Problem 37 - Calculus Early Transcendentals

Question

\(37-42\) Find all points of intersection of the given curves. $$ r=1+\sin \theta, \quad r=3 \sin \theta $$

Step-by-Step Solution

Answer

The curves intersect at points \( \left( \frac{3\sqrt{3}}{4}, \frac{3}{4} \right) \) and \( \left( -\frac{3\sqrt{3}}{4}, \frac{3}{4} \right) \).

Step 1: Convert to Cartesian Coordinates

First, we need to express both polar equations in Cartesian coordinates to make finding the intersection points easier. The conversion formulas we need are:\[ x = r \cos \theta \]\[ y = r \sin \theta \]For \( r = 1 + \sin \theta \):\[ x = (1 + \sin \theta) \cos \theta \]\[ y = (1 + \sin \theta) \sin \theta = \sin \theta + \sin^2 \theta \]For \( r = 3 \sin \theta \):\[ x = 3 \sin \theta \cos \theta \]\[ y = 3 \sin^2 \theta \]These represent the same curves in terms of \(x\) and \(y\) coordinates.

Step 2: Set Equations Equal

To find the points of intersection, set the right sides of the equations equal to each other.Set \( r \) values equal: \[ 1 + \sin \theta = 3 \sin \theta \] Simplify this equation to:\[ 1 = 2 \sin \theta \]\[ \sin \theta = \frac{1}{2} \]

Step 3: Solve for the values of θ

Solve \( \sin \theta = \frac{1}{2} \). The angles that satisfy this equation in the interval \([0, 2\pi]\) are:\[ \theta = \frac{\pi}{6}, \frac{5\pi}{6} \].

Step 4: Find r for each θ

Substitute back into one of the original equations to find \(r\) for each value of \(\theta\). Using \( r = 1 + \sin \theta \):For \( \theta = \frac{\pi}{6} \):\[ r = 1 + \frac{1}{2} = \frac{3}{2} \]For \( \theta = \frac{5\pi}{6} \):\[ r = 1 + \frac{1}{2} = \frac{3}{2} \]

Step 5: Convert Points to Cartesian Coordinates

Finally, convert the polar coordinates back to Cartesian coordinates to find the intersection points:For \( \theta = \frac{\pi}{6}, r = \frac{3}{2} \):\[ x = r \cos \theta = \frac{3}{2} \cos \frac{\pi}{6} = \frac{3}{2} \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{4} \]\[ y = r \sin \theta = \frac{3}{2} \sin \frac{\pi}{6} = \frac{3}{2} \cdot \frac{1}{2} = \frac{3}{4} \]For \( \theta = \frac{5\pi}{6}, r = \frac{3}{2} \):\[ x = r \cos \theta = \frac{3}{2} \cos \frac{5\pi}{6} = \frac{3}{2} \cdot -\frac{\sqrt{3}}{2} = - \frac{3\sqrt{3}}{4} \]\[ y = r \sin \theta = \frac{3}{2} \sin \frac{5\pi}{6} = \frac{3}{2} \cdot \frac{1}{2} = \frac{3}{4} \]

Step 6: Determine Points of Intersection

The points of intersection between the curves, given the calculations above, are:1. \( \left( \frac{3\sqrt{3}}{4}, \frac{3}{4} \right) \)2. \( \left( -\frac{3\sqrt{3}}{4}, \frac{3}{4} \right) \).

Key Concepts

Cartesian coordinates

Cartesian coordinates represent a point using two values: \((x, y)\). These coordinates are named after René Descartes, a French philosopher and mathematician. The Cartesian system uses a grid to plot points based on their horizontal and vertical positions. The \(x\) coordinate shows how far along the point is on the horizontal axis, while the \(y\) coordinate indicates its vertical position.

In the context of solving intersections in polar coordinates, converting them to Cartesian coordinates can simplify the calculations. By using the equations \(x = r \cos \theta\) and \(y = r \sin \theta\), we transform polar equations into the familiar \(x, y\) format, making it easier to determine where curves intersect.

intersection points

Intersection points occur where two curves cross each other. To find these points, we need to evaluate both curves and determine the coordinates where they meet.

In polar coordinates, intersection points can be found by setting the radius \(r\) expressions of the two curves equal to one another. This equates the distances from the origin to the curves and helps to solve for the angle \(\theta\) at these points of intersection. Once we know \(\theta\), we substitute it back into either of the original radius equations to find the corresponding \(r\).

• Determine \(\theta\) by equating the two \(r\) expressions.
• Use the resulting angle(s) to find corresponding \(r\) values.
• Convert the polar coordinates \((r, \theta)\) into Cartesian \((x, y)\) for each point.

conversion formulas

Converting between coordinate systems is a crucial skill in mathematics, specifically when working with polar and Cartesian systems. Conversion formulas allow us to switch between these systems smoothly.

For polar to Cartesian conversions, we use:

• \(x = r \cos \theta\)
• \(y = r \sin \theta\)

These equations use the radius \(r\)—the distance from the origin—and the angle \(\theta\)—the angle from the positive x-axis—to determine the Cartesian \(x\) and \(y\) values of a point.

Converting the equations between these systems helps to facilitate easier analysis, especially when solving for intersections or graphing complex shapes.

polar equation

A polar equation is expressed in terms of the radius \(r\) and the angle \(\theta\). This type of equation is particularly useful in scenarios where rotational symmetry is present or when describing curves like spirals and circles.

In polar coordinates, the value \(r\) represents how far the point is from the origin, and \(\theta\) is the angle measured counterclockwise from the positive x-axis. Polar equations like \(r = 1 + \sin \theta\) or \(r = 3 \sin \theta\) describe specific curves in polar space.

• These curves are plotted based on changing values of \(\theta\).
• Such equations often provide a more suitable representation for certain shapes, compared to Cartesian equations.
• Analyzing curves in polar form can unveil unique properties of symmetry and shape.

Understanding polar equations is key to grasping complex geometric concepts and their intersections.