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Asad0000 Group Title

Find a potential function for: F = y sin(z)i + x sin(z)j + xy cos(z)k

  • one year ago
  • one year ago

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  1. KingGeorge Group Title
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    By "potential function" you mean some function \(f(x,y,z)\) of three variables such that \[\frac{\partial}{\partial x}f(x,y,z)=y\sin(z)\]\[\frac{\partial}{\partial y}f(x,y,z)=x\sin(z)\]and\[\frac{\partial}{\partial z}f(x,y,z)=xy\cos(z)\]correct?

    • one year ago
  2. blackjesus Group Title
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    The potential of F is any function f such that del f = F.

    • one year ago
  3. KingGeorge Group Title
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    Good to know. Then, to solve this, notice that \[\frac{\partial}{\partial x}f(x,y,z)=y\sin(z)\]is a constant function of \(x\). Likewise, \[\frac{\partial}{\partial y}f(x,y,z)=x\sin(z)\]is a constant function of \(y\). What does that tell you about what the function \(f(x,y,z)\) has to look like?

    • one year ago
  4. Asad0000 Group Title
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    I got this: xycos(z), xycos(z), -xysin(z). Is this correct?

    • one year ago
  5. blackjesus Group Title
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    Don't quote me on this, but I believe George gave you part of the answer. Now if I remember correctly all you have to do integrate each part of his answer with respect to x, y, and z and then find the constant of integration. That is, intergrate the first part with respect to x, the second part with respect to y, and the last part with respect to z. Then find the constant of integration. I can't do it for you because I am not really sure how to go about it. I'ts been over 10 years since I graduate it from engineering school.

    • one year ago
  6. wio Group Title
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    There is a methodical way of doing this.

    • one year ago
  7. Asad0000 Group Title
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    I did the integration part and got the above answer but how do I find the constant of integration?

    • one year ago
  8. wio Group Title
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    \[ \frac{\partial}{\partial x}f(x,y,z)=y\sin(z) \implies f(x, y, z) = xy\sin(z)+g(y, z) \]

    • one year ago
  9. wio Group Title
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    So then you integrate another function to solve for \(g(y, z)\).

    • one year ago
  10. KingGeorge Group Title
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    Since you're asked for "a" potential function, you don't need to worry about a general constant C, since any constant will work, you can just choose 0 and be done with it.

    • one year ago
  11. blackjesus Group Title
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    If I remember correctly I think you can equate all our integrals to each other and then compare coefficients. May be you would be better off asking one of the resident geniuses.

    • one year ago
  12. KingGeorge Group Title
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    From what I'm seeing, wio's way of doing this will get you to a solution, rather easily.

    • one year ago
  13. wio Group Title
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    \[ \frac{\partial}{\partial y} xy\sin(z) +g(y, z) =x \sin(z) +g'(y, z) \]\[ \frac{\partial}{\partial y}f(x,y,z)=x\sin(z) \]This tells us what?

    • one year ago
  14. wio Group Title
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    It seems that \(g'(y, z) = 0\). So we know that \(g(y, z)\) is a constant with respect to \(y\).

    • one year ago
  15. blackjesus Group Title
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    wonder why none of the local residents just give you the answer

    • one year ago
  16. KingGeorge Group Title
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    ^^because giving answers for free is against the Code of Conduct.

    • one year ago
  17. wio Group Title
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    So we have: \[ f(x, y, z) =xy\sin(z) +h(z) \]\[ \frac{\partial}{\partial z} xy\sin(z) +h(z) = xy\cos(z) + h'(z) \]\[ \frac{\partial}{\partial z}f(x,y,z)=xy\cos(z) \]So \(h'(z)=0\). It's pretty obvious now that our potential function is just: \[ f(x, y, z) = xy\sin(z) +C \]And as @KingGeorge said, the \(C\) will work for any constant. In fact there just isn't anyway for us to know what \(C\) is.

    • one year ago
  18. blackjesus Group Title
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    what is "giving the answer for free?"

    • one year ago
  19. wio Group Title
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    @blackjesus It's giving them a solution without having them put forth any effort on their own part to solve it.

    • one year ago
  20. KingGeorge Group Title
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    ^^well said.

    • one year ago
  21. wio Group Title
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    Ideally you walk the through it. Sometimes people are so confused that they need a walk through of a problem to understand the method.

    • one year ago
  22. blackjesus Group Title
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    Oh! Without "giving" the answer away, if you take wio's answer and find the gradient of it. If you end up with the function F that you started with. You have the right answer.

    • one year ago
  23. blackjesus Group Title
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    Addendum: The only way to learn Math is to do lots of problems. However, answering a question with a question will not help the student learn anything. I have to go now. Good bye. Good luck.

    • one year ago
  24. wio Group Title
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    Yes, the methodology is as follows: 1) Integrate the function in terms of \(x\) (or some other variable). \( f(x, y, z) = \int F_xdx +g(y, z) \) 2) Take the derivative in terms of \(y\) (or some other variable). Then solve for \(g'(y, z)\). \(F_y = \frac{\partial }{\partial y}\int F_xdx +g'(y, z)\) \(g'(y, z) = F_y - \frac{\partial }{\partial y}\int F_xdx \) 3) Integrate in terms of \(y\). \(\int g'(y, z)dy = g(y, z) + h(z) \) 4) Solve for \(h(z)\) the same way we did for \(g(y, z)\).

    • one year ago
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