x(u,v),y(u,v) to u(x,y),v(x,y)

**waytogo** · December 16th 2010, 07:27 AM

Can anybody help with this:
I have $x=u+sin v, y=v+cos u$ . How to extract u,v as functions from x,y, i.e. to get u(x,y) and v(x,y)?

**HallsofIvy** · December 16th 2010, 11:38 AM

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.

**waytogo** · December 16th 2010, 11:56 AM

yes, i believe in that, so i am looking for some particular step which could make this task easier.

**snowtea** · December 16th 2010, 12:34 PM

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.

**waytogo** · December 16th 2010, 12:42 PM

No, the task actually is:
$x=u+sin(v)$
$y=v+cos(u)$
$z(u,v)=uv^2$
Find $\frac{\partial z}{\partial x} + \frac {\partial z}{\partial y}$ at point $(\frac {\pi +3}{3} , \frac {\pi +1}{3}).$
So I must find $\frac{\partial z}{\partial u} \frac{\partial u}{\partial x}$ , $\frac{\partial z}{\partial u} \frac{\partial u}{\partial y}$ , $\frac{\partial z}{\partial v} \frac{\partial v}{\partial x}$ , $\frac{\partial z}{\partial v} \frac{\partial v}{\partial y}$ .

**snowtea** · December 16th 2010, 12:51 PM

Right, so you don't actually need the explicit formula.

For example, to find $\frac{\partial u}{\partial x}$ and $\frac{\partial v}{\partial x}$ .

Take the partials of the equations directly (independent variables are $x,y$ ):
$<br /> \frac{\partial x}{\partial x} = \frac{\partial u}{\partial x} + \cos(v)\frac{\partial v}{\partial x}<br />$

$<br /> \frac{\partial y}{\partial x} = \frac{\partial v}{\partial x} - \sin(u)\frac{\partial u}{\partial x}<br />$

Note that $\frac{\partial x}{\partial x} = 1$ and $\frac{\partial y}{\partial x}=0$ .

Solve this system of linear equations at the given point for $\frac{\partial v}{\partial x}, \frac{\partial u}{\partial x}$ .

Similar process for other derivatives.

Note you do need to solve for u and v at the point $(\frac {\pi +3}{3} , \frac {\pi +1}{3})$ , and you use these values when evaluating things like sin(u).

**Danny** · December 16th 2010, 02:34 PM

Originally Posted by waytogo

No, the task actually is:
$x=u+sin(v)$
$y=v+cos(u)$
$z(u,v)=uv^2$
Find $\frac{\partial z}{\partial x} + \frac {\partial z}{\partial y}$ at point $(\frac {\pi +3}{3} , \frac {\pi +1}{3}).$
So I must find $\frac{\partial z}{\partial u} \frac{\partial u}{\partial x}$ , $\frac{\partial z}{\partial u} \frac{\partial u}{\partial y}$ , $\frac{\partial z}{\partial v} \frac{\partial v}{\partial x}$ , $\frac{\partial z}{\partial v} \frac{\partial v}{\partial y}$ .

Do you know about Jacobians? I would suggest you check the point. It looks a little dubious.

**HallsofIvy** · December 17th 2010, 01:37 AM

Is $\left(\frac{\pi+ 3}{3}, \frac{\pi+ 1}{3}\right)$ (x, y) or (u, v)?

**waytogo** · December 17th 2010, 03:54 AM

the point was incorrect, the actual point is $(\frac{\pi+3}{3},\frac{\pi+1}{2})$ .
If I have caught your idea, I should use theorem about the inverse Jacobian?

**Danny** · December 17th 2010, 05:03 AM

The point now makes sense. Yes, inverse Jacobians is the way I would do this.

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