Problem 75 - Algebra and Trigonometry Real Mathematics, Real People

Question

Convert the polar equation to rectangular form. $$r=2 \sin 3 \theta$$

Step-by-Step Solution

Answer

The rectangular form of the given polar equation \(r = 2 \sin 3\theta\) is \((\sqrt{x^2+y^2})^3 = 6y\sqrt{x^2+y^2} - 8y^3\).

Step 1: Write down the given equation

The given polar equation is \(r = 2 \sin 3\theta\).

Step 2: Express \(\sin \theta\) in terms of \(y\) and \(r\)

\(\sin \theta = \frac{y}{r}\). Substitute this into the given equation to obtain \(r = 2r \sin 3\theta\).

Step 3: Convert \(3\theta\) to rectangular coordinates

Using the double-angle formula, \(\sin 3\theta = 3\sin\theta - 4\sin^3\theta\), we obtain \(r = 2r(3\sin\theta - 4\sin^3\theta)\). This simplifies to \(r = 6r\sin\theta - 8r\sin^3\theta\).

Step 4: Substitute \(\sin \theta\) with \( \frac{y}{r}\)

Now substitute \(\sin\theta = \frac{y}{r}\) into the equation from step 3, to replace all instances of \(r\) and \(\theta\), and making sure that the equation is in terms of \(x\) and \(y\). This leads care to the equation \(r = 6y - 8y^3/r^2\), or in order to get the equation fully in terms of \(x\) and \(y\), rewrite the equation to \(r^3 = 6yr - 8y^3\).

Step 5: Substitute \(r\) with \(\sqrt{x^2+y^2}\)

Substitute \(r^3\) with \((\sqrt{x^2+y^2})^3\) and \(r\) with \(\sqrt{x^2+y^2}\), as substitution for the polar to rectangle conversion, leading us to the final presentation of the rectangle equation: \((\sqrt{x^2+y^2})^3 = 6y\sqrt{x^2+y^2} - 8y^3\).