anonymous
  • anonymous
Find a potential function for: F = y sin(z)i + x sin(z)j + xy cos(z)k
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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KingGeorge
  • KingGeorge
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?
anonymous
  • anonymous
The potential of F is any function f such that del f = F.
KingGeorge
  • KingGeorge
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?

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anonymous
  • anonymous
I got this: xycos(z), xycos(z), -xysin(z). Is this correct?
anonymous
  • anonymous
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.
anonymous
  • anonymous
There is a methodical way of doing this.
anonymous
  • anonymous
I did the integration part and got the above answer but how do I find the constant of integration?
anonymous
  • anonymous
\[ \frac{\partial}{\partial x}f(x,y,z)=y\sin(z) \implies f(x, y, z) = xy\sin(z)+g(y, z) \]
anonymous
  • anonymous
So then you integrate another function to solve for \(g(y, z)\).
KingGeorge
  • KingGeorge
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.
anonymous
  • anonymous
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.
KingGeorge
  • KingGeorge
From what I'm seeing, wio's way of doing this will get you to a solution, rather easily.
anonymous
  • anonymous
\[ \frac{\partial}{\partial y} xy\sin(z) +g(y, z) =x \sin(z) +g'(y, z) \]\[ \frac{\partial}{\partial y}f(x,y,z)=x\sin(z) \]This tells us what?
anonymous
  • anonymous
It seems that \(g'(y, z) = 0\). So we know that \(g(y, z)\) is a constant with respect to \(y\).
anonymous
  • anonymous
wonder why none of the local residents just give you the answer
KingGeorge
  • KingGeorge
^^because giving answers for free is against the Code of Conduct.
anonymous
  • anonymous
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.
anonymous
  • anonymous
what is "giving the answer for free?"
anonymous
  • anonymous
@blackjesus It's giving them a solution without having them put forth any effort on their own part to solve it.
KingGeorge
  • KingGeorge
^^well said.
anonymous
  • anonymous
Ideally you walk the through it. Sometimes people are so confused that they need a walk through of a problem to understand the method.
anonymous
  • anonymous
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.
anonymous
  • anonymous
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.
anonymous
  • anonymous
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 \(g(y, z)\).

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