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Question about line integral of F dot dr

  1. Dec 15, 2013 #1
    1. The problem statement, all variables and given/known data
    Evaluate the line integral of F dot dr where f(x,y)=<3x^2,2x+y> and C is a straight line segment from (1,2) to (5,4)


    2. Relevant equations
    Unfortunately I was out with family obligations when we covered line integrals and surface integrals so am stuck with the textbook for these so I will just jump to my 'attempt at a solution' since I have limited knowledge on these subjects.

    3. The attempt at a solution

    First I figured the equation of our line to be y=1/2x-1/2
    so if x=t then,
    y=1/2t-1/2
    so r(t)= ti+(1/2t+3/2)j
    and r'(t)= i+1/2j

    ∫<3t^2i+(5/2t+3/2j>dot<i+1/2j>dt
    ∫(3t^2+5/4t+6/2)dt

    Then we evaluate the integral from 1 to 5. How bad is it?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 15, 2013 #2

    LCKurtz

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    That should work; I didn't check your numbers. A nicer, and more standard, parameterizaton of the line would be ##\vec r(t) = \langle 1,2\rangle + t\langle 4,2\rangle## for ##0\le t\le 1##. I think you will find it simpler. Try it.
     
  4. Dec 15, 2013 #3
    Ah hah, a surprising result!

    If I have an integral of F dot T ds with F(x,y)=<4x,y^2> and C is the arc of a circle of radius 2 centered at the origin then do we just take,
    x=2cos t
    y=2sin t
    so we have the integral of <8cos t,4sin^2 t>dot<-2sin t,2cos t> with the bounds of 0 to pi/2?
     
  5. Dec 15, 2013 #4
    I thought I could figure this out, but...
    How did you get that parameterization?
     
  6. Dec 15, 2013 #5

    LCKurtz

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    Given points ##P_0,~P_1## the displacement vector from ##P_0## to## P_1## is ##D=P_1-P_0##. Then ##R= P_0 +tD## is at ##P_0## when ##t=0## and ##P_1## when ##t=1##.
     
  7. Dec 15, 2013 #6

    LCKurtz

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    Sure, if that's the part of the arc in which you are interested. You can use any parametrization and the idea is to use "nice" ones, like that one, for a circle. Note that going from one point to another over different curves may (but might not) give different results.
     
  8. Dec 15, 2013 #7
    Okay, so would r(t)=<5,4>+t<-4,-2> also be correct?
     
  9. Dec 15, 2013 #8
    Sorry, I forgot to mention 'in the first quadrant'...
    Nice to see the other part is correct.
     
  10. Dec 15, 2013 #9
    Another couple questions, how do we take the derivative of r(t) in this form? Is it just 4i+2j? and then what do we do with our f(x,y)=<3x^2,2x+y> to set it up for the dot product in the integral?
     
  11. Dec 15, 2013 #10

    LCKurtz

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    Yes and no. Yes, it's the same line segment, but no, it is in the other direction.

    Yes. The derivative of ##t## is ##1## so you just get the direction vector.

    Remember ##r(t) = \langle x,y\rangle##. Your parameterization gives ##x## and ##y## in terms of ##t##.
     
  12. Dec 15, 2013 #11
    Okay so we would have the integral of <48t^2,10t>dot<4,2> and the new bounds will be from 0 to 1 and then just calculus as usual. Is that right?
     
    Last edited: Dec 15, 2013
  13. Dec 15, 2013 #12

    LCKurtz

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    I haven't worked it out. But you could check by comparing the answer against your original calculation of the integral.
     
  14. Dec 15, 2013 #13
    Very good, although I will rejoice when the semester is finished and this material can be covered properly (hopefully my copy of Anton will resurface soon as well, Stewart is a pain).

    The only other question I have is if we have multiple 'segments' then we simply evaluate each one with their respective bounds and add the up, correct?
     
  15. Dec 15, 2013 #14

    LCKurtz

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  16. Dec 15, 2013 #15
    Thank you LCKurtz, that should do it.
     
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