Question

Find the values of \(\theta\) in the given interval where the graph of the polar function has horizontal and vertical tangent lines. \(r=\cos (2 \theta) ; \quad[0,2 \pi]\)

Short answer

Horizontal tangents occur at \(\theta = \frac{ ",

Step-by-step solution

Understand Horizontal and Vertical Tangents

A curve in polar coordinates has a horizontal tangent when \( \frac{dy}{d\theta} = 0 \) and has a vertical tangent when \( \frac{dx}{d\theta} = 0 \). For polar coordinates, \( r = f(\theta) \), \( x = r\cos\theta \) and \( y = r\sin\theta \).

Find Expressions for dx/dθ and dy/dθ

First, find \( x \) and \( y \) in terms of \( \theta \):\[ x = r \cos \theta = \cos(2\theta) \cos \theta \]\[ y = r \sin \theta = \cos(2\theta) \sin \theta \]Next, perform the derivatives using the product rule:\[ \frac{dx}{d\theta} = \frac{d}{d\theta}(\cos(2\theta) \cos \theta) = -2\sin(2\theta) \cos \theta - \cos(2\theta) \sin \theta \]\[ \frac{dy}{d\theta} = \frac{d}{d\theta}(\cos(2\theta) \sin \theta) = -2\sin(2\theta) \sin \theta + \cos(2\theta) \cos \theta \]

Find Horizontal Tangents

Set \( \frac{dy}{d\theta} = 0 \):\[ -2\sin(2\theta) \sin \theta + \cos(2\theta) \cos \theta = 0 \]This can be factored to give:\[ \cos(2\theta) \cos \theta = 2\sin(2\theta) \sin \theta \]Using the identity \( \sin(2A) = 2\sin(A)\cos(A) \), this simplifies to:\[ \tan(\theta) = \frac{1}{2}\tan(2\theta) \]Solve this equation for \( \theta \) in the interval \([0, 2\pi]\).

Solve for Vertical Tangents

Set \( \frac{dx}{d\theta} = 0 \):\[ -2\sin(2\theta) \cos \theta - \cos(2\theta) \sin \theta = 0 \]Similarly solve:\[ \sin(2\theta) \cos \theta = \frac{1}{2} \cos(2\theta) \sin \theta \]Divide through by \( \cos \theta \sin \theta \) (assuming \( \theta eq k\frac{\pi}{2} \)):\[ \tan(2\theta) = 2\tan(\theta) \]Solve this equation for \( \theta \) in the interval \( [0, 2\pi] \).

Verify Solutions

Check solutions from Steps 3 and 4 to ensure they fall within the interval \([0, 2\pi]\). Consider extraneous solutions which might arise during simplification, particularly via division by zero or trigonometric identity application.

Key concepts

Horizontal Tangent

When examining polar graphs, understanding where horizontal tangents occur is crucial. A horizontal tangent happens when the rate of change of the y-coordinate with respect to the angle \( \theta \), denoted as \( \frac{dy}{d\theta} \), equals zero.
In the context of the given exercise, you find a horizontal tangent by setting the derivative \( \frac{dy}{d\theta} \) to zero and solving for \( \theta \). For the function \( r = \cos(2\theta) \), we substitute into the formulas for polar coordinates:

• \( x = r \cos \theta = \cos(2\theta) \cos \theta \)
• \( y = r \sin \theta = \cos(2\theta) \sin \theta \)

Then, differentiate \( y \) with respect to \( \theta \), set it to zero, and solve for \( \theta \). This process helps identify the points where the curve flattens horizontally.

Vertical Tangent

A vertical tangent occurs when the rate of change of the x-coordinate with respect to \( \theta \), noted as \( \frac{dx}{d\theta} \), becomes zero. This scenario is integral for understanding parts of the polar graph where no slope in the x-direction is observed.
In our case, to locate vertical tangents for \( r = \cos(2\theta) \), you set \( \frac{dx}{d\theta} \) to zero and solve for \( \theta \). The expressions derived during differentiation using both product and chain rules will guide you. This step ensures you find the \( \theta \) values correlating to the peaks or valleys in the graph, where direction shifts vertically.

Derivative in Polar Coordinates

Derivatives in polar coordinates help describe how the coordinates change as the angle \( \theta \) varies. For a curve in polar form, \( x \) and \( y\) depend on both \( r \) and \( \theta \). Understanding derivatives in this context is essential for finding tangents as shown in the exercise.
Use the following relationships:

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

The derivatives \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \) are calculated using product and chain rules, factoring in changes in both \( r \) and \( \theta \). Through calculation, you identify values of \( \theta \) that meet the conditions for horizontal and vertical tangents, crafting a more complete picture of the function's graph.

Trigonometric Identities

Trigonometric identities serve as tools simplifying complex expressions within polar coordinates, especially when looking for tangent lines. They allow rewriting and solving equations that result from differentiating expressions.
One key identity from the exercise is \( \sin(2A) = 2\sin(A)\cos(A) \). Using this, expressions like \( \tan(2\theta) \) can be rewritten to aid in solving equations. These identities reduce algebraic complexity, turning derivatives into simpler forms as seen in the transformations used in finding tangents. Remember to leverage these identities; they aren't just tricks but essential methods that create pathways to solving these coordinate problems efficiently.