Problem 61
Question
Replace the Cartesian equations in Exercises \(49-62\) by equivalent polar equations. $$ (x-3)^{2}+(y+1)^{2}=4 $$
Step-by-Step Solution
Verified Answer
The polar equation is \(r^2 - 6r\cos\theta + 2r\sin\theta = -6\).
1Step 1: 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:\[(r\cos\theta - 3)^2 + (r\sin\theta + 1)^2 = 4\]
2Step 2: Expand the Expression
Expand the equation from the previous step:\[(r\cos\theta - 3)^2 = r^2\cos^2\theta - 6r\cos\theta + 9\]\[(r\sin\theta + 1)^2 = r^2\sin^2\theta + 2r\sin\theta + 1\]
3Step 3: 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:\[r^2 - 6r\cos\theta + 2r\sin\theta + 10 = 4\]
4Step 4: Isolate Terms
Subtract 10 from both sides to isolate terms involving \(r\):\[r^2 - 6r\cos\theta + 2r\sin\theta = -6\]
5Step 5: 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.
Key Concepts
Cartesian CoordinatesPolar CoordinatesEquation Conversion
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.
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 \( y = r\sin\theta \).
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.
Other exercises in this chapter
Problem 60
Replace the Cartesian equations in Exercises \(49-62\) by equivalent polar equations. $$ (x-5)^{2}+y^{2}=25 $$
View solution Problem 61
Find the center, foci, vertices, asymptotes, and radius, as appropriate, of the conic sections in Exercises \(57-68 .\) $$ x^{2}+5 y^{2}+4 x=1 $$
View solution Problem 61
Graph the lines and conic sections in Exercises \(47-56\) Find a polar equation for the parabola with focus \((0,0)\) and direc- \(\operatorname{trix} r \cos \t
View solution Problem 62
Find a polar equation for the parabola with focus \((0,0)\) and direc\(\operatorname{trix} r \cos (\theta-\pi / 2)=2\)
View solution