# 1. Polar Coordinates

## c. Polar Coordinates - 2D

## 3. Polar Equations

A rectangular equation is an equation which involves the rectangular coordinates, \(x\) and \(y\). For example, \(y=x^2\) and \(\left(x^2+y^2\right)^2=2xy\) are rectangular equations. A solution of a rectangular equation is a point \(P=(x,y)\) for which its coordinates \((x,y)\) satisfy the equation. For example, \((-2,4)\) is a solution of \(y=x^2\) and \((-2,-2)\) is a solution of \(\left(x^2+y^2\right)^2=2xy\).

A polar equation is an equation which
involves the polar coordinates, \(r\) and \(\theta\). For example,
\(r^2=1+r\sin 2\theta\) and \(r=-e^\theta\) are polar equations.
A solution of a polar equation is
a point \(P=(r,\theta)\) for which *some* pair of coordinates
\((r,\theta)\) satisfy the equation. For example, \(P=(1,\pi)\) is a
solution of \(r^2=1+r\sin 2\theta\). In addition, \(P=(1,\pi)\) is a
solution of \(r=-e^\theta\) because we can also write \(P\) as
\(P=(-1,0)\) and \((-1,0)\) satisfies \(r=-e^\theta\). This is
weird but true!

On this page, we discuss converting rectangular equations into polar equations and vice versa.

### Rectangular to Polar

To convert a rectangular equation into a polar equation, simply substitute \(x=r\cos\theta\) and \(y=r\sin\theta\) and simplify. As a shortcut, you may use the identity \(x^2+y^2=r^2\). You may also cancel factors of \(r\) provided the origin remains a solution.

Convert the rectangular equation \(\left(x^2+y^2\right)^2=2xy\) to polar form.

Substitute \(x=r\cos\theta\) and \(y=r\sin\theta\) and simplify: \[\begin{aligned} \left(x^2+y^2\right)^2&=2xy \\ \left(r^2\right)^2&=2(r\cos\theta)(r\sin\theta) \\ r^4&=2r^2\sin\theta\cos\theta \\ r^2&=2\sin\theta\cos\theta \end{aligned}\] We can cancel the \(r^2\) because \((r,\theta) =(0,0)\) is still a solution. Finally, we use the identity \(\sin(2\theta)=2\sin\theta\cos\theta\) to conclude: \[ r^2=\sin(2\theta) \]

Obtain a polar equation for the hyperbola, \(x^2-y^2=1\).

\(r^2\cos 2\theta =1\)

or \(r^2=\sec 2\theta\)

or \(r=\pm\sqrt{\sec 2\theta }\)

\[\begin{aligned} x^2-y^2&=1 \\ (r\cos\theta)^2-(r\sin\theta)^2&=1 \\ r^2\left(\cos^2\theta-\sin^2\theta\right)&=1 \\ r^2\cos(2\theta)&=1 \end{aligned}\] We have used the identity \(\cos(2\theta)=\cos^2\theta-\sin^2\theta\). We can also rewrite the equation as \(r^2=\sec(2\theta)\) or \(r=\pm\sqrt{\sec(2\theta)}\).

### Polar to Rectangular

To convert a polar equation into a rectangular equation, substitute \(\cos\theta=\dfrac{x}{r}\) and \(\sin\theta=\dfrac{y}{r}\) or if useful \(\tan\theta=\dfrac{y}{x}\). Then substitute \(r^2=x^2+y^2\).

Convert the polar equation \(r^2=\cos 2\theta\) to rectangular form.

First use the identity \(\cos(2\theta)=\cos^2\theta-\sin^2\theta\) to express everything in terms of \(\sin\theta\) and \(\cos\theta\): \[\begin{aligned} r^2&=\cos 2\theta \\ r^2&=\cos^2\theta-\sin^2\theta \\ r^2&=\left(\dfrac{x}{r}\right)^2-\left(\dfrac{y}{r}\right)^2 \end{aligned}\] Now clear the denominator: \[\begin{aligned} r^4&=x^2-y^2 \\ \left(x^2+y^2\right)^2&=x^2-y^2 \end{aligned}\]

Obtain a Cartesian equation for the polar equation, \(r=2\cos\theta\). Identify the graph by looking at the rectangular equation.

To identfy the graph, first complete the square.

\(x^2+y^2=2x\)
or \((x-1)^2+y^2=1\)

Circle of radius \(1\) centered at \((1,0)\).

\[\begin{aligned} r&=2\cos\theta \\ r&=2\dfrac{x}{r} \\ r^2&=2x \\ x^2+y^2&=2x \end{aligned}\] To identify the graph, we first complete the square by adding \(-2x+1\) to both sides: \[\begin{aligned} x^2-2x+1+y^2&=1 \\ (x-1)^2+y^2&=1 \end{aligned}\]

This is a circle of radius \(1\) centered at \((1,0)\).

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