Problem 31 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 23-36, name the curve with the given polar equation. If it is a conic, give its eccentricity. Sketch the graph. \(r=\frac{6}{2+\sin \theta}\)

Step-by-Step Solution

Answer

The curve is an ellipse with eccentricity \(0.5\).

Step 1: Identify the Polar Equation Type

The given polar equation is of the form \( r = \frac{ed}{1 + e \sin \theta} \). In this specific equation, \( r = \frac{6}{2 + \sin \theta} \). We can rewrite it in the standard conic form \( r = \frac{6/2}{1 + \frac{1}{2} \sin \theta} \). Here, the constant term associated with \( \sin \theta \) is \( \frac{1}{2} \).

Step 2: Recognize the Equation of a Conic in Polar Coordinates

A conic described in polar coordinates with an equation of the form \( r = \frac{ed}{1 + e \sin \theta} \) indicates a conic section. The conic's eccentricity is given by \( e = \frac{1}{2} \). Since \( e < 1 \), it indicates that the conic is an ellipse.

Step 3: Determine the Eccentricity and Conic Type

Since we have already determined that the equation can be expressed as \( r = \frac{3}{1 + \frac{1}{2} \sin \theta} \), the eccentricity \( e = \frac{1}{2} \) implies that the conic section is an ellipse, as eccentricity values less than 1 correspond to ellipses.

Step 4: Sketch the Graph

The graph of this polar equation represents an ellipse centered at the pole (origin). Its size and orientation will depend on the polar coefficients. With \( e = \frac{1}{2} \), this ellipse will be oriented such that its major axis is perpendicular to the direction of maximum or minimum radius found when \( \theta = \frac{\pi}{2} \) and \( \theta = \frac{3\pi}{2} \). Wherever \( \sin \theta = 0 \), the radius is minimized, so this ellipse will extend symmetrically around the pole.

Key Concepts

Eccentricity

Eccentricity is a crucial concept when analyzing and identifying different types of conic sections in geometry. In simple terms, eccentricity tells us how much a conic section deviates from being circular. The eccentricity is denoted by the symbol \( e \).

• If \( e = 0 \), the conic is a circle, which is perfectly round.
• If \( 0 < e < 1 \), the conic is an ellipse, which is oval-shaped.
• If \( e = 1 \), we have a parabola.
• If \( e > 1 \), the conic is a hyperbola, which opens outward.

In the given example with the polar equation \( r = \frac{6}{2 + \sin \theta} \), the eccentricity is calculated to be \( e = \frac{1}{2} \). This means the shape of our conic is an ellipse, as \( 0 < e < 1 \). The lower the eccentricity, the more circular the ellipse appears.

Conic Sections

Conic sections are shapes that you get when you slice a cone with a plane. They are very important in mathematics because they describe so many real-world structures and phenomena, from planetary orbits to architectural curves.
There are four main types of conic sections:

• Circle - formed when the cutting plane is perpendicular to the cone's axis.
• Ellipse - obtained when a plane cuts through a cone at an angle, but does not touch the base.
• Parabola - formed when the plane is parallel to the edge of the cone.
• Hyperbola - occurs when the plane cuts through both parts of the cone.

In polar coordinates, certain equations represent these conics, allowing us to quickly identify their type and properties, such as eccentricity. In our given equation \( r = \frac{6}{2 + \sin \theta} \), it's been determined that the conic section is an ellipse because the eccentricity is \( \frac{1}{2} \).

Ellipse

Ellipses are fascinating curves, often described as 'stretched circles'. They have two focal points, and the sum of the distances from any point on the ellipse to these two points is constant. This unique property defines an ellipse.
An ellipse in polar coordinates is commonly expressed in the form \( r = \frac{ed}{1 + e\cos\theta} \) or \( r = \frac{ed}{1 + e\sin\theta} \). The parameters of the equation elaborate on:

• The eccentricity \( e \), which dictates how stretched the ellipse is.
• \( d \), a constant that impacts the overall size of the ellipse.

In our equation \( r = \frac{6}{2 + \sin \theta} \), the eccentricity \( e \) is \( 0.5 \) which informs us about the elliptical nature of the graph. The formula's structure gives it properties where the ellipse's major axis appears perpendicular to the angle of the sine term in the equation.

Graph Sketching

Graph sketching is a valuable skill that allows you to visualize mathematical equations. Particularly in polar coordinates, it involves understanding how equations translate into curves like ellipses.
For the polar equation \( r = \frac{6}{2 + \sin \theta} \):

• Identify critical angles such as \( \theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2} \) to better understand the ellipse's scope.
• Plot points by evaluating the equation for different \( \theta \) values, noticing how \( r \), the radius, changes.
• The graph of this equation is an ellipse centered at the origin, and it’s helpful to know that its orientation is based on the sine term, which tends to stretch along the perpendicular axis.

Using the determined eccentricity \( e = 0.5 \), sketching becomes easier, as we understand the nature of the ellipse involves a balance between stretching and maintaining a rounded shape. This results in a symmetrical eye-shaped curve positioned around the pole.