Can anybody help with this:

I have . How to extract u,v as functions from x,y, i.e. to get u(x,y) and v(x,y)?

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- December 16th 2010, 07:27 AMwaytogox(u,v),y(u,v) to u(x,y),v(x,y)
Can anybody help with this:

I have . How to extract u,v as functions from x,y, i.e. to get u(x,y) and v(x,y)? - December 16th 2010, 11:38 AMHallsofIvy
Essentially, you want to solve the two

**non-linear**equations x= u+ sin v, y= v+ cos u for u and v in terms of x and y. I doubt that there will be any elementary solution for that. - December 16th 2010, 11:56 AMwaytogo
yes, i believe in that, so i am looking for some particular step which could make this task easier.

- December 16th 2010, 12:34 PMsnowtea
Does a problem specifically ask for u(x,y) and v(x,y)?

Because if it is only asking for derivatives at certain points, you don't actually need the explicit formulas. - December 16th 2010, 12:42 PMwaytogo
No, the task actually is:

Find at point

So I must find , , , . - December 16th 2010, 12:51 PMsnowtea
Right, so you don't actually need the explicit formula.

For example, to find and .

Take the partials of the equations directly (independent variables are ):

Note that and .

Solve this system of linear equations at the given point for .

Similar process for other derivatives.

Note you do need to solve for u and v at the point , and you use these values when evaluating things like sin(u). - December 16th 2010, 02:34 PMJester
- December 17th 2010, 01:37 AMHallsofIvy
Is (x, y) or (u, v)?

- December 17th 2010, 03:54 AMwaytogo
the point was incorrect, the actual point is .

If I have caught your idea, I should use theorem about the inverse Jacobian? - December 17th 2010, 05:03 AMJester
The point now makes sense. Yes, inverse Jacobians is the way I would do this.