Problem 41 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(F\) and \(F^{\prime}\) be fixed points with polar coordinates \((a, 0)\) and \((-a, 0)\), respectively. Show that the set of points \(P\) satisfying \(|P F|\left|P F^{\prime}\right|=a^{2}\) is a lemniscate by finding its polar equation.

Step-by-Step Solution

Answer

The polar equation is \(r^2 = 2a^2 \cos^2 \theta\).

Step 1: Understanding the Problem

We are asked to find the polar equation of a lemniscate for which the product of the distances from any point \(P\) to two fixed points \(F\) and \(F'\) is constant and equal to \(a^2\). The coordinates of \(F\) and \(F'\) are \((a, 0)\) and \((-a, 0)\), respectively.

Step 2: Write the Distance in Polar Form

For a point \(P\) with polar coordinates \((r, \theta)\), the distance from \(P\) to \(F = (a, 0)\) is \(\sqrt{r^2 - 2ar \cos \theta + a^2}\). Similarly, the distance from \(P\) to \(F' = (-a, 0)\) is \(\sqrt{r^2 + 2ar \cos \theta + a^2}\).

Step 3: Set Up the Equation

According to the problem, \(|PF| \times |PF'| = a^2\). Substitute the expressions for \(|PF|\) and \(|PF'|\) into this equation:\[\sqrt{r^2 - 2ar \cos \theta + a^2} \times \sqrt{r^2 + 2ar \cos \theta + a^2} = a^2\]

Step 4: Square Both Sides

Square both sides to eliminate the square roots:\[(r^2 - 2ar \cos \theta + a^2)(r^2 + 2ar \cos \theta + a^2) = a^4\]

Step 5: Expand and Simplify

Expand the left-hand side:\[ r^4 + a^4 + 2r^2a^2 - 4a^2r^2 \cos^2 \theta = a^4 \]Combine like terms:\[ r^4 - 2a^2r^2 \cos^2 \theta = 0 \]

Step 6: Factor the Equation

Factor the expression by recognizing it as a difference of squares:\[ r^2(r^2 - 2a^2 \cos^2 \theta) = 0 \]

Step 7: Solve for Polar Equation

Since \(r = 0\) corresponds to the origin, the non-trivial solution here is:\[ r^2 = 2a^2 \cos^2 \theta \]This is the polar equation of the lemniscate.

Key Concepts

Polar Coordinates

Polar coordinates are a way of describing the position of a point in a plane. Instead of using the usual x and y coordinates, polar coordinates use a radius and an angle. The radius, often denoted as \(r\), is the distance from the origin to the point, while the angle, \(\theta\), measures the counterclockwise angle from the positive x-axis to the line segment connecting the origin to the point.

• A point in polar coordinates is written as \((r, \theta)\).
• The angle \(\theta\) is usually given in radians.
• If the point is located on the positive x-axis, \(\theta = 0\).

For example, the point \((3, \pi/4)\) means the point is 3 units away from the origin, located at an angle of \(\pi/4\) radians from the positive x-axis. Polar coordinates are especially useful in scenarios where there's symmetry around a central point, like circles or shapes such as the lemniscate.

Distance Formula

The distance formula in polar coordinates helps us calculate the distance between points when their position is specified using polar coordinates. If you have a fixed point \(F = (a, 0)\) and a movable point \(P = (r, \theta)\), we can determine the distance \(|PF|\) using the formula:

• \( |PF| = \sqrt{r^2 - 2ar \cos \theta + a^2} \)
• This formula is derived using the law of cosines, which connects distances and angles in triangles.

The idea is similar when you compute distance to the point \(-a, 0\):

• \( |PF'| = \sqrt{r^2 + 2ar \cos \theta + a^2} \)
• Remember, these expressions simplify our understanding of how \(P\) relates to \(F\) and \(F'\).

These distances help us express more complex relationships and conditions in problems like finding the equations of certain curves.

Polar Equation

A polar equation represents a curve on a plane using polar coordinates. In our exercise, the main goal was to find the polar equation of a lemniscate, defined uniquely by the condition that the product of distances from two focal points remains constant.

• For the lemniscate: \( |PF| \times |PF'| = a^2\).
• This led to the polar equation \( r^2 = 2a^2 \cos^2 \theta\).

Polar equations are instrumental in understanding more than linear relationships—they reveal how varied forms like curves and spirals develop.

• The polar equation \( r^2 = 2a^2 \cos^2 \theta \) highlights the symmetry and unique loop shapes of lemniscates.
• This insight shows us the power of polar forms to describe more intricate shapes.

Trigonometric Identity

Trigonometric identities simplify expressions, especially when we're working with polar equations that involve angles. In our specific equation, \(\cos^2 \theta\) played a fundamental part.

• Identities like \(\sin^2 \theta + \cos^2 \theta = 1\) are foundational tools.
• They allow us to transform and simplify equations efficiently.

By recognizing and applying these identities, expressions can often be refactored to more manageable forms, crucial in solving for specific equations like our lemniscate example.

• Using \(\cos\theta\) is key because it directly aligned with our exercise's symmetric property.
• Mastering these identities prepares you for deeper exploration into wave functions, oscillations, and many more in calculus and beyond.