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Chapter 10: Problem 61

Replace the Cartesian equations in Exercises \(49-62\) by equivalent polar equations. $$ (x-3)^{2}+(y+1)^{2}=4 $$

Short Answer

Expert verified
The polar equation is \(r^2 - 6r\cos\theta + 2r\sin\theta = -6\).

Step by step solution

01

Convert Cartesian to Polar Coordinates

The given Cartesian equation is \((x-3)^2 + (y+1)^2 = 4\). In polar coordinates, we have \(x = r\cos\theta\) and \(y = r\sin\theta\). Substituting these into the equation results in:\[(r\cos\theta - 3)^2 + (r\sin\theta + 1)^2 = 4\]
02

Expand the Expression

Expand the equation from the previous step:\[(r\cos\theta - 3)^2 = r^2\cos^2\theta - 6r\cos\theta + 9\]\[(r\sin\theta + 1)^2 = r^2\sin^2\theta + 2r\sin\theta + 1\]
03

Combine and Simplify

Combine the expanded expressions:\[r^2\cos^2\theta - 6r\cos\theta + 9 + r^2\sin^2\theta + 2r\sin\theta + 1 = 4\]Using the identity \(r^2(\cos^2\theta + \sin^2\theta) = r^2\), we simplify the equation:\[r^2 - 6r\cos\theta + 2r\sin\theta + 10 = 4\]
04

Isolate Terms

Subtract 10 from both sides to isolate terms involving \(r\):\[r^2 - 6r\cos\theta + 2r\sin\theta = -6\]
05

Final Polar Equation

The resulting polar equation is:\[r^2 - 6r\cos\theta + 2r\sin\theta = -6\] This is an equivalent polar representation of the original Cartesian equation.

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Key Concepts

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

Cartesian Coordinates
Cartesian coordinates are a method of representing points in a plane using two numbers, essentially mapping points on a grid. This system uses two perpendicular lines called axes: the x-axis (horizontal) and the y-axis (vertical).
  • A point in the Cartesian system is identified by a pair \( (x, y) \), where \( x \) is the distance from the y-axis and \( y \) is the distance from the x-axis.
  • This coordinate system is very useful for plotting equations and understanding spatial relationships in a straightforward, intuitive way.
  • Geometric shapes such as lines, circles, and parabolas can be easily described using Cartesian coordinates and equations.
In the provided exercise, we start with a Cartesian equation of a circle centered at the point \( (3, -1) \) with a radius of \( 2 \) since \( 4 = 2^2 \). This will be converted into polar coordinates, offering an alternative representation.
Polar Coordinates
Polar coordinates offer another way to locate a point on a plane, emphasizing the direction and distance from a fixed point, the origin. Unlike the Cartesian coordinates, which depend on perpendicular axes, polar coordinates use a radial system.
  • A point in polar coordinates is represented by \( (r, \theta) \), where \( r \) is the radial distance from the origin and \( \theta \) is the angle measured from the positive x-axis.
  • This system is particularly useful for dealing with problems involving circular and rotational symmetry.
  • The relationship between Cartesian and polar coordinates is given by \( x = r\cos\theta \) and \( y = r\sin\theta \).
In the exercise solution, polar coordinates are employed to transform the given circle's Cartesian equation into a form that uses these radial and angular measurements, facilitating certain types of analysis and problem-solving in mathematics, especially calculus and trigonometry.
Equation Conversion
Equation conversion between coordinate systems can sometimes simplify the problem or provide a more insightful form of the equation. Converting from Cartesian to Polar coordinates involves substituting the expressions for \( x \) and \( y \) in terms of \( r \) and \( \theta \).
  • Begin by substituting \( x = r\cos\theta \) and \( y = r\sin\theta \) into the Cartesian equation.
  • Expand and simplify the resulting expression by applying trigonometric identities where possible.
  • The aim is to express the equation solely in terms of \( r \) and \( \theta \), which can then depict geometric properties differently than their Cartesian counterparts.
The solution provided demonstrates the conversion process, resulting in a polar equation: \( r^2 - 6r\cos\theta + 2r\sin\theta = -6 \). This offers a deeper understanding of the circle's nature in terms of its radius and angle from the origin, a perspective that is often simpler for studying periodic functions or phenomena exhibiting rotational symmetry.

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Most popular questions from this chapter

Exercises \(49-52\) give equations for ellipses and tell how many units up or down and to the right or left each ellipse is to be shifted. Find an equation for the new ellipse, and find the new foci, vertices, and center. $$ \frac{x^{2}}{3}+\frac{y^{2}}{2}=1, \quad \text { right } 2, \text { up } 3 $$

The hyperbola \(\left(x^{2} / 16\right)-\left(y^{2} / 9\right)=1\) is shifted 2 units to the right to generate the hyperbola $$\frac{(x-2)^{2}}{16}-\frac{y^{2}}{9}=1$$ a. Find the center, foci, vertices, and asymptotes of the new hyperbola. b. Plot the new center, foci, vertices, and asymptotes, and sketch in the hyperbola.

Which of the following has the same graph as \(r=\cos 2 \theta ?\) $$ \text { a. }r=-\sin (2 \theta+\pi / 2) \quad \text { b. } r=-\cos (\theta / 2) $$

Exercises \(45-48\) give equations for parabolas and tell how many units up or down and to the right or left each parabola is to be shifted. Find an equation for the new parabola, and find the new vertex, focus, and directrix. $$ x^{2}=8 y, \quad \text { right } 1, \text { down } 7 $$

The hyperbola \(\left(y^{2} / 4\right)-\left(x^{2} / 5\right)=1\) is shifted 2 units down to generate the hyperbola $$\frac{(y+2)^{2}}{4}-\frac{x^{2}}{5}=1$$ a. Find the center, foci, vertices, and asymptotes of the new hyperbola. b. Plot the new center, foci, vertices, and asymptotes, and sketch in the hyperbola.
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