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Chapter 11: Problem 49

Replace the polar equations in Exercises \(27-52\) with equivalent Cartesian Replace the polar equations in Exercises \(27-52\) with equivalent Cartesian equations. Then describe or identify the graph. $$r=2 \cos \theta+2 \sin \theta$$

Short Answer

Expert verified
The Cartesian equation is \((x-1)^2 + (y-1)^2 = 2\); it is a circle centered at \((1,1)\) with radius \(\sqrt{2}\).

Step by step solution

01

Understand the Polar Equation

We are given the polar equation \(r = 2\cos\theta + 2\sin\theta\). We need to express this equation in Cartesian coordinates \(x\) and \(y\). Remember that \(r\) represents the radius from the origin to a point, while \(\theta\) is the angle from the positive \(x\)-axis.
02

Use Polar to Cartesian Conversion Formulas

Recall the conversion formulas: \(x = r\cos\theta\) and \(y = r\sin\theta\). Additionally, \(r^2 = x^2 + y^2\). These will help us transform the equation into Cartesian coordinates.
03

Express \(r\) in Terms of \(x\) and \(y\)

Since \(r\cos\theta = x\) and \(r\sin\theta = y\), rewrite the equation \(r = 2\cos\theta + 2\sin\theta\) as \(r = 2\left(\frac{x}{r}\right) + 2\left(\frac{y}{r}\right)\), which simplifies to \(r = \frac{2x}{r} + \frac{2y}{r}\). Multiplying through by \(r\) to clear the fractions, we get \(r^2 = 2x + 2y\).
04

Replace \(r^2\) with \(x^2 + y^2\)

Substitute \(r^2 = x^2 + y^2\) into the equation from Step 3: \[ x^2 + y^2 = 2x + 2y \] This is now a Cartesian equation.
05

Identify the Graph

Rewrite the equation as: \[ x^2 - 2x + y^2 - 2y = 0 \] Complete the square for both \(x\) and \(y\): \[ (x-1)^2 - 1 + (y-1)^2 - 1 = 0 \]Adding the 1's to both sides, it becomes: \[ (x-1)^2 + (y-1)^2 = 2 \]This is the equation of a circle with center \((1,1)\) and radius \(\sqrt{2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polar Coordinates
Polar coordinates are a method of expressing points in a two-dimensional plane through a radius and an angle. These coordinates are especially useful for problems involving rotation or circular motion. In polar coordinates, a point is represented as \((r, \theta)\):
  • \(r\) is the radius, or the distance from the origin to the point.
  • \(\theta\) is the angle measured from the positive x-axis.
The angle \(\theta\) is typically expressed in radians. Polar coordinates can offer a simpler approach sometimes, especially for equations involving circles or other curves that originate from a single point and radiate outward.
However, for other uses, like straight lines or more complex curves, cartesian coordinates might be simpler or preferable.
Cartesian Coordinates
Cartesian coordinates are a standard method for describing a position in a plane through two numerical coordinates that are distances from two intersected perpendicular axes, typically labeled as the x-axis and y-axis.
  • The x-coordinate measures the distance to the point along the horizontal x-axis.
  • The y-coordinate measures the distance to the point along the vertical y-axis.
Each point is represented by \((x, y)\). The Cartesian system is beneficial for solving algebraic equations and graphing linear relationships. In this exercise, we were tasked with converting a polar equation into a Cartesian form to facilitate understanding and to identify its geometrical representation on the Cartesian plane.
Circle Equation
The equation of a circle in the Cartesian coordinate system is derived from its definition: it is the set of all points that are equidistant from a fixed point, known as the center. The general form of a circle's equation is \((x - h)^2 + (y - k)^2 = r^2\), where:
  • \((h, k)\) is the center of the circle.
  • \(r\) is the radius.
In the example provided, from converting the polar equation into Cartesian form, we identified the circle's equation as \((x-1)^2 + (y-1)^2 = 2\). Here, the center is located at \((1, 1)\) and the radius is \(\sqrt{2}\). This form makes it easier to visualize the exact size and location of the circle in a plane.

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Most popular questions from this chapter

Replace the Cartesian equations in Exercises \(53-66\) with equivalent polar equations. $$x^{2}+(y-2)^{2}=4$$

Exercises \(25-28\) give the eccentricities and the vertices or foci of hyperbolas centered at the origin of the \(x y\) -plane. In each case, find the hyperbola's standard-form equation in Cartesian coordinates. $$\begin{array}{l}{\text { Eccentricity: } 3} \\ {\text { Foci: }( \pm 3,0)}\end{array}$$

Find the center, foci, vertices, asymptotes, and radius, as appropriate, of the conic sections in Exercises \(57-68 .\) $$ x^{2}-y^{2}+4 x-6 y=6 $$

Exercises \(35-38\) give information about the foci, vertices, and asymptotes of hyperbolas centered at the origin of the \(x y\) -plane. In each case, find the hyperbola's standard-form equation from the information given. $$ \begin{array}{l}{\text { Vertices: }( \pm 3,0)} \\ {\text { Asymptotes: } y=\pm \frac{4}{3} x}\end{array} $$

Graph the lines and conic sections in Exercises \(65-74.\) $$r=-2 \cos \theta$$
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