Problem 39 - Thomas Calculus

Question

Find the points of intersection of the pairs of curves in Exercises \(39-42\) $$ r^{2}=\sin 2 \theta, \quad r^{2}=\cos 2 \theta $$

Step-by-Step Solution

Answer

The curves intersect at several points determined by \(\theta = \frac{\pi}{8} + n\frac{\pi}{2}\) with \(r = \pm \sqrt{\frac{\sqrt{2}}{2}}\).

Step 1: Equate the Curves

Since both equations involve \(r^2\), we equate them to find the intersections: \(\sin 2\theta = \cos 2\theta\).

Step 2: Solve the Angle Equation

We know that \(\sin 2\theta = \cos 2\theta\) can be rewritten as \(\tan 2\theta = 1\). This gives \(2\theta = \frac{\pi}{4} + n\pi\), where \(n\) is an integer.

Step 3: Solve for \(\theta\)

Divide the equation from Step 2 by 2: \(\theta = \frac{\pi}{8} + n\frac{\pi}{2}\). This gives the possible values of \(\theta\).

Step 4: Calculate \(r\) for Each \(\theta\)

Substitute \(\theta\) back into one of the original equations, say \(r^2 = \sin 2\theta\), to find \(r\). For \(\theta = \frac{\pi}{8}\) (and similar values), calculate \(r^2 = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\). So \(r = \pm \sqrt{\frac{\sqrt{2}}{2}}\).

Step 5: List Points of Intersection

The points of intersection are \(\left(r,\theta\right)\) for each solution. For \(\theta = \frac{\pi}{8}\), \(r = \pm \sqrt{\frac{\sqrt{2}}{2}} \). Repeat this process for each \(\theta = \frac{\pi}{8} + n\frac{\pi}{2}\) to list all intersection points.

Key Concepts

Curve Intersection

When studying polar coordinates, curve intersection involves finding common points where two curves meet in a polar plane. We achieve this by identifying polar coordinates \( (r, \theta) \) that satisfy both equations. In our case, we have \( r^2 = \sin 2\theta \) and \( r^2 = \cos 2\theta \). By equating these expressions, \( \sin 2\theta = \cos 2\theta \), we find conditions where the sine and cosine are equal.
This process simplifies our work to solving a trigonometric equation in terms of \( \theta \). Once we find \( \theta \), we substitute it back into one of the equations to solve for the radial distance \( r \).
The intersection points are of the form \( (r, \theta) \), and these coordinates can be plotted in a polar graph to visualize the curve intersections.

Trigonometric Equations

Trigonometric equations play a crucial role in solving problems involving polar coordinates. In the exercise, after equating \( \sin 2\theta = \cos 2\theta \), it can be transformed using trigonometric identities. For instance, can rewrite the equation as \( \tan 2\theta = 1 \). This step is crucial in simplifying our problem into a more manageable form.
We know from standard trigonometry that \( \tan \alpha = 1 \) at angles like \( \alpha = \frac{\pi}{4}, \frac{5\pi}{4}, \) and so on. Hence, we find the general solutions for \( 2\theta \) which are \( \frac{\pi}{4} + n\pi \) where \( n \) is an integer. Finally, dividing by two gives us the specific values for \( \theta \).
These solutions guide us to calculate the precise angles \( \theta \) where both curve equations are met, leading us to our points of intersection.

Analytic Geometry

Analytic geometry connects algebra and geometry through equations and graphs. In polar coordinates, instead of the \'x\' and \'y\' we are accustomed to from Cartesian coordinates, we work with \'r\' and \( \theta \).
Using polar equations, like our exercise given by \( r^2 = \sin 2\theta \) and \( r^2 = \cos 2\theta \), we navigate through polar systems' complexities to find intersection points. \( r \) represents the radial distance from the origin to a point on the plane, while \( \theta \) indicates the angle from the positive x-axis.
This approach involves converting algebraic solutions into geometric interpretations. Solving for \( r \) and \( \theta \) helps us in sketching and understanding beautiful and often intricate polar curves and their intersections. Hence, this exercise not only strengthens our algebraic skills but also enhances our ability to visualize and interpret geometrical phenomena.