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Line integral problems

  1. Feb 20, 2016 #1
    • Thread moved from the technical forums, so no HH Template is shown.
    I'm used to parameterizing however I'm not sure how to solve these types of problems, any help would be much appreciated.

    1) Calculate the line integral ∫v⋅dr along the curve y=x3 in the xy-plane when -1≤x≤2 and v=xyi+x2j

    2) a) Find the work that the force F = (y2+5)i+(2xy-8)j carries out along two paths ABC and ADC which are composed of perpendicular lines between the points A,B,C,D. A,B,C,D are corners in a square and the corners have the coordinates (0,0),(1,0),(1,1),(0,1) respectively

    b) Calculate the work done along a straight line from A to C.

    c) Since the work done appears to be independent of the path taken, we expect that the force can be written as the gradient to a potential V. Find the potential-function and show that the difference between the potential at A and C are equivalent to the work done.


    Attempt at reaching a solution:

    1) I substituted y=x3 into the equation then did a normal integral with respect to x. It doesn't seem right to me however I'm used to dS and t so these problems just messed my thinking up.

    2) Similar thinking to 1) is messing me up I think. Did normal integrals with respect to x and y.
     
  2. jcsd
  3. Feb 20, 2016 #2

    SteamKing

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    Here, r is a vector between the origin and a point (x,y) on the curve y = x3.

    Do you know how to set up the components of the vector r to describe y = x3 ?

    Once you find r, you can calculate dr and substitute into the line integral expression. After calculating v ⋅ dr, then you can start to treat the line integral like a regular integral.
     
  4. Feb 20, 2016 #3
    No I have no idea how to split it up, the dot product should be easy enough but I'm more used to x(t), y(t) being given.. How would you split it up?
     
  5. Feb 20, 2016 #4

    SteamKing

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    Do you know how to decompose a vector into its components?
     
  6. Feb 20, 2016 #5
    It's been a while... I don't quite remember
     
  7. Feb 20, 2016 #6

    SteamKing

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    You can look up how on the internet. That's what it's for.

    Remember, i, j , and k are the unit vectors for a general vector r, such that r = x i + y j + z k

    You have a curve described by y = x3. Care to take a guess as to how this curve could be turned into a vector expression?
     
  8. Feb 20, 2016 #7
    r=x^3i+yj ?
     
  9. Feb 20, 2016 #8

    SteamKing

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    What's the independent variable in y = x3 ?
     
  10. Feb 20, 2016 #9
    x would be the independent variable, so it would be x^3j?
     
  11. Feb 20, 2016 #10

    SteamKing

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    Correct. r = x i + x3 j + 0 k

    Can you calculate dr now?
     
  12. Feb 20, 2016 #11
    dr = (i + 3x2) dx

    The dot product would be v⋅dr = xy+3x4 dx

    The integral would then be: ## \int_{-1}^2 xy+3x^4 \, dx ## ?
     
  13. Feb 20, 2016 #12

    SteamKing

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    Don't forget the j component for the 3x2.
    What does y equal when traversing the curve y = x3 ?
     
  14. Feb 20, 2016 #13
    Of course, the integral becomes x^4+3x^4 dx. thank you!
     
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