x + y wrx you get y left over

Answer 29

\[\int\limits\limits_{}^{}f.dr= x^2/2 + 2xy -2 yz + z^2/2=I\]
now do the limits on I

Answer 30

show me your method later princess :)

I am going to bed. i'll look later

Answer 31

gotta go nw. will show it later.:)

Answer 32

OP did you find the answer?

Answer 33

nope. I dnt knw to work this out using line integrals. cos I havn't learnt t yet. I tooin binary's method bt was unabl to get the final answer.

Answer 34

It is not going to work. BInaries method is wrong formula. It doesn't even have a derivitive.

Answer 35

why do you want a derivative in between

Answer 36

u can use this formula ven the question is like this
Find the work done by the force F=(3x^2+6yz)i+(2y+3xz)j+(1+5xyz^2)k in moving a particle from (0 0 0) to (1 1 1)
(i)along the straight line x=y=z, which joins these two points.

Answer 37

Here is why

Answer 38

you cant just say dx

you have to say dx= x'(t) dt

Answer 39

only when i want them to be the function of t

Answer 40

You have to be in terms of t

also as I said before

Another problem is this

you have int(x+y) wrx from 1-0 would evaluate to -1/2-y

You can tell it is instantly wrong because it is supposse to be a number with no variables.

Answer 41

you see that I up there.
put substitute (0,-2,3) and subtract with I substituted with (1,0,-1).
u'll get it.

Answer 42

Wow, thank you everyone for the response! I have had internet problems since posting but am back online :) Thanks for the help. Timo the answer is 19/2 in the back of the book- so well done. I am still having trouble understanding how to do it as my lecturer didnt cover the material necessary for your approach but will look at the examples youve posted. Thanks again all.

Answer 43

See I was right :D

Answer 44

use the following parameters
x = 1-t
y = -2*t
z = -1+4*t

they were gotten by
(1-t)*r0+t*r1
r0 is intial r1 is end point

Gives you the parametrization of a line segment.
----------
now do this
r(t)= 1- t i + -2t j + -1+4t z
r'(t)=-i-2*j+4*k
F(x,y,z)= (x+y)i + (x-z)j +(z-y)k the original equation
Use your parameters to plug into z y and z

as in ((1-t) + ( -2*t)) i +.... and so on

take dot product of F(x,y,z) dot r'(t)

I got
-7+15*t


then just intigrat that from 0 to 1 with respect to x :)

Answer 45

Your parametrisation is beyond me lol. we havent learnt that- is there any other approach?

Answer 46

well ajppreincess reached the right answer 2 using a method I am not sure about.

Answer 47

Well we learned the parametric form of line in ch12.5 or 12.6


then we used that in ch 15.2

Answer 48

So (1-t)*r0+t*r1 is the general equation of how to parametrise ? im sorry im not understanding very well haha. Could you show me some steps if u have time?

Answer 49

where r0 is (x0,y0,z0) initial point and r1 is just (x1,y1,z1) end point
from (1,0,-1) to (0,-2,3)


to get x you have

x=(1-t)*(x0)+t*(x1)
...
(1-t) (1)+t*(0)

and you just simplify rearrange and such
x=1-t

Answer 50

do same for y and z

Answer 51

great! i got them now. I had just never seen that equation before! very handy. Is it only for straight lines?

Answer 52

Yup only for straigh line.

It is for a line segment :)

Answer 53

but the question didnt really state it was a line segment- how do we know it wasnt a weird curve? I am really struggling in this topic- can you recommend a really good online resource that explains line and surface integrals really clearly? Our lecturer rushed through it and the exam is soon!

Answer 54

there is something called path independance If the Force field is conservative :)

Answer 55

ok- if the curl F is zero?

Answer 56

there is a teste for it I think if you take the cross product of it and it =0

Answer 57

like walking up stairs.

if you wanna get to 5th floor

you waste same amnt of work aka energy

If you take ladder straight up

curling stairs

zig zaging stair ...

Answer 58

Also why did we integrate from 0 to 1?

Answer 59

that is due to the paremeter formula I gave you

t goes fro 0 to 1 :)

Answer 60

Always?

Answer 61

try it out

x = 1-t
y = -2*t
z = -1+4*t

plug in t as 0 than 1 and you should get
(1,0,-1) to (0,-2,3)

Answer 62

Yes always

Answer 63

brilliant! thanks

Answer 64

It is just part of the formula
(1-t)*r0+t*r1 where 0<t<1

Answer 65

great! you've been a great help thanks. can you recommend any good websites for line and surface integrals?

Answer 66

Oh sorry I don know :(

but I love khanacademy.org IDK if he has them. WHat he does have tho is great :)

Answer 67

Oh great thanks alot!

Answer 68

Ive heard that website is good- but had forgotten!

Answer 69

It is very good but I dont remember seeing no line integrals there . I think I did but i was surprised on how he usually explains it well but not that time. It was so long ago tho.

Answer 70

Cool. Ill dig around for something! Thanks again. Good luck with your studies.