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# How to convert polar to cartesian and vice versa?

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how to convert polar coordinates to cartesian and vice versa?

Sep 9, 2014

### Best Answer

#2
+20850
+5

How to convert polar to cartesian and vice versa ?

I. polar to cartesian : $$\boxed{\begin{array}{rcl} x=r*\cos{(\alpha)} \\ y=r*\sin{(\alpha)} \end{array} }$$

II. cartesian  to polar : $$r: \begin{array}{rcl} x^2+y^2&=&r^2\cos^2{(\alpha)}+r^2\sin^2{(\alpha)} \\ x^2+y^2&=& r^2( \underbrace{ \sin^2{(\alpha)}+\cos^2{(\alpha)} }_{=1} ) \\ x^2+y^2&=&r^2 \\ \end{array} \boxed{r=\sqrt{x^2+y^2}}$$

$$\alpha: \begin{array}{rcl} \frac{y}{x}&=& \frac { r\sin{(\alpha)} } { r\cos{(\alpha)} }\\ \frac{y}{x}&=& \tan{(\alpha)} \end{array} \boxed{ \alpha=\tan^{-1}\left( \frac{y}{x} \right) }$$

$$\begin{array}{rcll} y>0 &and& x>0 & \quad \alpha \\ y>0 &and& x<0 & \quad \alpha +180\ensurement{^{\circ}}\\ y<0 &and& x<0 & \quad \alpha +180\ensurement{^{\circ}}\\ y<0 &and& x>0 & \quad \alpha +360\ensurement{^{\circ}}\\ \hline y=0 &and& x>0 & \quad \alpha = 0\ensurement{^{\circ}}\\ y>0 &and& x=0 & \quad \alpha = 90\ensurement{^{\circ}}\\ y=0 &and& x<0 & \quad \alpha = 180\ensurement{^{\circ}}\\ y<0 &and& x=0 & \quad \alpha = 270\ensurement{^{\circ}} \end{array}$$

y = 0   and  x = 0   $$\alpha$$  undefined!

Sep 9, 2014

### 2+0 Answers

#1
+27377
+5

cartesian to polar:

$$r=\sqrt{x^2+y^2}$$

$$\theta=\tan^{-1}(\frac{y}{x})$$

polar to cartesian:

$$x=r\cos{\theta}$$

$$y=r\sin{\theta}$$

.
Sep 9, 2014
#2
+20850
+5
Best Answer

How to convert polar to cartesian and vice versa ?

I. polar to cartesian : $$\boxed{\begin{array}{rcl} x=r*\cos{(\alpha)} \\ y=r*\sin{(\alpha)} \end{array} }$$

II. cartesian  to polar : $$r: \begin{array}{rcl} x^2+y^2&=&r^2\cos^2{(\alpha)}+r^2\sin^2{(\alpha)} \\ x^2+y^2&=& r^2( \underbrace{ \sin^2{(\alpha)}+\cos^2{(\alpha)} }_{=1} ) \\ x^2+y^2&=&r^2 \\ \end{array} \boxed{r=\sqrt{x^2+y^2}}$$

$$\alpha: \begin{array}{rcl} \frac{y}{x}&=& \frac { r\sin{(\alpha)} } { r\cos{(\alpha)} }\\ \frac{y}{x}&=& \tan{(\alpha)} \end{array} \boxed{ \alpha=\tan^{-1}\left( \frac{y}{x} \right) }$$

$$\begin{array}{rcll} y>0 &and& x>0 & \quad \alpha \\ y>0 &and& x<0 & \quad \alpha +180\ensurement{^{\circ}}\\ y<0 &and& x<0 & \quad \alpha +180\ensurement{^{\circ}}\\ y<0 &and& x>0 & \quad \alpha +360\ensurement{^{\circ}}\\ \hline y=0 &and& x>0 & \quad \alpha = 0\ensurement{^{\circ}}\\ y>0 &and& x=0 & \quad \alpha = 90\ensurement{^{\circ}}\\ y=0 &and& x<0 & \quad \alpha = 180\ensurement{^{\circ}}\\ y<0 &and& x=0 & \quad \alpha = 270\ensurement{^{\circ}} \end{array}$$

y = 0   and  x = 0   $$\alpha$$  undefined!

heureka Sep 9, 2014

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