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Chapter 8: Problem 26

Convert the rectangular equation to polar form and sketch its graph. $$ \left(x^{2}+y^{2}\right)^{2}-9\left(x^{2}-y^{2}\right)=0 $$

Short Answer

Expert verified
The polar representation of the given rectangular equation is \( r = 0 \) and \( r = \pm 3\sqrt{\cos(2\theta)} \). The graph represents a lemniscate with a maximum distance of 3 from the origin along the x-axis.

Step by step solution

01

Convert to Polar Coordinates

Substitute \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\) into the given equation. This gives us \( \left(r\cos(\theta)^{2}+r\sin(\theta)^{2}\right)^{2}-9\left(r\cos(\theta)^{2}-r\sin(\theta)^{2}\right)=0 \)
02

Simplify using Trigonometric Identities

Knowing that \( \cos^2(\theta) + \sin^2 (\theta) = 1 \) and \( \cos^2 (\theta) - \sin^2 (\theta) = \cos (2\theta) \), the equation simplifies to \( r^4 - 9r^2\cos(2\theta) = 0 \)
03

Factor Out Common Term

Factor out the common term \(r^2\), which simplifies the equation to \( r^2(r^2 - 9\cos(2\theta)) = 0 \). The solutions are \( r = 0 \), and \( r^2 = 9\cos(2\theta) \) or \( r = \pm 3\sqrt{\cos(2\theta)} \)
04

Sketch the Graph

We know that \( r = 0 \) represents the origin. The other equation represents a lemniscate. As \( \cos(2\theta) \) varies between -1 and 1, the graph will extend from -3 to 3 in both the x and y-directions. The graph has two loops, one on either side of the origin, each with a maximum distance of 3 from the origin along the x-axis

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Most popular questions from this chapter

Sketch the curve represented by the parametric equations (indicate the orientation of the curve), and write the corresponding rectangular equation by eliminating the parameter. $$ x=t^{2}+t, \quad y=t^{2}-t $$

Which integral yields the arc length of \(r=3(1-\cos 2 \theta)\) ? State why the other integrals are incorrect. (a) \(3 \int_{0}^{2 \pi} \sqrt{(1-\cos 2 \theta)^{2}+4 \sin ^{2} 2 \theta} d \theta\) (b) \(12 \int_{0}^{\pi / 4} \sqrt{(1-\cos 2 \theta)^{2}+4 \sin ^{2} 2 \theta} d \theta\) (c) \(3 \int_{0}^{\pi} \sqrt{(1-\cos 2 \theta)^{2}+4 \sin ^{2} 2 \theta} d \theta\) (d) \(6 \int_{0}^{\pi / 2} \sqrt{(1-\cos 2 \theta)^{2}+4 \sin ^{2} 2 \theta} d \theta\)

Give the integral formulas for area and arc length in polar coordinates.

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In Exercises \(27-38,\) find a polar equation for the conic with its focus at the pole. (For convenience, the equation for the directrix is given in rectangular form.) \(\frac{\text { Conic }}{\text { Parabola }} \quad \frac{\text { Eccentricity }}{e=1} \quad \frac{\text { Directrix }}{x=-1}\)
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