15. Polar Coordinates

a. Polar Coordinate System

3. Polar Equations

A rectangular equation is an equation which involves the rectangular coordinates, \(x\) and \(y\). For example, \(y=x^2\) and \((x^2+y^2)^2=2xy\) are rectangular equations.

A polar equation is an equation which involves the polar coordinates, \(r\) and \(\theta\). For example, \(r=e^\theta\) and \(r^2=1+r\sin2\theta\) are polar equations.

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 \((x^2+y^2)^2=2xy\) to polar form.

Substitute \(x=r\cos\theta\) and \(y=r\sin\theta\) and simplify: \[\begin{aligned} (x^2+y^2)^2&=2xy \\ (r^2)^2&=2(r\cos\theta)(r\sin\theta) \\ r^4&=2r^2\sin\theta\cos\theta \\ r^2&=2\sin\theta\cos\theta \end{aligned}\] We can cancel the \(r^2\) because \((r,\theta)=(0,0)\) is still a solution. Finally, we optionally 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, \(x^2-y^2=1\).

\(r^2\cos2\theta=1\)
or \(r^2=\sec2\theta\)
or \(r=\pm\sqrt{\sec2\theta }\)

\[\begin{aligned} x^2-y^2&=1 \\ (r\cos\theta)^2-(r\sin\theta)^2&=1 \\ r^2(\cos^2\theta-\sin^2\theta)&=1 \\ r^2\cos(2\theta)&=1 \end{aligned}\] 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.

We first use the identity \(\cos(2\theta)=\cos^2\theta-\sin^2\theta\) to express everything in terms of \(\sin\theta\) and \(\cos\theta\): \[\begin{aligned} r^2&=\cos2\theta \\ r^2&=\cos^2\theta-\sin^2\theta \\ r^2&=\left(\dfrac{x}{r}\right)^2-\left(\dfrac{y}{r}\right)^2 \end{aligned}\] We now clear the denominator and substitute for \(r^2\): \[\begin{aligned} r^4&=x^2-y^2 \\ (x^2+y^2)^2&=x^2-y^2 \end{aligned}\]

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 \((1,0)\).

r=2costheta

\[\begin{aligned} r&=2\cos\theta \\ r&=2\dfrac{x}{r} \\ r^2&=2x \\ x^2+y^2&=2x \end{aligned}\]

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 \((1,0)\).

r=2costheta

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