Problem 22 - College Trigonometry

Question

In Exercises \(21-30\), find the exact polar coordinates of the points of intersection of graphs of the polar equations. Remember to check for intersection at the pole (origin). $$ r=1+\sin (\theta) \text { and } r=1-\cos (\theta) $$

Step-by-Step Solution

Answer

Intersection point is \( \left(1+\frac{\sqrt{2}}{2}, \frac{3\pi}{4}\right) \). No intersection at the pole.

Step 1: Understand the problem

We are given two polar equations: \( r = 1 + \sin(\theta) \) and \( r = 1 - \cos(\theta) \). Our task is to find their points of intersection in polar coordinates. This involves equating the two equations for \( r \) and finding the corresponding \( \theta \) values.

Step 2: Set equations equal

Set the equations equal to find relationship between \( \sin(\theta) \) and \( \cos(\theta) \):\[ 1 + \sin(\theta) = 1 - \cos(\theta) \]Simplify this equation to obtain:\[ \sin(\theta) + \cos(\theta) = 0 \]

Step 3: Solve for \(\theta\)

Solve \( \sin(\theta) + \cos(\theta) = 0 \):\[ \sin(\theta) = -\cos(\theta) \]Dividing both sides by \( \cos(\theta) \), we get:\[ \tan(\theta) = -1 \]\( \tan(\theta) = -1 \) implies that \( \theta = \frac{3\pi}{4} + n\pi \) for \( n \in \mathbb{Z} \). We will check if these intersections occur in the given equations.

Step 4: Check intersection at \(\theta = \frac{3\pi}{4}\)

For \( \theta = \frac{3\pi}{4} \), check \( r \r=1+\sin(\frac{3\pi}{4}) = 1 + \frac{\sqrt{2}}{2} \ \r=1-\cos(\frac{3\pi}{4}) = 1 - (-\frac{\sqrt{2}}{2}) \)Both are equal, leading to an intersection point at \( \left(1+\frac{\sqrt{2}}{2}, \frac{3\pi}{4}\right) \).

Step 5: Check pole (origin) intersection

Point is at pole if both equations equal zero simultaneously.* For \( r = 1 + \sin(\theta) = 0 \), \( \sin(\theta) = -1 \), so \( \theta = \frac{3\pi}{2} \).* For \( r = 1 - \cos(\theta) = 0 \), \( \cos(\theta) = 1 \), so \( \theta = 0 \).\( \theta \) does not match; thus, there's no intersection at pole.

Key Concepts

Polar Equations

Polar equations help us represent curves and shapes in a plane using polar coordinates. Unlike the traditional Cartesian coordinates that use x and y, polar coordinates describe a point by its distance from the origin (r) and the angle it makes with the positive x-axis (\( \theta \)).
The polar equation \( r = 1 + \sin(\theta) \) draws a lime-shaped curve known as a limaçon, while \( r = 1 - \cos(\theta) \) creates a similar curve on a different orientation. These equations use trigonometric functions to dictate r's variances with \( \theta \).

• The first equation involves \( \sin(\theta) \), which peaks and dips across a cycle.
• The second equation uses \( \cos(\theta) \), shifting its curve differently.

Polar equations let us explore complex relationships in a straightforward manner.

Intersection of Graphs

Finding where two polar graphs intersect is crucial in understanding how these curves relate spatially. To find their intersections, the values of \( r \) given by both equations must be equal for the same value of \( \theta \).
We set the two equations equal to each other: \( 1 + \sin(\theta) = 1 - \cos(\theta) \). This simplifies to \( \sin(\theta) + \cos(\theta) = 0 \), our intersection condition.
Solving leads us to valid points of intersection, which tells us where the shapes cross.

• These points reveal telling features about overlap and symmetry.
• We've found an intersection point where \( \theta = \frac{3\pi}{4} \).

This critical step shows the power of equations in revealing the geometrical intersections.

Trigonometric Identities

Trigonometric identities simplify the equations at hand, making complex expressions more manageable. In our problem, we used \( \sin(\theta) + \cos(\theta) = 0 \), leading us to critical angle solutions.
The identity links two prominent trigonometric functions, helping us combine and reduce mathematical expressions. Trig identities reveal inherent relationships between angles and help solve angular equations efficiently.
Consider how \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \). Here, this correspondence is revealed in the transition from \( \sin(\theta) = -\cos(\theta) \) to \( \tan(\theta) = -1 \).

• Using tangent, we derive \( \theta \) values signaling graph intersections.
• Identities deepen our understanding of trigonometric function behavior.

Employing these identities is instrumental in resolving angle-based problems.

Angle Solutions

Solving for \( \theta \) in polar intersections involves finding specific angle solutions where functions equate. In our problem, \( \tan(\theta) = -1 \) gives \( \theta = \frac{3\pi}{4} + n\pi \), which describes a set of angles where the graphs meet.
Importantly, angle solutions often repeat at consistent intervals, indicated by \( n\pi \), a cyclic behavior intrinsic to many trigonometric problems.
We check specific solutions like \( \theta = \frac{3\pi}{4} \) and verify both equations yield equal values for r, confirming intersecting points.

• Other solutions might not lie within desired ranges, necessitating specific verifications.
• Each angle solution provides possible points to explore intersections graphically.

Solving these angles offers clearer insights into the graph's intersecting behaviors.