Question

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

Answer 1

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?

Answer 2

The potential of F is any function f such that del f = F.

Answer 3

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?

Answer 4

I got this:
xycos(z), xycos(z), -xysin(z).
Is this correct?

Answer 5

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.

Answer 6

There is a methodical way of doing this.

Answer 7

I did the integration part and got the above answer but how do I find the constant of integration?

Answer 8

\[
\frac{\partial}{\partial x}f(x,y,z)=y\sin(z) \implies f(x, y, z) = xy\sin(z)+g(y, z)
\]

Answer 9

So then you integrate another function to solve for \(g(y, z)\).

Answer 10

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.

Answer 11

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.

Answer 12

From what I'm seeing, wio's way of doing this will get you to a solution, rather easily.

Answer 13

\[
\frac{\partial}{\partial y} xy\sin(z) +g(y, z) =x
\sin(z) +g'(y, z) \]\[
\frac{\partial}{\partial y}f(x,y,z)=x\sin(z)
\]This tells us what?

Answer 14

It seems that \(g'(y, z) = 0\). So we know that \(g(y, z)\) is a constant with respect to \(y\).

Answer 15

wonder why none of the local residents just give you the answer

Answer 16

^^because giving answers for free is against the Code of Conduct.

Answer 17

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.

Answer 18

what is "giving the answer for free?"

Answer 19

@blackjesus It's giving them a solution without having them put forth any effort on their own part to solve it.

Answer 20

^^well said.

Answer 21

Ideally you walk the through it. Sometimes people are so confused that they need a walk through of a problem to understand the method.

Answer 22

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.

Answer 23

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.

Answer 24

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 \(g(y, z)\).