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anonymous

  • one year ago

Medal****Please Help Given the vectorfield F(x,y)= (x+2xsiny) i + (x^2cosy +2y) j Determine whether F is conservative. If it is, find a potential function

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  1. anonymous
    • one year ago
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    @ganeshie8

  2. ganeshie8
    • one year ago
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    Hint : curl must be 0 for the vector field to be conservative

  3. anonymous
    • one year ago
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    what formula would i apply here?

  4. ganeshie8
    • one year ago
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    \(F = Mi + Nj\) \(\text{curl(F)} = N_x - M_y\)

  5. ganeshie8
    • one year ago
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    F(x,y)= (x+2xsiny) i + (x^2cosy +2y) j M = x + 2xsiny N = x^2cosy + 2y N_x = ? M_y = ?

  6. anonymous
    • one year ago
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    i take the derivative?

  7. ganeshie8
    • one year ago
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    N_x means the partial of N with respect to x

  8. anonymous
    • one year ago
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    2xcosy?

  9. ganeshie8
    • one year ago
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    what is 2xcosy ?

  10. anonymous
    • one year ago
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    Nx

  11. ganeshie8
    • one year ago
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    OK, work the curl

  12. anonymous
    • one year ago
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    2xcosy-My....i dont know the My

  13. Loser66
    • one year ago
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    .

  14. ganeshie8
    • one year ago
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    You need to work M_y too

  15. anonymous
    • one year ago
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    -2xsiny+2 for My?

  16. ganeshie8
    • one year ago
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    M = x + 2xsiny M_y = ?

  17. anonymous
    • one year ago
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    1+2cosy

  18. ganeshie8
    • one year ago
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    try again

  19. anonymous
    • one year ago
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    2xcosy

  20. ganeshie8
    • one year ago
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    Yes, curl = ?

  21. anonymous
    • one year ago
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    2xcosy-2xcosy=0

  22. ganeshie8
    • one year ago
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    Since the curl is 0, the given vector field is conservative and a potential function exists

  23. ganeshie8
    • one year ago
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    Let \(f(x,y)\) be the potential function, then this must satisfy : \(f_x = M\) \(f_y = N\)

  24. anonymous
    • one year ago
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    what if it wasnt 0, it would be non-conservative but still a potential function?

  25. ganeshie8
    • one year ago
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    If the curl is not 0, then the vector field is not conservative and consequently there will not be a potential function. You cannot find a potential function.

  26. ganeshie8
    • one year ago
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    potential function for a vector field exists if and only if the curl of vector field is 0

  27. anonymous
    • one year ago
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    oki! from fx=m fy=n....whats the next step?

  28. ganeshie8
    • one year ago
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    plug in m and n

  29. ganeshie8
    • one year ago
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    \[f_x = x + 2x\sin y\] simply integrate both sides with respect to \(x\) to get \(f\)

  30. anonymous
    • one year ago
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    x^2/2 +2x^3/3siny?

  31. ganeshie8
    • one year ago
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    try again

  32. ganeshie8
    • one year ago
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    and what about the integration constant ?

  33. anonymous
    • one year ago
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    x.... x^2/2+x^2siny

  34. anonymous
    • one year ago
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    x^2/2+2x^2/2siny

  35. ganeshie8
    • one year ago
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    Easy, \[f_x = x + 2x\sin y\] integrating both sides with respect to \(x\) gives \[f = \frac{x^2}{2} + x\sin y + \color{red}{g(y)}\]

  36. ganeshie8
    • one year ago
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    that \(\color{red}{g(y)}\) is the arbitrary constant, it shows up everytime you integrate

  37. ganeshie8
    • one year ago
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    you need to find \(\color{red}{g(y)}\)

  38. anonymous
    • one year ago
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    xsiny why isnt it x^2siny

  39. ganeshie8
    • one year ago
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    Oops! my mistake... see if below looks fine \[f_x = x + 2x\sin y\] integrating both sides with respect to \(x\) gives \[f = \frac{x^2}{2} + x^2\sin y + \color{red}{g(y)}\]

  40. anonymous
    • one year ago
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    i take the third derviative to find g(Y)

  41. ganeshie8
    • one year ago
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    There is an easy way take the derivative of \(f\) with respect to \(y\) and compare it with the equation \(f_y = N\)

  42. ganeshie8
    • one year ago
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    \[f = \frac{x^2}{2} + x^2\sin y + \color{red}{g(y)}\] \[f_y = ?\]

  43. anonymous
    • one year ago
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    Y^2

  44. ganeshie8
    • one year ago
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    how

  45. anonymous
    • one year ago
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    nevermind...is it x^2cosy

  46. ganeshie8
    • one year ago
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    \[f = \frac{x^2}{2} + x^2\sin y + \color{red}{g(y)}\] \[f_y = x^2\cos y + \color{red}{g'(y)}\]

  47. ganeshie8
    • one year ago
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    compare this with the other equation \(f_y = N\)

  48. anonymous
    • one year ago
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    g'(y) is 2y?

  49. ganeshie8
    • one year ago
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    Yes, integrate and solve \(g(y)\)

  50. anonymous
    • one year ago
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    y^2

  51. ganeshie8
    • one year ago
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    Looks good! plug that in the potential function and you're done!

  52. ganeshie8
    • one year ago
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    \[f = \frac{x^2}{2} + x^2\sin y + \color{red}{g(y)} = \frac{x^2}{2} + x^2\sin y + \color{red}{y^2} \]

  53. anonymous
    • one year ago
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    thank you so much! x^2/2+x^2siny+Y^2

  54. ganeshie8
    • one year ago
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    just a sanity check : find the partials and see if you really get M and N

  55. ganeshie8
    • one year ago
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    you must get \(f_x = M\) and \(f_y = N\)

  56. anonymous
    • one year ago
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    yes i do

  57. anonymous
    • one year ago
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    thank you again ganeshie8!!

  58. IrishBoy123
    • one year ago
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    for completeness, there's a really handy way around the final step that avoids fiddling around with the integration constants. because the field is conservative, it is path independent so you can do a line integral. \(W = \int \vec F \bullet \vec dr = \int <x+2xsiny, x^2cosy +2y> \bullet <dx, dy>\) from (0,0) to (x,y) using, for example, these steps: |dw:1436724170339:dw| the line integral is straightforward

  59. ganeshie8
    • one year ago
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    I prefer the line integral too (ofcourse when eyeballing method fails :))

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