Problem 34 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the given curves and find their points of intersection. $$ r=1-\cos \theta, r=1+\cos \theta $$

Step-by-Step Solution

Answer

The curves intersect at \((1, \pi/2)\) and \((1, 3\pi/2)\).

Step 1: Recognize the Polar Equations

The given equations are in polar form: \( r = 1 - \cos \theta \) and \( r = 1 + \cos \theta \). Our task is to sketch these curves and determine their points of intersection.

Step 2: Sketch the Curve for \( r = 1 - \cos \theta \)

This is the equation of a limaçon with an inner loop. To sketch it, note that when \( \theta = 0 \), \( r = 0 \); when \( \theta = \pi/2 \), \( r = 1 \); when \( \theta = \pi \), \( r = 2 \). The curve loops back to \( r = 0 \) at \( \theta = 2\pi \). Plot these points and draw a smooth curve to get the limaçon with an inner loop.

Step 3: Sketch the Curve for \( r = 1 + \cos \theta \)

This is also a limaçon curve, but without an inner loop. To sketch it, observe that when \( \theta = 0 \), \( r = 2 \); when \( \theta = \pi/2 \), \( r = 1 \); when \( \theta = \pi \), \( r = 0 \); and back to \( r = 2 \) at \( \theta = 2\pi \). Plot these points and connect them smoothly.

Step 4: Set Equations Equal to Find Intersection Points

Set \( 1 - \cos \theta = 1 + \cos \theta \). Solving this equation, we find: \( -\cos \theta = \cos \theta \) gives \( 2\cos \theta = 0 \), leading to \( \cos \theta = 0 \). So, \( \theta = \frac{\pi}{2}, \frac{3\pi}{2} \).

Step 5: Find Corresponding \( r \) Values

Substitute \( \theta = \pi/2 \) into either equation to find \( r = 1 \) and \( \theta = 3\pi/2 \) to find \( r = 1 \). The points of intersection are thus: \((1, \pi/2)\) and \((1, 3\pi/2)\) in polar coordinates.

Step 6: Verify and Interpret the Results

Finally, verify the points of intersection on the sketch. Both curves are symmetric, and the points \((1, \pi/2)\) and \((1, 3\pi/2)\) should make sense as intersections of the two curves. The curves intersect where the radial distance is the same for the given angles in both curves.

Key Concepts

Intersection of Curves

When examining the intersection of curves in polar coordinates, the goal is to find points where two curves meet. This involves setting their equations equal to each other. For the given problem, we have two curves:

• The curve described by the equation: \( r = 1 - \cos \theta \)
• The curve described by the equation: \( r = 1 + \cos \theta \)

At points of intersection, the radial distance, \( r \), must be the same for both curves at particular angles \( \theta \). To find \( \theta \), equate the two equations and solve for \( \theta \). In the step-by-step solution, we see that setting \( 1 - \cos \theta = 1 + \cos \theta \) leads to finding the intersections of \( \theta = \frac{\pi}{2} \) and \( \theta = \frac{3\pi}{2} \). Once you have these values, you substitute them back to find the corresponding \( r \) values.

Limaçon

The term 'limaçon' refers to a type of polar curve that can take various shapes based on the equation constants. The limaçon is defined by equations of the form \( r = a + b \cos \theta \) or \( r = a + b \sin \theta \). Its shape is highly dependent on the ratio of \( a \) to \( b \). For our exercise, the curves are:

• \( r = 1 - \cos \theta \)
• \( r = 1 + \cos \theta \)

The difference between the limaçons is due to the signs. The first equation, \( r = 1 - \cos \theta \), forms a limaçon with an inner loop, while the second, \( r = 1 + \cos \theta \), forms a limaçon without an inner loop.

• For \( r = 1 - \cos \theta \), values reach from outer points toward the origin and then create an inner loop.
• In \( r = 1 + \cos \theta \), the curve bulges out, never looping back across the origin.

Equation Solving

Solving equations in polar coordinates, particularly when dealing with intersections, involves setting equations equal and solving for angles. This challenges students to understand manipulations specific to trigonometric identities. In our task, the equations \( 1 - \cos \theta = 1 + \cos \theta \) simplify by moving terms and using underlying trigonometric identities.

• Start with \( -\cos \theta = \cos \theta \).
• This can be rewritten and solved: \( 2\cos \theta = 0 \).
• From here, solve for \( \theta \): \( \cos \theta = 0 \). This gives us the standard angles \( \theta = \frac{\pi}{2} \) and \( \frac{3\pi}{2} \).

These angles correspond to intersections when substituted back into either polar curve equation.

Curve Sketching

Sketching curves in polar coordinates often begins by understanding the behavior of the equation at different angles. This includes noting values at key angles like \( \theta = 0 \), \( \frac{\pi}{2} \), and \( \pi \), which highlight critical points on the curve.

• For \( r = 1 - \cos \theta \), start at \( r = 0 \) when \( \theta = 0 \), then move to \( r = 1 \) at \( \theta = \frac{\pi}{2} \), \( r = 2 \) at \( \theta = \pi \), and finally back to \( r = 0 \) at \( \theta = 2\pi \).
• For \( r = 1 + \cos \theta \), identify points starting at \( r = 2 \) when \( \theta = 0 \), down to \( r = 1 \) at \( \theta = \frac{\pi}{2} \), then \( r = 0 \) at \( \theta = \pi \), coming back to \( r = 2 \) at \( \theta = 2\pi \).

Connecting these points smoothly, understanding the symmetrical nature of the equations, results in accurately sketched limaçons. Observing the behavior at each quadrant provides a guided way to ensure precision and accuracy in your sketches.