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Sketch the given curves and find their points of intersection. $$ r^{2}=4 \cos 2 \theta, r=2 \sqrt{2} \sin \theta $$

Short Answer

Expert verified
The points of intersection are the solutions of the equations \( x^2 - y^2 = 4 \) and \( x^2 + y^2 = 2\sqrt{2}y \).

Step by step solution

01

Convert to Cartesian Coordinates

We begin by converting the polar equations into Cartesian form. For the first equation, \[ r^2 = 4 \ \cos 2\theta = 4(\cos^2\theta - \sin^2\theta) \] leads to \[ x^2 - y^2 = 4 \] when converted.For the second equation, \[ r = 2\sqrt{2}\sin\theta \] translates to \[ r = 2\sqrt{2} (\frac{y}{r}) \] implying \[ r^2 = 2\sqrt{2}y \]. Since \( r^2 = x^2 + y^2 \), we get \[ x^2 + y^2 = 2\sqrt{2}y \].
02

Set Equations Equal

We have two equations:1. \( x^2 - y^2 = 4 \)2. \( x^2 + y^2 = 2\sqrt{2}y \)Add these two equations to eliminate one of the variables:\[ (x^2 - y^2) + (x^2 + y^2) = 4 + 2\sqrt{2}y \]This simplifies to \[ 2x^2 = 4 + 2\sqrt{2}y \].
03

Solve for x

The equation \[ 2x^2 = 4 + 2\sqrt{2}y \] can be rearranged as:\[ x^2 = 2 + \sqrt{2}y \].
04

Substitute x back

From the earlier equation, we know \[ x^2 - y^2 = 4 \] and \[ x^2 = 2 + \sqrt{2}y \].Substitute for \(x^2\) in the first equation to obtain:\[ 2 + \sqrt{2}y - y^2 = 4 \].Simplify to:\[ y^2 - \sqrt{2}y + 2 = 0 \].
05

Solve Quadratic Equation for y

Solve the quadratic \( y^2 - \sqrt{2}y + 2 = 0 \) using the quadratic formula:\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 1 \), \( b = -\sqrt{2} \), and \( c = 2 \).Substitute these values to find the roots for \( y \).
06

Solve for Corresponding x

For each solution for \( y \), substitute back into the equation \( x^2 = 2 + \sqrt{2}y \) to find corresponding values of \( x \). Remember that \( x^2 \) has two possible solutions, \( x = \pm\sqrt{x^2} \).
07

Convert Back to Polar for Intersection Points

For each \((x,y)\), convert back to polar coordinates using \( r = \sqrt{x^2 + y^2} \) and \( \theta = \tan^{-1}\left(\frac{y}{x}\right) \) to find the points of intersection in polar form.

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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 an essential system in mathematics for representing points on a plane. They consist of two values: the radial distance from a fixed origin and the angle from a fixed direction, usually the positive x-axis. This system is particularly useful in cases where relationships between points exhibit circular or rotational symmetry.

In polar coordinates, any point is defined by \(r, \theta\), where \(r\) is the radius or distance from the origin and \(\theta\) is the angle.
  • The angle \(\theta\) measures the direction of the line connecting the origin to the point, typically in radians.
  • The radius \(r\) signifies how far this line extends.
Using polar coordinates is advantageous when dealing with curves like circles, spirals, and ellipses. It simplifies understanding and calculating intersections of such curves, as is the case in the given problem where two curves are expressed in terms of \(r\) and \(\theta\).
Cartesian Coordinates
Cartesian coordinates are the most common coordinate system and involve two perpendicular axes: the x-axis (horizontal) and the y-axis (vertical). Every point on the plane is represented by a pair of numbers \(x, y\), which indicate horizontal and vertical distances from the origin.

This system is particularly useful for solving many types of mathematical problems because of its straightforward, metric nature that fits our intuitive understanding of movement along straight lines. When working with problems involving curves' intersections, converting polar coordinates to Cartesian coordinates can simplify calculations and solution finding.

In the problem above, converting from polar to Cartesian coordinates helps in visualizing intersections more clearly. This is achieved by using the known relations:
  • \(x = r \cdot \cos\theta\)
  • \(y = r \cdot \sin\theta\)
These conversions allow the initial polar equations to be expressed in terms of x and y, making them suitable for algebraic manipulation and solving.
Quadratic Equation
Quadratic equations are polynomial equations of the second degree, usually in the form \( ax^2 + bx + c = 0 \). Solving these equations is a fundamental skill in algebra, involving finding the values of the variable that make the equation true.

These equations can be solved by several methods, including:
  • Factoring
  • Completing the square
  • Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
In the given problem, solving the equivalent quadratic equation for y helps in identifying key intersection points of the curves. The solution provides values that are substituted back to find corresponding x-values, ultimately aiding in finding the points where the curves intersect in Cartesian form.
Coordinate Conversion
Coordinate conversion between polar and Cartesian systems is a vital skill in mathematics, especially when analyzing curves and their intersections. This involves transforming one coordinate representation into another using known relationships.

Given a point \(r, \theta\) in polar coordinates, the conversion to Cartesian coordinates \(x, y\) involves:
  • \(x = r \cdot \cos\theta\)
  • \(y = r \cdot \sin\theta\)
Conversely, converting from Cartesian to polar coordinates is done using:
  • \(r = \sqrt{x^2 + y^2}\)
  • \(\theta = \tan^{-1}\left(\frac{y}{x}\right)\)
This process is essential in the problem's final step, converting the intersecting points back to polar form to fully answer the prompt. Understanding coordinate conversion allows for versatility in analyzing geometrical problems, making it easier to interpret and solve complex issues across different representation systems.

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Most popular questions from this chapter

Find the polar equation for the curve described by the following Cartesian equations. (a) \(y=45\) (b) \(x^{2}+y^{2}=36\) (c) \(x^{2}-y^{2}=1\) (d) \(4 x y=1\) (e) \(y=3 x+2\) (f) \(3 x^{2}+4 y=2\) (g) \(x^{2}+2 x+y^{2}-4 y-25=0\) Computers and graphing calculators offer a wonderful opportunity to experiment with the graphing of polar equations of the form \(r=f(\theta) .\) In some cases these aids require that the equations be recast in a parametric form. Since \(x=r \cos \theta=f(\theta) \cos \theta\) and \(y=r \sin \theta=f(\theta) \sin \theta\), you can use the parametric graphing capabilities to graph \(x=f(t) \cos t\) and \(y=f(t) \sin t\) as a set of parametric equations.

Sketch the graph of the given equation. \(\frac{(x+3)^{2}}{4}-\frac{(y+2)^{2}}{16}=1\)

Let \(r_{1}\) and \(r_{2}\) be the minimum and maximum distances (perihelion and aphelion, respectively) of the ellipse \(r=\) \(e d /\left[1+e \cos \left(\theta-\theta_{0}\right)\right]\) from a focus. Show that (a) \(r_{1}=e d /(1+e), r_{2}=e d /(1-e)\), (b) major diameter \(=2 e d /\left(1-e^{2}\right)\) and minor diameter \(=\) \(2 e d / \sqrt{1-e^{2}}\)

Show that the polar equation of the circle with center \((c, \alpha)\) and radius \(a\) is \(r^{2}+c^{2}-2 r c \cos (\theta-\alpha)=a^{2} .\)

Let \(F\) and \(F^{\prime}\) be fixed points with polar coordinates \((a, 0)\) and \((-a, 0)\), respectively. Show that the set of points \(P\) satisfying \(|P F|\left|P F^{\prime}\right|=a^{2}\) is a lemniscate by finding its polar equation.

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