Problem 34 - Precalculus Mathematics for Calculus

Question

Sketch a graph of the polar equation. $$r=2+\sin \theta$$

Step-by-Step Solution

Answer

The graph is a dimpled limaçon.

Step 1: Convert to Cartesian

The polar equation is given by \( r = 2 + \sin \theta \). First, we convert the polar equation to Cartesian coordinates for easier analysis. Using the polar to Cartesian transformations \( x = r \cos \theta \) and \( y = r \sin \theta \), we know that \( r^2 = x^2 + y^2 \) and \( an \theta = \frac{y}{x} \). However, since \sin \theta = \frac{y}{r}\, we substitute this into the original equation to get \( r = 2 + \frac{y}{r} \). Multiply both sides by \( r \) to clear the fraction: \( r^2 = 2r + y \). Substitute \( r^2 = x^2 + y^2 \) to get: \( x^2 + y^2 = 2r + y \).

Step 2: Simplify the Equation

Replace \( r \) with \( \sqrt{x^2 + y^2} \) in the equation so that it entirely consists of Cartesian coordinates: \( x^2 + y^2 = 2\sqrt{x^2 + y^2} + y \). Rearrange the equation by moving all terms to one side: \( x^2 + (y - 1)^2 - 1 = 2\sqrt{x^2 + y^2} \). This is a complex form to handle algebraically, but we can identify general properties of the curve from its polar form.

Step 3: Identify the Curve Type

The equation \( r = 2 + \sin \theta \) represents a limaçon with a dimple in polar coordinates because it has the form \( r = a + b\sin \theta \) where \( 0 < b < a \). Here \; a = 2 \; and \; b = 1. Since \( 0 < b < a \), it is a 'dimpled limaçon'.

Step 4: Analyze Key Points

Find key values of \( r \) at specific angles \( \theta \) to plot some crucial points: - At \( \theta = 0\), \( r = 2 + \sin 0 = 2 \)- At \( \theta = \frac{\pi}{2} \), \( r = 2 + \sin \frac{\pi}{2} = 3 \)- At \( \theta = \pi \), \( r = 2 + \sin \pi = 2 \)- At \( \theta = \frac{3\pi}{2} \), \( r = 2 + \sin \frac{3\pi}{2} = 1 \)

Step 5: Sketch the Graph

Using the identified key points and the fact that the curve is a 'dimpled limaçon', sketch the curve. Start at \( \theta = 0 \) with \( r = 2 \), then increase to \( r = 3 \) at \( \theta = \frac{\pi}{2} \), back to \( r = 2 \) at \( \theta = \pi \), and finally down to \( r = 1 \) at \( \theta = \frac{3\pi}{2} \). Draw a smooth, continuous curve, noting the symmetry about the horizontal axis.

Key Concepts

Limaçon

A limaçon is a type of curve that appears in polar coordinate systems and can be identified by equations of the form \( r = a + b\sin \theta \) or \( r = a + b\cos \theta \). These equations feature two key parameters: \( a \) and \( b \), which determine the shape of the curve.

Depending on the values of \( a \) and \( b \), limaçon curves can appear with or without loops and dimples. For the equation \( r = 2 + \sin \theta \), we have \( a = 2 \) and \( b = 1 \). Since \( 0 < b < a \), this results in a limaçon with a dimple. This specific type is sometimes called a "dimpled limaçon."

The characteristics of these curves vary widely:

• If \( b = a \), the limaçon will have a cardioid shape, resembling a heart.
• If \( b > a \), the curve exhibits an inner loop.
• If \( 0 < b < a \), as in our case, the curve is dimpled, showing a gentle indentation.

Such properties help in predicting the curve's shape given its polar equation. Analytical solutions like these ensure clear understanding before moving on to more visual representations.

Polar to Cartesian Conversion

Converting from polar to Cartesian coordinates is a critical skill in understanding and analyzing polar curves. Polar coordinates express a point in terms of its distance \( r \) from the origin and the angle \( \theta \) relative to the positive x-axis.

To convert these into the Cartesian system, which uses x and y coordinates:

• The x-coordinate is obtained by \( x = r \cos \theta \).
• The y-coordinate is obtained by \( y = r \sin \theta \).

This transformation allows us to express complex polar equations in a more familiar Cartesian form.

Using this conversion, we see that the original equation \( r = 2 + \sin \theta \) can be transformed into Cartesian form by first multiplying through by \( r \) (resulting in \( r^2 = 2r + y \)) and then substituting \( r^2 = x^2 + y^2 \). The result is an equation involving \( x \) and \( y \), such as \( x^2 + y^2 = 2\sqrt{x^2 + y^2} + y \).

Although the Cartesian equation might appear challenging, polar to Cartesian conversions offer valuable insights into the relationships and characteristics of polar curves.

Graph Sketching in Polar Coordinates

Sketching graphs in polar coordinates can be a fascinating exercise, involving a blend of algebraic manipulation and geometric insight. It begins with evaluating the behavior of the function at specific angles \( \theta \).

For \( r = 2 + \sin \theta \), key points are vital for sketching the curve. Consider:

• At \( \theta = 0 \), \( r = 2 + \sin 0 = 2 \).
• At \( \theta = \frac{\pi}{2} \), \( r = 2 + \sin \frac{\pi}{2} = 3 \).
• At \( \theta = \pi \), \( r = 2 + \sin \pi = 2 \).
• At \( \theta = \frac{3\pi}{2} \), \( r = 2 + \sin \frac{3\pi}{2} = 1 \).

These points help outline the basic shape on the polar coordinate system. Starting from \( r = 2 \) at \( \theta = 0 \), moving through \( r = 3 \) at \( \theta = \frac{\pi}{2} \), returning to \( r = 2 \) at \( \theta = \pi \), and dropping to \( r = 1 \) at \( \theta = \frac{3\pi}{2} \).

By drawing these points and smoothly connecting them, one can achieve the curve of the dimpled limaçon. Understanding the symmetry and progression offers a satisfying visual conclusion to the analytical process of plotting polar equations.