MATH 251 Lecture 15

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Derivatives of Polar Coordinates

$\begin{array}{rl}\mathrm{d}r& =r_x\mathrm{d}x+r_y\mathrm{d}y\\ \mathrm{d}\theta & ={\theta }_x\mathrm{d}x+{\theta }_y\mathrm{d}y\end{array}$

$\left(\begin{array}{c}\mathrm{d}x\\ \mathrm{d}y\end{array}\right)=\left(\begin{array}{cc}\cos \theta & -r\sin \theta \\ \sin \theta & r\cos \theta \end{array}\right)\left(\begin{array}{c}\mathrm{d}r\\ \mathrm{d}\theta \end{array}\right)$

Inverse Matrix

$\begin{array}{rl}A& =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\\ A^{-1}& =\frac{1}{\mathrm{d}\mathrm{e}\mathrm{t}\left(A\right)}\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)\end{array}$

Solution

${\left(\begin{array}{cc}\cos \theta & -r\sin \theta \\ \sin \theta & r\cos \theta \end{array}\right)}^{-1}=\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\frac{\sin \theta }{r}& \frac{\cos \theta }{r}\end{array}\right)$