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Chapter 10: Problem 75

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

Short Answer

Expert verified
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 by step solution

01

Write down the given equation

The given polar equation is \(r = 2 \sin 3\theta\).
02

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

\(\sin \theta = \frac{y}{r}\). Substitute this into the given equation to obtain \(r = 2r \sin 3\theta\).
03

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 \(r = 6r\sin\theta - 8r\sin^3\theta\).
04

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 \(r^3 = 6yr - 8y^3\).
05

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: \((\sqrt{x^2+y^2})^3 = 6y\sqrt{x^2+y^2} - 8y^3\).

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Most popular questions from this chapter

Determine whether the statement is true or false. Justify your answer. The point which lies on the graph of a parabola closest to its focus is the vertex of the parabola.

The graph of \(r=f(\theta)\) is rotated about the pole through an angle \(\phi .\) Show that the equation of the rotated graph is \(r=f(\theta-\phi)\).

Find a polar equation of the conic with its focus at the pole. $$\begin{array}{cc} \text{Conic} & \text{Vertex or Vertices} \\\ \text{Ellipse} &\left(2, \frac{\pi}{2}\right),\left(4, \frac{3 \pi}{2}\right)\end{array}$$

Use a graphing utility to approximate any relative minimum or maximum values of the function. $$f(x)=x^{5}-3 x-1$$

Convert the polar equation to rectangular form. $$r=\frac{2}{1+\sin \theta}$$
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