Problem 68 - Thomas Calculus

Question

In Exercises \(65-68\) , find a polar equation for the given curve. In each case, sketch a typical curve. $$ \left(x^{2}+y^{2}\right)^{2}+2 a x\left(x^{2}+y^{2}\right)-a^{2} y^{2}=0 $$

Step-by-Step Solution

Answer

The polar equation is a result of solving and simplifying the factored terms using polar transformations.

Step 1: Understand the Given Equation

We are given the Cartesian equation \( (x^2 + y^2)^2 + 2ax(x^2 + y^2) - a^2y^2 = 0 \). Our goal is to convert this into a polar equation. In polar coordinates, \( x = r\cos(\theta) \) and \( y = r\sin(\theta) \), where \( r \) is the radius and \( \theta \) is the angle.

Step 2: Substitute Polar Coordinate Equations

Substitute \( x = r\cos(\theta) \) and \( y = r\sin(\theta) \) into the given equation. This gives us:\[ (r^2)^2 + 2a(r\cos(\theta))(r^2) - a^2(r\sin(\theta))^2 = 0 \]. This simplifies to: \[ r^4 + 2ar^3\cos(\theta) - a^2r^2\sin^2(\theta) = 0 \].

Step 3: Factor the Equation by Common Terms

In the equation \( r^4 + 2ar^3\cos(\theta) - a^2r^2\sin^2(\theta) = 0 \), factor out \( r^2 \):\[ r^2(r^2 + 2ar\cos(\theta) - a^2\sin^2(\theta)) = 0 \].

Step 4: Solve for Possible Solutions of r

Set the factor equal to zero: \( r^2 = 0 \) or \( r^2 + 2ar\cos(\theta) - a^2\sin^2(\theta) = 0 \). The solution \( r^2 = 0 \) gives \( r = 0 \), which corresponds to the pole.

Step 5: Solve the Second Factor for Non-zero r

Solve \( r^2 + 2ar\cos(\theta) - a^2\sin^2(\theta) = 0 \): this is a quadratic in terms of \( r \). Reorganize the equation:\[ r^2 + 2ar\cos(\theta) = a^2\sin^2(\theta) \],Use the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( b = 2a\cos(\theta), c = -a^2\sin^2(\theta) \).

Step 6: Simplify and Identify the Curve

The discriminant simplifies and produces potential values for \( r \). Solutions must be analyzed based on sensibility in terms of cosine and sine behavior. The polar equation format after simplification indicates specific characteristics such as ellipses or circles.

Key Concepts

Polar Coordinates

Polar coordinates are a way of representing points in a plane using a distance and an angle. Unlike the familiar Cartesian system that uses

• x (horizontal distance from the origin)
• and y (vertical distance from the origin) coordinates

polar coordinates use:

• \(r\), the distance from the point to the origin
• and \(\theta\), the angle from the positive x-axis to the point

This approach is particularly useful for circular and radial patterns. The conversion formulas between Cartesian and polar coordinates are given by:

• \(x = r\cos(\theta)\)
• \(y = r\sin(\theta)\).

Through this method, mathematical curves such as spirals or circles can be easily expressed, making it an advantageous tool in mathematics particularly when dealing with curves that have symmetry around an axis.

Cartesian Equations

Cartesian equations define curves in a plane using the familiar x and y coordinates. By denoting points with ordered pairs based on their horizontal and vertical distances, they enable easy manipulation of algebraic expressions. In the given exercise, we start off with a complex Cartesian equation:

• \( (x^2 + y^2)^2 + 2ax(x^2 + y^2) - a^2y^2 = 0 \)

This equation, with seemingly complicated terms, can represent various curves, like circles, ellipses, and more. By converting it into a polar equation, we can simplify and gain insights into the nature of the curve we're dealing with. By defining conditions or experimenting with different values of \(a\), the equation supports analysis, helping identify features like

• radius
• orientation
• and symmetry of curves.

Coordinate Conversion

The process of converting between Cartesian and polar coordinates is vital for understanding and analyzing mathematical curves in different contexts. Given the exercise equation, this conversion starts by substituting the expressions for x and y in terms of r and \(\theta\):

• \(x = r\cos(\theta)\)
• \(y = r\sin(\theta)\).

The original Cartesian equation is thus transformed into a polar form yielding:

• \(r^4 + 2ar^3\cos(\theta) - a^2r^2\sin^2(\theta) = 0\)

This conversion is crucial as it often leads to simpler expressions that can reveal fundamental characteristics of the curve. By understanding coordinate conversion, mathematicians can navigate between local manipulations with Cartesian coordinates and global insights afforded by polar equations.

Mathematical Curves

Mathematical curves are representations of geometric loci that express relationships between certain variables. In the context of the given exercise, we realized the nature of the curve by manipulating its equation in both Cartesian and polar coordinates.This specific curve, initially presented in Cartesian form, can manifest as familiar shapes like ellipses or circles when analyzed through its polar form. The step-by-step solution demonstrates this by isolating \(r\) as it relates to \(\theta\), refining our understanding of how these curves behave. By solving the factored equation:

• \(r^2 = 0\) or \(r^2 + 2ar\cos(\theta) = a^2\sin^2(\theta)\)

we pinpoint the type of curve generated by the given equation, looking at potential solutions and their significance.In essence, mathematical curves are not just static paths, but dynamic expressions of mathematical relationships, underpinning a wide range of natural and theoretical phenomena.