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Chapter 10: Problem 66

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

Short Answer

Expert verified
The rectangular form of the given polar equation is \(x^2 - 2x + y^2 = 0\).

Step by step solution

01

Apply the Polar to Rectangular Conversion Formulas

Start by applying the formula for conversion from polar to rectangular coordinates. Specifically, replace r in the given equation with \(x^2 + y^2\). The polar equation is \(r = 2 \cos \theta\). Replacing r gives: \(x^2 + y^2 = 2 \cos \theta\).
02

Simplify \(\cos \theta\) in rectangular terms

Next, the goal is to express \(\cos \theta\) in terms of x and y. To this end, recall that \(\cos \theta = \frac{x}{r}\), and replace \(\cos \theta\) in the equation obtained in Step 1 by \(\frac{x}{r}\): \(x^2 + y^2 = 2 \frac{x}{\sqrt{x^2 + y^2}}\).
03

Conversion to rectangular form

Now, simplify the equation from Step 2 to arrive at the equation in rectangular form. First, cross-multiply to eliminate the denominator giving: \(x^2 + y^2 = 2x\). Then, subtract 2x from both sides to isolate the terms: \(x^2 - 2x + y^2 = 0\).

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Most popular questions from this chapter

Check for symmetry with respect to both axes and to the origin. Then determine whether the function is even, odd, or neither. $$y=e^{x}$$

Find a polar equation of the conic with its focus at the pole. $$\begin{array}{cc} \text{Conic} & \text{Vertex or Vertices} \\\ \text{Ellipse} &(2,0),(10, \pi)\end{array}$$

Find a polar equation of the conic with its focus at the pole. $$\begin{array}{cc} \text{Conic} & \text{Vertex or Vertices} \\\ \text{Hyperbola} &(2,0),(-8, \pi)\end{array}$$

Use a graphing utility to graph the rotated conic. $$r=\frac{8}{4+3 \sin (\theta+\pi / 6)}$$

Use the Law of sines or the Law of cosines to solve the triangle. $$A=56^{\circ}, C=38^{\circ}, c=12$$
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