AP Calculus BC glossary

Polar to Cartesian Conversion

Also called: Polar to rectangular, Converting polar coordinates

Polar to Cartesian conversion rewrites a polar point or curve in rectangular form using x equals r cos theta and y equals r sin theta. Going the other way, r squared equals x squared plus y squared. For a curve r equals f of theta, it turns theta into a parameter.

x = r\cos\theta, \quad y = r\sin\theta, \quad x^2 + y^2 = r^2

Substituting $r=f(\theta)$ into the two conversion equations produces $x=f(\theta)\cos\theta$ and $y=f(\theta)\sin\theta$ , a parametric curve with $\theta$ as the parameter, which is why slopes on a polar curve need no new machinery. Converting an equation rather than a point usually turns on multiplying through by $r$ , so that $r^2$ and $r\cos\theta$ can each be replaced at once.

r = 2\cos\theta \ \Longrightarrow\ r^2 = 2r\cos\theta \ \Longrightarrow\ x^2 + y^2 = 2x \ \Longrightarrow\ (x-1)^2 + y^2 = 1

That polar equation is a circle of radius $1$ centred at $(1,0)$ , far easier to recognise once it is rectangular. Coming back the other way, $r^2 = x^2 + y^2$ and $\tan\theta = y/x$ . On the vertical axis, where $x = 0$ , the tangent equation says nothing and $\theta$ is $\pi/2$ or $3\pi/2$ directly. Polar names are also not unique: the same point is described by $(r,\theta)$ and by $(-r,\theta+\pi)$ , which is why $r=0$ and negative $r$ have to be checked separately when curves are intersected.

The mistake

A calculator's $\arctan$ returns only angles between $-\pi/2$ and $\pi/2$ , so it cannot tell $(-3,-4)$ from $(3,4)$ . Both give $\tan\theta = 4/3$ and the same output, and reading that output as the angle drops a third quadrant point into the first, reflected through the origin. Check the signs of $x$ and $y$ , and add $\pi$ whenever the point lies left of the $y$ axis.

Appears in: Unit 9: Parametric, Polar, and Vector (BC)