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psk981

  • 3 years ago

Evaluate: Please look at the integral

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  1. psk981
    • 3 years ago
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    \[\int\limits_{C}^{} F.T ds \] for a vector field F=x ^{2}i-yj

  2. psk981
    • 3 years ago
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    from (2,4) to (1.1)

  3. psk981
    • 3 years ago
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    |dw:1352896610110:dw|

  4. experimentX
    • 3 years ago
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    along the straight line? what is T ?

  5. psk981
    • 3 years ago
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    you have to find flow

  6. experimentX
    • 3 years ago
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    |dw:1352897206497:dw|

  7. psk981
    • 3 years ago
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    how did u get r(t) i know you used the points but why did u pick t for the second set of points

  8. amistre64
    • 3 years ago
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    the t used is just a variable scalar to stretch the vector to all point along the line from the point used to anchor it to the line

  9. psk981
    • 3 years ago
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    so how would i find F. dr/dt where T=dr/dt?

  10. amistre64
    • 3 years ago
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    i dont recall the flow stuff to clearly, but does this look familiar? \[\frac{dr}{dt}=\frac{dr}{dx}\frac{dx}{dt}+\frac{dr}{dy}\frac{dy}{dt}\]

  11. psk981
    • 3 years ago
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    looks like chain rule

  12. amistre64
    • 3 years ago
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    this is a line integral right?

  13. amistre64
    • 3 years ago
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    http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsVectorFields.aspx ive always found this to be a rather good read. so im going over it

  14. psk981
    • 3 years ago
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    r(t)= <2-t, 4-3t>

  15. psk981
    • 3 years ago
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    F=(2-t)^2i-(4-3t)j

  16. amistre64
    • 3 years ago
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    i believe we also need r'

  17. psk981
    • 3 years ago
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    how would i get it i know its a derivative would it be r(t)= -i-3j

  18. psk981
    • 3 years ago
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    r'(t)=-i-3j

  19. amistre64
    • 3 years ago
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    then thats its; dot F and r' together to get a scalar equation to integrate right?

  20. psk981
    • 3 years ago
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    yes its F. dr/dt

  21. psk981
    • 3 years ago
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    i am not sure abt the dr/dt part

  22. amistre64
    • 3 years ago
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    r= <2-t, 4-3t> r'= <(2-t)', (4-3t)'> r'= <-1, -3>

  23. amistre64
    • 3 years ago
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    F=<(2-t)^2 ,-4+3t> dot r'=< -1 , -3 > ----------------------- -(2-t^2)+12-9t

  24. amistre64
    • 3 years ago
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    got me ^2 in the wrong side ... :/

  25. amistre64
    • 3 years ago
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    the line is from t=0 to t=1 giving us\[\int_{0}^{1}-4-t^2+4t-9t+12~dt\]\[\int_{0}^{1}-t^2-5t+8~dt\]

  26. amistre64
    • 3 years ago
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    do you have an answer key to check with by chance?

  27. psk981
    • 3 years ago
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    nope

  28. amistre64
    • 3 years ago
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    well, weve followed the simple directions from pauls site; so it should be good ;)

  29. psk981
    • 3 years ago
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    these vector fields so confusing

  30. amistre64
    • 3 years ago
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    indeed they are

  31. psk981
    • 3 years ago
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    i have another vector question

  32. amistre64
    • 3 years ago
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    the Force equation gives us the force at each for each x,y point along the line the lines vector equation gives us the x and y values along the path F(r) is just defining the forces along the path; dotting with r' tho has me a little baffled at an explanation at the moment tho ;)

  33. psk981
    • 3 years ago
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    how do i draw a vector field

  34. amistre64
    • 3 years ago
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    for each lattice point on a graph you draw a little arrow indicating the direction and magnitude of the vector associated with the values of x and y (or whatever reference frame your using) at that point

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