Problem 252 - Calculus Volume 3

Question

Find the points at which the following polar curves have a horizontal or vertical tangent line. $$ r=2 \sin (2 \theta) $$

Step-by-Step Solution

Answer

Determine \(\theta\) for zero derivatives, then find points using \(r=2\sin(2\theta)\).

Step 1: Convert the Polar Equation to Cartesian Coordinates

In polar coordinates, a point is given as \((r, \theta)\). To find tangents, convert the polar equation \(r=2\sin(2\theta)\) into Cartesian coordinates using the relations \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\).

Step 2: Differentiate the Cartesian Equations

Substitute the expressions for \(x\) and \(y\) in terms of \(r\) and \(\theta\). Then, differentiate these equations with respect to \(\theta\) to find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\).

Step 3: Find Conditions for Horizontal Tangents

Horizontal tangents occur when \(\frac{dy}{d\theta} = 0\). Solve for \(\theta\) when the derivative \(\frac{dy}{d\theta}\) is equal to zero, ignoring any values where \(\frac{dx}{d\theta}\) is also zero as those would be points of indefiniteness.

Step 4: Find Conditions for Vertical Tangents

Vertical tangents occur when \(\frac{dx}{d\theta} = 0\). Solve for \(\theta\) when the derivative \(\frac{dx}{d\theta}\) is equal to zero, ignoring any values where \(\frac{dy}{d\theta}\) is also zero.

Step 5: Calcuate Points of Tangents

After finding values of \(\theta\) for both horizontal and vertical tangents, substitute them back into the original polar equation to get the points in polar form \((r, \theta)\).

Key Concepts

Horizontal Tangents

In polar coordinates, a horizontal tangent occurs where the slope of the tangent line is zero. For polar curves represented by the functions \(r(\theta)\), we need to find where the derivative of the radial component with respect to \(\theta\) results in the angle-dependent y-component of the curve being momentarily stationary.

To find horizontal tangents, analyze the polar function by converting it into Cartesian form and then differentiate the resulting y-component, \(y = r(\theta) \sin(\theta)\), with respect to \(\theta\). Set \(\frac{dy}{d\theta} = 0\) and solve for \(\theta\). These values, when substituted into the polar equation, give the angles at which the polar curve has horizontal tangents.

Remember that these tangents occur when the rate of change of the y-value per unit change in angle \(\theta\) becomes zero, marking points of parallel alignment to the x-axis.

Vertical Tangents

Vertical tangents in polar coordinates are points where the slope of the tangent line becomes infinite. At these points, the x-component is momentarily stationary while the y-component could still be changing, leading to a tangent line that is vertical.

To find these vertical tangents, convert your polar equation into Cartesian coordinates and differentiate the x-component, \(x = r(\theta) \cos(\theta)\), with respect to \(\theta\). Investigate where \(\frac{dx}{d\theta} = 0\). Solving this equation will reveal the critical values of \(\theta\) where the curve has a vertical tangent.

Keep in mind that truly vertical slopes occur with undefined values of slope \( \left( \frac{dy}{dx} \right) \), emphasizing purely x-slope dependency without y-value changes at that exact moment.

Polar to Cartesian Conversion

Converting from polar equations to Cartesian (rectangular) coordinates is essential in analyzing curves for tangents and other geometric properties. The fundamental conversion formulas are \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\).

When needing to convert a polar equation such as \(r = 2 \sin(2\theta)\), these formulas allow a transformation into terms of x and y, essentially mapping circular-based coordinates onto the more familiar grid-like structure.

• Multiply the given expression for \(r\) by both \(\cos(\theta)\) and \(\sin(\theta)\) to get two Cartesian equations.
• These conversions will yield new expressions for \(x\) and \(y\) in terms of polar data, aiding in further derivative operations.

This step drastically simplifies differentiation, making it easier to analyze such aspects as tangent lines and curve behavior in a plane.

Differentiation in Polar Coordinates

Differentiation in polar coordinates involves calculating derivatives of expressions \(x(\theta)\) and \(y(\theta)\) obtained from the original polar equation. After conversion, these expressions are differentiated with respect to \(\theta\) to find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\).

These derivatives are crucial in determining intervals and angles where unique features like tangents appear. The process involves applying standard differentiation rules as if the polar functions were regular parametric equations.

• Compute \(\frac{dy}{d\theta}\) for analyzing changes in the vertical component.
• Compute \(\frac{dx}{d\theta}\) for insights into horizontal movements needed to identify vertical tangents.

The differential relationships between these converted forms facilitate finding significant points of the curve, clearly epitomized by tangent calculations.