Question

How do you compute the line integral of a field along a parametrized curve?

Answer 1

i believe its\[F(r)\dot{}r' \]

Answer 2

it would help with more specifics to the problem :)

Answer 3

i can give you the problem verbatum if that would help

Answer 4

it would be better than me trying to guess at it, yes ;)

Answer 5

Compute the line integral of the field F(x, y) = ⟨2x sin y, x^2 cos y⟩ along
the curve with parametrization r(t) = ⟨e^(cos t+sin t), e^(cos t-sin t)⟩, 0 ≤ t ≤ 2.

Answer 6

the boxes are ( ) ok

Answer 7

so lets start by taking the derivative of r

Answer 8

with respect to t?

Answer 9

yes, since there really is no other variables in r .. lets try it with wrt t :)

Answer 10

\[\Large x(t)=e^{cos t+sin t}\\ \Large y(t)= e^{cos t-sin t}\]
\[\Large x'(t)=(-sint+cost)e^{cos t+sin t}\\ \Large y'(t)= (-sint-cost)e^{cos t-sin t}\]

agree?

Answer 11

(cos t - sin t)e^(cos t + sin t), (-cos t - sin t)e^(cos t - sin t)

Answer 12

thats what i got

Answer 13

good

now dot that to F and what do we get?

Answer 14

2sin y , -x^2sin y

Answer 15

oy this can get messy but with any luck it simplifies nicely

if you wanna swap out the xs and ys in F first with their r part equivalents that something we will have to do eventually

Answer 16

<a,b> dot <c,d> = ac + bd just in case youve forgotten how to dot :)

Answer 17

i missread what you typed i thought you said to DO it to F

Answer 18

x=rcos\[x = r \cos \theta, y = r \sin \]

Answer 19

yes?

Answer 20

no, lets do it by components so its simpler to work thru

given F(x,y) and the parametric r(x(t),y(t))

we want to determine the dot product of F(r) and r'; r' = (x'(t),y'(t))

so lets dot them and see if it cleans up any; recall that x = e^(cos t+sin t)
and y= e^(cos t-sin t)

taking the "x" component of F we have: 2x sin y

replacing x and y we get

2e^(cos t+sin t) sin e^(cos t-sin t)

and we determined the r' "x'" components to be:(cos t - sin t)e^(cos t + sin t)

so in effect, multiply those together

Answer 21

ok gimme a sec........maybe more

Answer 22

:) i can already tell you its a pain to do without the wolf :)

Answer 23

ill work out the y parts and then we can add the 2 together

Answer 24

F, y component is: x^2 cos y
x = e^(cos t+sin t)
y = e^(cos t-sin t)

replace xy parts in the F

e^(2cos t+2sin t) cos(e^(cos t-sin t))

multiply this with the y' from the r' ... the (-cos t - sin t)e^(cos t - sin t)

( e^(2cos t+2sin t) cos(e^(cos t-sin t)))((-cos t - sin t)e^(cos t - sin t))

and the wolf gives me ...

- e^(sint+3cost) cos(e^(cost-sint)) (sint+cost)

http://www.wolframalpha.com/input/?i=%28+e%5E%282cos+t%2B2sin+t%29+cos%28e%5E%28cos+t-sin+t%29%29%29%28%28-cos+t+-+sin+t%29e%5E%28cos+t+-+sin+t%29%29

Answer 25

for x i ended up with
(cos t - sin t) 2e^(cos t + sin t)(sin e^(cos t - sin t)

Answer 26

does wolfram agree?

Answer 27

(2e^(cos t+sin t) sin(e^(cos t-sin t)))((cos t - sin t)e^(cos t + sin t)); close

http://www.wolframalpha.com/input/?i=%282e%5E%28cos+t%2Bsin+t%29++sin%28e%5E%28cos+t-sin+t%29%29%29%28%28cos+t+-+sin+t%29e%5E%28cos+t+%2B+sin+t%29%29

your 2e part needs some "2s" in it :)

Answer 28

its on my paper just not on my screen....oops

Answer 29

(cos t - sin t) 2e^(2cos t + 2sin t)(sin e^(cos t - sin t)- e^(sint+3cost) cos(e^(cost-sint)) (sint+cost)

now lets see of the wolf can clean this up any ... crosses finger

the alternate forms dont look any better

http://www.wolframalpha.com/input/?i=2e%5E%282cos+t+%2B+2sin+t%29sin%28e%5E%28cos+t+-+sin+t%29%29%28cos+t+-+sin+t%29+-+e%5E%28sint%2B3cost%29+cos%28e%5E%28cost-sint%29%29+%28sint%2Bcost%29+

you think we can clean it up with a usubstitution? or do you feel confident enough to muddle thru it?

Answer 30

either or

Answer 31

next i need to integrate that monster correct?

Answer 32

yep

and just in case we might have gonna astray .. we havent. that dot is good

http://www.wolframalpha.com/input/?i=%5B2e%5E%28cos+t%2Bsin+t%29+sin%28e%5E%28cos+t-sin+t%29%29+%2C%28e%5E%28cos+t%2Bsin+t%29%29%5E2+cos%28e%5E%28cos+t-sin+t%29%29%5D+dot+%5B%28cos+t+-+sin+t%29e%5E%28cos+t+%2B+sin+t%29%2C+%28-cos+t+-+sin+t%29e%5E%28cos+t+-+sin+t%29%5D

Answer 33

and in case you need to dbl chk your answer :)

http://www.wolframalpha.com/input/?i=integrate+%282e%5E%282cos+t+%2B+2sin+t%29*sin%28e%5E%28cos+t+-+sin+t%29%29*%28cos+t+-+sin+t%29+-+e%5E%28sint%2B3cost%29+*cos%28e%5E%28cost-sint%29%29+*%28sint%2Bcost%29%29%2C+t%3D0..1

Answer 34

thank you for your help

Answer 35

youre welcome.

If you were expected to do this all by hand, the person who assigned it to you ought to be fired. The concepts are important to know .. not the mess

Answer 36

everything else that he assigned was much more simplistic. i wonder if there is a theorem to simplify this problem? if t goes from 0 to 2pi would that simplify this?

Answer 37

there might be some obscure thrm out there, but the ugliness is in the integral that results ... which of course a computer can do with ease

Answer 38

even splitting it up into its different sum parts doesnt help clean it up ... maybe a few substitutions would work out but oy!!

Answer 39

its just too cluttered for me to see anything doable it it :)

Answer 40

would it help to change it to polar coordinates?

Answer 41

dunno, a change doesnt necessarily equate to simpler

Answer 42

ok thank you for your time

Answer 43

good luck ;)