Problem 50 - Thomas Calculus

Question

Find a polar equation in the form \(r \cos \left(\theta-\theta_{0}\right)=r_{0}\) for each of the lines in Exercises \(49-52 .\) $$\sqrt{3} x-y=1$$

Step-by-Step Solution

Answer

The polar equation is \( r \cos(\theta - \frac{\pi}{6}) = \frac{1}{2} \).

Step 1: Identify the Standard Equation

The given equation is \( \sqrt{3}x - y = 1 \). This is a linear equation in Cartesian coordinates of the form \( ax + by = c \).

Step 2: Compare with Polar Equation Form

The desired polar equation is \( r \cos(\theta - \theta_0) = r_0 \), which implies converting Cartesian format to polar format.

Step 3: Recap of Polar-Cartesian Relations

Remember that \( x = r \cos \theta \) and \( y = r \sin \theta \) are the relations between Cartesian and polar coordinates.

Step 4: Substitute Cartesian Variables with Polar Equivalents

Replace \( x \) with \( r\cos\theta \) and \( y \) with \( r\sin\theta \) in the equation \( \sqrt{3}x - y = 1 \) to get: \( \sqrt{3}(r\cos\theta) - (r\sin\theta) = 1 \).

Step 5: Factor Out the Common \( r \) Term

Factor out \( r \): \[ r(\sqrt{3}\cos\theta - \sin\theta) = 1 \]

Step 6: Transform to Desired Form

The expression inside the parenthesis resembles the form for cosine expansion \( \cos(\theta - \theta_0) = \cos\theta\cos\theta_0 + \sin\theta\sin\theta_0 \). Use \( \cos\theta_0 = \frac{\sqrt{3}}{2}, \sin\theta_0 = -\frac{1}{2} \) such that: \( \sqrt{3}\cos\theta - \sin\theta = 2(\frac{\sqrt{3}}{2}\cos\theta - \frac{1}{2}\sin\theta) = 2\cos(\theta - \frac{\pi}{6}) \)

Step 7: Express in Polar Form

Replace the transformed expression with cosine form: \[ r(2\cos(\theta - \frac{\pi}{6})) = 1 \] Simplify it to get:\[ r\cos(\theta - \frac{\pi}{6}) = \frac{1}{2} \] This is the polar equation in the requested form.

Key Concepts

Cartesian to Polar Conversion

In mathematics, converting between coordinate systems is an essential skill needed to describe points in different ways. The relationship between Cartesian and polar coordinates is fundamental.In Cartesian coordinates, a point in a plane can be described as \(x, y\), where \(x\) and \(y\) are its horizontal and vertical distances from the origin, respectively. On the other hand, in polar coordinates, that same point is expressed as \(r, \theta\), where \(r\) is the distance from the origin, and \(\theta\) is the angle from the positive x-axis.To convert equations between these systems, you can use key relationships derived from the geometry of a right triangle:

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

By substituting these relationships into a Cartesian equation, it is possible to rewrite it in polar form. For example, given an equation like \( \sqrt{3}x - y = 1 \), replacing \( x \) and \( y \) using the above trigonometric relationships allows you to translate the equation to the polar format.Understanding this conversion helps visualize how lines, curves, and other geometric figures look when plotted on polar graphs.

Linear Equations

Linear equations represent straight lines in a coordinate plane. These can be expressed in the standard Cartesian form \( ax + by = c \), where \(a\), \(b\) are coefficients, and \(c\) is a constant. The components \(a\) and \(b\) determine the slope and orientation of the line.In polar coordinates, linear equations assume a different form, often expressed as \(r \cos(\theta - \theta_0) = r_0\), highlighting their radial attributes.

Understanding Standard Forms

The manipulation and transformation of linear equations aim at finding expressions that simplify calculations and reveal deeper geometric insights. When converting a linear equation like \( \sqrt{3}x - y = 1\) from Cartesian to polar form, recognizing patterns such as cosine expressions (

\( \cos(\theta - \theta_0) \)

comes in handy because these identities align coordinates to make equations easier to understand and graph.Thus, linear equations serve as a cornerstone for interpreting arrangements of lines across various mathematical contexts, whether in familiar Cartesian systems or the more radial-focused polar layouts.

Trigonometric Identities

Trigonometric identities are mathematical equations that relate angles and side ratios in triangles. They are pivotal for transforming and simplifying equations, especially when converting between coordinate systems.Some important identities include:

• \( \cos^2\theta + \sin^2\theta = 1 \)
• \( \cos(\theta - \theta_0) = \cos\theta\cos\theta_0 + \sin\theta\sin\theta_0 \)

This latter identity was used in the example to match and transform terms into a familiar cosine expansion, aiding in the conversion of the Cartesian equation \( \sqrt{3}x - y = 1\) into the polar form \( r \cos(\theta - \frac{\pi}{6}) = \frac{1}{2} \).

Using Identities in Conversion

Utilizing these identities allows us to express complex mathematical conditions in an elegantly simplified manner. By aligning coefficient values with known trigonometric constants, such as \(\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}\) and \(\sin\frac{\pi}{6} = \frac{1}{2} \), it becomes possible to reform complex equations, unveiling their trigonometric structure.These identities thus enable a deeper comprehension of equations and assist in analytical geometry, enhancing our ability to interpret mathematical narratives with ease.