# simultaneous equation problem

• Jan 8th 2011, 07:23 PM
Hefotos
simultaneous equation problem
I have these two equations

$1 = 2e^{2r}cos\theta\frac{\partial r}{\partial x} -e^{2r}sin\theta\frac{\partial \theta}{\partial x}$

$0 = 3e^{3r}sin\theta\frac{\partial r}{\partial x} +e^{3r}cos\theta\frac{\partial \theta}{\partial x}$

and I have to solve them simultaneously to obtain $\frac{\partial r}{\partial x}$ and $\frac{\partial \theta}{\partial x}$. I hope this was the correct forum to post this question in.

Any hints would be helpful, thanks.
• Jan 8th 2011, 07:39 PM
Prove It
$\displaystyle \left[\begin{matrix}2e^{2r}\cos{\theta} & -e^{2r}\sin{\theta}\\ 3e^{3r}\sin{\theta} & \phantom{-}e^{3r}\cos{\theta}\end{matrix}\right]\left[\begin{matrix}\frac{\partial r}{\partial x}\\ \frac{\partial \theta}{\partial x}\end{matrix}\right] = \left[\begin{matrix}1\\ 0\end{matrix}\right]$

$\displaystyle \left[\begin{matrix}2e^{2r}\cos{\theta} & -e^{2r}\sin{\theta}\\ 3e^{3r}\sin{\theta} & \phantom{-}e^{3r}\cos{\theta}\end{matrix}\right]^{-1} \left[\begin{matrix}2e^{2r}\cos{\theta} & -e^{2r}\sin{\theta}\\ 3e^{3r}\sin{\theta} & \phantom{-}e^{3r}\cos{\theta}\end{matrix}\right]\left[\begin{matrix}\frac{\partial r}{\partial x}\\ \frac{\partial \theta}{\partial x}\end{matrix}\right] = \left[\begin{matrix}2e^{2r}\cos{\theta} & -e^{2r}\sin{\theta}\\ 3e^{3r}\sin{\theta} & \phantom{-}e^{3r}\cos{\theta}\end{matrix}\right]^{-1}\left[\begin{matrix}1\\ 0\end{matrix}\right]$

$\displaystyle \mathbf{I}\left[\begin{matrix}\frac{\partial r}{\partial x}\\ \frac{\partial \theta}{\partial x}\end{matrix}\right] = \frac{1}{2e^{2r}\cos{\theta}\cdot e^{3r}\cos{\theta} - (-e^{2r}\sin{\theta})\cdot 3e^{3r}\sin{\theta}}\left[\begin{matrix}\phantom{-3}e^{3r}\cos{\theta} & \phantom{2}e^{2r}\sin{\theta}\\-3e^{3r}\sin{\theta} & 2e^{2r}\cos{\theta}\end{matrix}\right]\left[\begin{matrix}1\\ 0\end{matrix}\right]$

$\displaystyle \left[\begin{matrix}\frac{\partial r}{\partial x}\\ \frac{\partial \theta}{\partial x}\end{matrix}\right] = \frac{1}{2e^{5r}\cos^2{\theta} + 3e^{5r}\sin^2{\theta}}\left[\begin{matrix}\phantom{-3}e^{3r}\cos{\theta}\\-3e^{3r}\sin{\theta}\end{matrix}\right]$

$\displaystyle \left[\begin{matrix}\frac{\partial r}{\partial x}\\ \frac{\partial \theta}{\partial x}\end{matrix}\right] = \frac{1}{2e^{5r} + e^{5r}\sin^2{\theta}}\left[\begin{matrix}\phantom{-3}e^{3r}\cos{\theta}\\-3e^{3r}\sin{\theta}\end{matrix}\right]$

So $\displaystyle \frac{\partial r}{\partial x} = \frac{e^{3r}\cos{\theta}}{2e^{5r}+ e^{5r}\sin^2{\theta}}$ and $\displaystyle \frac{\partial \theta}{\partial x} = -\frac{3e^{3r}\sin{\theta}}{2e^{5r} + e^{5r}\sin^2{\theta}}$.
• Jan 8th 2011, 08:11 PM
Hefotos
Thank you very much. This makes absolute sense.