Converting between polar and rectangular
Converting Between Polar and Rectangular
Converting Between Polar and Rectangular is a topic in Parametric & Polar in the California Common Core State Standards. It is aligned to Standard G-GPE, which requires students to convert between polar and rectangular coordinates.
Converting between polar and rectangular uses \(x=r\cos\theta\), \(y=r\sin\theta\), \(r^2=x^2+y^2\), and \(\tan\theta=\dfrac{y}{x}\).
Theory
The same point has both rectangular coordinates \((x,y)\) and polar coordinates \((r,\theta)\), linked by a right triangle:
Use the first pair to go polar \(\to\) rectangular, and the second to go rectangular \(\to\) polar (minding the quadrant for \(\theta\)).
The conversion relationships:
How to convert coordinates
- Polar \(\to\) rectangular: \(x=r\cos\theta,\ y=r\sin\theta\).
- Rectangular \(\to\) polar: \(r=\sqrt{x^2+y^2}\), \(\theta=\arctan\dfrac{y}{x}\) (fix the quadrant).
- Equations: multiply by \(r\) or substitute to swap \(r\cos\theta\leftrightarrow x\), \(r^2\leftrightarrow x^2+y^2\).
Use \(x=r\cos\theta,\ y=r\sin\theta\).
| \(x\) | \(=\) | \(4\cos 60^\circ=2\) |
| \(y\) | \(=\) | \(4\sin 60^\circ=2\sqrt3\) |
Find \(r\) and \(\theta\).
| \(r\) | \(=\) | \(\sqrt{1^2+1^2}=\sqrt2\) |
| \(\theta\) | \(=\) | \(\arctan\dfrac{1}{1}=45^\circ\) |
So \((\sqrt2,\ 45^\circ)\).
Multiply by \(r\) and substitute \(r^2=x^2+y^2\), \(r\cos\theta=x\).
| \(r^2\) | \(=\) | \(2r\cos\theta\) |
| \(x^2+y^2\) | \(=\) | \(2x\) |
A circle: \((x-1)^2+y^2=1\).
Substitute \(x=r\cos\theta\).
| \(r\cos\theta\) | \(=\) | \(3\) |
| \(r\) | \(=\) | \(\dfrac{3}{\cos\theta}\) |
Common pitfalls
Frequently asked questions
How do you convert polar to rectangular coordinates?
Use \(x=r\cos\theta\) and \(y=r\sin\theta\).
How do you convert rectangular to polar coordinates?
\(r=\sqrt{x^2+y^2}\) and \(\theta=\arctan\dfrac{y}{x}\), adjusting for the quadrant.
How do you convert a polar equation to rectangular form?
Multiply through by \(r\) or substitute so that \(r\cos\theta\to x\), \(r\sin\theta\to y\), and \(r^2\to x^2+y^2\).
Why check the quadrant when finding theta?
Because \(\arctan\) only returns angles in two quadrants; the actual point may require adding \(180^\circ\).