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Chapter 6: Problem 60

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

Short Answer

Expert verified
The rectangular form of the given polar equation is \(x^2 + y^2 = 3 \sqrt{x^2 + y^2} - 3y^2\).

Step by step solution

01

Expressing r in terms of x and y

Start by using the substitution identity for \(r\) in rectangular coordinates, \(r = \sqrt{x^2 + y^2}\). This gives the equation: \(\sqrt{x^2 + y^2} = 3 \cos 2 \theta\).
02

Substitution of cos 2 theta

Using the double angle identity for cosine, one can replace \(\cos 2 \theta\) by \(1 - 2 \sin^2 \theta\). This allows prevention of having to use the value of \(r\) twice in substitution for cos(\( \theta\)) and sin(\(\theta\)). The equation becomes: \(\sqrt{x^2 + y^2} = 3(1 - 2 \sin^2 \theta)\).
03

Expressing sin theta in terms of y and r

The sine function can be expressed in rectangular coordinates as \(\sin \theta = y / r\), by the identity substitution. Substituting this into equation from step 2 gives: \(\sqrt{x^2 + y^2} = 3(1 - 2 (y/r)^2)\).
04

Substituting r back into the equation

Replace \(r\) by \(\sqrt{x^2 + y^2}\) and simplify to get the equation in rectangular form: \(\sqrt{x^2 + y^2} = 3(1 - 2 (y/\sqrt{x^2 + y^2})^2}\).
05

Simplification

Simplify the equation obtained in step 4 which results in the rectangular form of the equation: \(x^2 + y^2 = 3 \sqrt{x^2 + y^2} - 3y^2\).

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Most popular questions from this chapter

\(r=\frac{3}{2+\cos \theta}\)

Sketch the graph of each equation. (a) \(r=1-\sin \theta\) (b) \(r=1-\sin \left(\theta-\frac{\pi}{4}\right)\)

\(r^{2}=4 \sin \theta\)

Exploration Sketch the graph of \(r=6 \cos \theta\) over each interval. Describe the part of the graph obtained in each case. (a) \(0 \leq \theta \leq \frac{\pi}{2}\) (b) \(\frac{\pi}{2} \leq \theta \leq \pi\) (c) \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\) (d) \(\frac{\pi}{4} \leq \theta \leq \frac{3 \pi}{4}\)

\(r=3 \sin 2 \theta\)
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