Problem 57 - College Algebra

Question

Express each equation in polar form with \(r\) as a function of \(\theta\). $$ x^{2}+x y+y^{2}=4 $$

Step-by-Step Solution

Answer

\(r = \sqrt{\frac{4}{1 + \cos\theta\sin\theta}}\).

Step 1: Convert Cartesian to Polar Coordinates

In polar coordinates, the relationships are given by \( x = r\cos\theta \) and \( y = r\sin\theta \). We substitute these into the given equation:\[x^2 + xy + y^2 = 4 \]becomes:\[(r\cos\theta)^2 + (r\cos\theta)(r\sin\theta) + (r\sin\theta)^2 = 4\]

Step 2: Simplify the Polar Equation

Expand and combine like terms:\[ r^2\cos^2\theta + r^2\cos\theta\sin\theta + r^2\sin^2\theta = 4\]Factor out \(r^2\):\[ r^2(\cos^2\theta + \sin^2\theta + \cos\theta\sin\theta) = 4\]Recall the trigonometric identity \(\cos^2\theta + \sin^2\theta = 1\), which simplifies the equation to:\[ r^2(1 + \cos\theta\sin\theta) = 4\]

Step 3: Express r as a Function of \(\theta\)

To express \(r^2\) in terms of \(\theta\), rearrange the equation:\[ r^2 = \frac{4}{1 + \cos\theta\sin\theta}\]Take the square root of both sides to solve for \(r\):\[ r = \sqrt{\frac{4}{1 + \cos\theta\sin\theta}}\]

Key Concepts

Trigonometric Identity

In mathematics, a trigonometric identity is an equation that involves trigonometric functions and is true for every value of the involved variables for which both sides of the equality are defined. These identities can be useful for simplifying expressions and solving trigonometric equations.

One fundamental trigonometric identity often used is \[\cos^2\theta + \sin^2\theta = 1\]This particular identity comes in handy when transforming expressions that involve squares of sine and cosine into a simpler form, such as unifying various terms under one umbrella.

In our exercise, we applied this identity to simplify \( r^2\cos^2\theta + r^2\sin^2\theta \), allowing us to replace the combined terms with the number 1. This significantly reduces complexity when handling mathematical problems, especially when transitioning between coordinate systems.

Cartesian to Polar Conversion

Converting from Cartesian coordinates (what you might call the usual \(x\) and \(y\) grid) to polar coordinates is about shifting our perspective. In Cartesian coordinates, a point is represented by \((x, y)\). In polar coordinates, it's represented by \((r, \theta)\), where \(r\) is the radial distance from the origin, and \(\theta\) is the angle from the positive x-axis.

To transition between these systems, we use the formulas:

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

Thus, given an equation in \(x\) and \(y\), we can substitute these expressions to convert it to a function of \(r\) and \(\theta\).

For our exercise, substituting \((x = r\cos\theta)\) and \((y = r\sin\theta)\) into the equation \(x^2 + xy + y^2 = 4\) allowed us to start working in the polar coordinate system.

Polar Equation Simplification

Once a problem is set in polar coordinates, the next step is to simplify it as much as possible. After making substitutions for \(x\) and \(y\), we get an equation with terms involving \(r\) and trigonometric functions of \(\theta\).

Our goal is to tidy these terms up so they express \(r\) as cleanly as possible. To achieve this in our problem, we expanded all products and collected like terms: \[r^2\cos^2\theta + r^2\cos\theta\sin\theta + r^2\sin^2\theta\]

Factoring out \(r^2\) from the expression grouped everything under the function \(\cos^2\theta + \sin^2\theta\), which we know already simplifies to 1 through our trigonometric identity. This simplification reduces unwieldy polynomials into comprehensible expressions.

Expressing r as a Function of Theta

Having simplified the polar equation, the final goal typically is to express \(r\) as a function of \(\theta\). This involves isolating \(r\) on one side of the equation. Through manipulation and algebraic operations, we find an expression that defines how \(r\) changes based on \(\theta\).

In our example, after removing common factors and applying identities, we reached \[r^2 = \frac{4}{1 + \cos\theta\sin\theta}\]From here, taking the square root of both sides provided us the relationship we're often looking for, \(r\) in terms of \(\theta\): \[r = \sqrt{\frac{4}{1 + \cos\theta\sin\theta}}\]

This equation tells us the radial distance \(r\) at any angle \(\theta\), offering a complete picture in polar terms. Expressing equations in such a form often makes them easier to understand, analyze, and graph in the polar plane.