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)?

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- Dec 16th 2010, 07:27 AM #1

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- Dec 16th 2010, 11:38 AM #2

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- Dec 16th 2010, 11:56 AM #3

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- Dec 16th 2010, 12:34 PM #4

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- Dec 16th 2010, 12:42 PM #5

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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 #6

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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 #7

- Dec 17th 2010, 01:37 AM #8

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- Dec 17th 2010, 03:54 AM #9

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- Dec 17th 2010, 05:03 AM #10