Problem 24 - The Calculus with Analytic Geometry

Question

Draw a sketch of the graph of the given equation.\(r=3+3 \cos \theta\) (cardioid)

Step-by-Step Solution

Answer

The graph is a cardioid with key points at \((6, 0)\), \((3, \frac{\pi}{2})\), \((0, \pi)\), and \((3, \frac{3\pi}{2})\).

Step 1: Identify the type of graph

The equation given is in polar coordinates, indicating that the graph is a cardioid: \(r = 3 + 3 \, \cos \theta\).

Step 2: Determine the key points

To sketch the graph, determine the key points for \(\theta\) values. Specifically, evaluate the equation for \(\theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}\). This will help plot critical points.

Step 3: Evaluate for \(\theta = 0\)

When \(\theta = 0\), \[r = 3 + 3 \cos 0 = 3 + 3 \cdot 1 = 6\] This gives the point \((6, 0)\).

Step 4: Evaluate for \(\theta = \frac{\pi}{2}\)

When \(\theta = \frac{\pi}{2}\), \[r = 3 + 3 \cos \left( \frac{\pi}{2} \right) = 3 + 3 \cdot 0 = 3\] This gives the point \((3, \frac{\pi}{2})\).

Step 5: Evaluate for \(\theta = \pi\)

When \(\theta = \pi\), \[r = 3 + 3 \cos \pi = 3 + 3 \cdot (-1) = 3 - 3 = 0\] This gives the point \((0, \pi)\).

Step 6: Evaluate for \(\theta = \frac{3\pi}{2}\)

When \(\theta = \frac{3\pi}{2}\), \[r = 3 + 3 \cos \left( \frac{3\pi}{2} \right) = 3 + 3 \cdot 0 = 3\] This gives the point \((3, \frac{3\pi}{2})\).

Step 7: Plot and sketch the graph

Using the points found: \((6, 0)\), \((3, \frac{\pi}{2})\), \((0, \pi)\), and \((3, \frac{3\pi}{2})\), sketch the shape of a cardioid. Generally, the graph will look like a heart shape centered around the pole.

Key Concepts

polar coordinates

Polar coordinates are an alternative to Cartesian coordinates for specifying the position of a point. In the polar coordinate system, each point is determined by a distance from a central point (called the origin or pole) and an angle from a reference direction (usually the positive x-axis). The distance is denoted as \( r \) and the angle as \( \theta \). This system is particularly useful for problems involving circular or rotational symmetry.

• The distance \( r \) can be positive or negative. Positive values mean the point is in the direction of \( \theta \), while negative values mean it is in the opposite direction.
• The angle \( \theta \) is usually measured in radians or degrees, and it can be positive (counterclockwise) or negative (clockwise).
• Equations in polar coordinates often have the form \( r = f(\theta) \), where \( f \) is some function.

By understanding polar coordinates, we can better visualize and solve problems that are naturally circular or rotational, such as those involving waves, orbits, and circular motion.

cardioid

A cardioid is a special type of curve that resembles a heart shape. It gets its name from the Greek word 'kardia', meaning 'heart'. In polar coordinates, a cardioid can be represented by equations of the form \( r = a \, (1 + \, \text{cos} \, \theta) \) or \( r = a \, (1 + \, \text{sin} \, \theta) \), where \( a \) is a constant.

• The cardioid has an important property: it has a single cusp at the pole (origin).
• It is symmetric with respect to the line where the cosine or sine term is zero.
• The size and orientation of a cardioid depend on the constant \( a \). For example, the larger the value of \( a \), the larger the cardioid.

Cardioids appear in various applications, from the design of certain microphones to the study of reflective properties in optics. The equation given in the exercise, \( r = 3 + 3 \, \text{cos} \, \theta \), indicates a cardioid that is symmetric about the polar axis.

plotting key points

Plotting key points is essential for accurately sketching graphs, especially in polar coordinates where the shape of the graph can be less intuitive. By evaluating the equation at specific angles \( \theta \), we can find crucial points that help us understand the overall shape and structure of the graph.

• Let's evaluate \( \theta \) at typical values: \( 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2} \).
• At \( \theta = 0 \), the equation \( r = 3 + 3 \, \text{cos} \, 0 \) yields \( r = 6 \), giving the point (6, 0).
• At \( \theta = \frac{\pi}{2} \), the equation becomes \( r = 3 + 3 \, \text{cos} \, \left( \frac{\pi}{2} \right) \) and yields \( r = 3 \), giving the point (3, \frac{\pi}{2}).
• At \( \theta = \pi \), we get \( r = 3 + 3 \, \text{cos} \, \pi \), resulting in \( r = 0 \). This gives the point (0, \pi).
• At \( \theta = \frac{3\pi}{2} \), \( r = 3 + 3 \, \text{cos} \, \left( \frac{3\pi}{2} \right) \) returns \( r = 3 \), yielding the point (3, \frac{3\pi}{2}).

By plotting these key points on a polar grid and connecting them smoothly, we can sketch the cardioid shape accurately. This methodical approach helps ensure that our graph remains both accurate and easy to understand.