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

• Dec 16th 2010, 07:27 AM
waytogo
x(u,v),y(u,v) to u(x,y),v(x,y)
Can anybody help with this:
I have $\displaystyle x=u+sin v, y=v+cos u$. How to extract u,v as functions from x,y, i.e. to get u(x,y) and v(x,y)?
• Dec 16th 2010, 11:38 AM
HallsofIvy
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.
• Dec 16th 2010, 11:56 AM
waytogo
yes, i believe in that, so i am looking for some particular step which could make this task easier.
• Dec 16th 2010, 12:34 PM
snowtea
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.
• Dec 16th 2010, 12:42 PM
waytogo
No, the task actually is:
$\displaystyle x=u+sin(v)$
$\displaystyle y=v+cos(u)$
$\displaystyle z(u,v)=uv^2$
Find $\displaystyle \frac{\partial z}{\partial x} + \frac {\partial z}{\partial y}$ at point $\displaystyle (\frac {\pi +3}{3} , \frac {\pi +1}{3}).$
So I must find $\displaystyle \frac{\partial z}{\partial u} \frac{\partial u}{\partial x}$,$\displaystyle \frac{\partial z}{\partial u} \frac{\partial u}{\partial y}$,$\displaystyle \frac{\partial z}{\partial v} \frac{\partial v}{\partial x}$,$\displaystyle \frac{\partial z}{\partial v} \frac{\partial v}{\partial y}$.
• Dec 16th 2010, 12:51 PM
snowtea
Right, so you don't actually need the explicit formula.

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

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

$\displaystyle \frac{\partial y}{\partial x} = \frac{\partial v}{\partial x} - \sin(u)\frac{\partial u}{\partial x}$

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

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

Similar process for other derivatives.

Note you do need to solve for u and v at the point $\displaystyle (\frac {\pi +3}{3} , \frac {\pi +1}{3})$, and you use these values when evaluating things like sin(u).
• Dec 16th 2010, 02:34 PM
Jester
Quote:

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

Do you know about Jacobians? I would suggest you check the point. It looks a little dubious.
• Dec 17th 2010, 01:37 AM
HallsofIvy
Is $\displaystyle \left(\frac{\pi+ 3}{3}, \frac{\pi+ 1}{3}\right)$ (x, y) or (u, v)?
• Dec 17th 2010, 03:54 AM
waytogo
the point was incorrect, the actual point is $\displaystyle (\frac{\pi+3}{3},\frac{\pi+1}{2})$.
If I have caught your idea, I should use theorem about the inverse Jacobian?
• Dec 17th 2010, 05:03 AM
Jester
The point now makes sense. Yes, inverse Jacobians is the way I would do this.