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how to convert polar coordinates to cartesian and vice versa?

Guest Sep 9, 2014

Best Answer 

 #2
avatar+18833 
+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!

heureka  Sep 9, 2014
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2+0 Answers

 #1
avatar+26406 
+5

cartesian to polar:

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

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

 

polar to cartesian:

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

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

Alan  Sep 9, 2014
 #2
avatar+18833 
+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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