Problem 16
Question
Exer. 13-26: Find a polar equation that has the same graph as the equation in \(x\) and \(y\). $$ x^{2}+y^{2}=2 $$
Step-by-Step Solution
Verified Answer
The polar equation is \(r = \sqrt{2}\).
1Step 1: Understand the Equation Type
The given Cartesian equation is a circle equation: \(x^2 + y^2 = 2\). This represents a circle centered at the origin with a radius of \(\sqrt{2}\).
2Step 2: Recall Polar to Cartesian Transformations
In polar coordinates, the transformations are: \(x = r\cos\theta\) and \(y = r\sin\theta\). Moreover, \(r^2 = x^2 + y^2\).
3Step 3: Substitute Cartesian Terms with Polar Equivalents
Substitute \(x^2 + y^2\) with \(r^2\) in the polar equation to transform the given Cartesian equation. Thus, the equation becomes \(r^2 = 2\).
4Step 4: Solve for the Polar Equation
Since \(r^2 = 2\), taking the square root of both sides gives \(r = \sqrt{2}\). This is the polar equation for the given Cartesian circle.
Key Concepts
Cartesian to Polar TransformationCircle EquationPolar Coordinates
Cartesian to Polar Transformation
Transforming equations from Cartesian coordinates to polar coordinates involves a few straightforward substitutions. To establish a connection between Cartesian and polar coordinates:
In the case of our example, the Cartesian circle equation \( x^2 + y^2 = 2 \) can seamlessly transform to the polar form \( r^2 = 2 \). This transformation is possible because polar coordinates naturally describe circles with \( r \) referring to the radius, emphasizing the symmetry and simplicity polar coordinates can bring.
- Replace the Cartesian coordinates \( x \) and \( y \) with the polar coordinates' equivalent forms: \( x = r\cos\theta \) and \( y = r\sin\theta \).
- Use \( r^2 = x^2 + y^2 \) to link the coordinates.
In the case of our example, the Cartesian circle equation \( x^2 + y^2 = 2 \) can seamlessly transform to the polar form \( r^2 = 2 \). This transformation is possible because polar coordinates naturally describe circles with \( r \) referring to the radius, emphasizing the symmetry and simplicity polar coordinates can bring.
Circle Equation
Circle equations are a prime example of how to model symmetrical shapes. In Cartesian coordinates, circles are usually expressed as \( x^2 + y^2 = r^2 \), where \( r \) is the radius of the circle. This form indicates that every point on the circle is equidistant from the center, determined by the radius.
For the given example \( x^2 + y^2 = 2 \), the equation implies a circle centered at the origin with a radius of \( \sqrt{2} \).
For the given example \( x^2 + y^2 = 2 \), the equation implies a circle centered at the origin with a radius of \( \sqrt{2} \).
- This equation style is not only helpful for plotting but also offers insights into the circle's characteristics, like its radius and center.
- Being familiar with these properties can aid in converting the circle from one coordinate system to another.
Polar Coordinates
Polar coordinates can describe a point in the plane using a distance and an angle, rather than two perpendicular distances.
In the exercise at hand, converting the circle equation to polar form resulted in the equation \( r = \sqrt{2} \). Here, \( r \) is constant, indicating every point on this circle lies at a steady radius from the origin. This characteristic displays the elegance and utility of polar coordinates in presenting circular geometries with simplicity, especially in scenarios where the origin serves as a natural focal point.
- The distance from the origin is called \( r \) (the radial coordinate).
- The angle \( \theta \), measured from the positive x-axis, is the angular coordinate.
In the exercise at hand, converting the circle equation to polar form resulted in the equation \( r = \sqrt{2} \). Here, \( r \) is constant, indicating every point on this circle lies at a steady radius from the origin. This characteristic displays the elegance and utility of polar coordinates in presenting circular geometries with simplicity, especially in scenarios where the origin serves as a natural focal point.
Other exercises in this chapter
Problem 15
Exer. 13-26: Find a polar equation that has the same graph as the equation in \(x\) and \(y\). $$ x^{2}+y^{2}=16 $$
View solution Problem 15
Find the vertices, the foci, and the equations of the asymptotes of the hyperbola. Sketch its graph, showing the asymptotes and the foci. $$4 y^{2}-x^{2}+40 y-4
View solution Problem 16
Find the vertices, the foci, and the equations of the asymptotes of the hyperbola. Sketch its graph, showing the asymptotes and the foci. $$25 x^{2}-9 y^{2}+100
View solution Problem 17
Exer. 13-26: Find a polar equation that has the same graph as the equation in \(x\) and \(y\). $$ y^{2}=6 x $$
View solution