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Mathematics 17 Online
OpenStudy (mendicant_bias):

Finding the potential function for a given vector field, original prompt posted below:

OpenStudy (mendicant_bias):

http://i.imgur.com/m52GKAP.png Not sure how to do this, looking up how atm.

OpenStudy (mendicant_bias):

\[\int\limits_{}^{}F \cdot dr = f(B)-f(A)\]

OpenStudy (mendicant_bias):

@wio (lol I promise I'll stop badgering you soon, test is in two hours)

OpenStudy (mendicant_bias):

I'm guessing in order to do this, I have to know that the vector field is conservative, and I can tell that the curl is zero without computing it, so there;s that; Now I'm trying to find the potential function of a conservative vector field.

OpenStudy (mendicant_bias):

\[\triangledown f = F\]

OpenStudy (mendicant_bias):

I think I need to sort-of integrate some of the components of the given function?

OpenStudy (mendicant_bias):

\[x^2+\frac{3}{2}y^2+2z^2=f(x,y,z) \ ?\]

OpenStudy (mendicant_bias):

Oh yeah, plus C.

OpenStudy (mendicant_bias):

Alright, solved this one right, going to move on to a new type, something about solvinv exact differential forms

OpenStudy (anonymous):

Well, you can't always just integrate the components separately.

OpenStudy (anonymous):

In this case it works simply due to the fact that you don't have something like: \[ \mathbf F =g(x,y,z)\mathbf i+h(x,y,z)\mathbf j+m(x,y,z)\mathbf k \]

OpenStudy (anonymous):

It's very easy when: \[ \mathbf F = g(x)\mathbf i+h(y)\mathbf j+m(z)\mathbf k \]

OpenStudy (mendicant_bias):

Oh, is this the thing, uh, with, oh god, I really didn't like these.....lol,

OpenStudy (mendicant_bias):

The constants that get integrated to become mystery functions of other variables, something like that? I think I know what you're talking about.

OpenStudy (mendicant_bias):

Thank you for mentioning that; I'm going to close this one and work on the other one I've said up, which I'm guessing is implicitly asking that I evaluate a line integral in the form of its potential function, and what you said I think will pop up there.

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