Problem 253 - Calculus Volume 3

Question

Find the points at which the following polar curves have a horizontal or vertical tangent line. $$ \text { The cardioid } r=1+\sin \theta $$

Step-by-Step Solution

Answer

Horizontal tangents at \( \theta = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6}\). Vertical tangents require further solving of \( \cos^2 \theta = \sin\theta + \sin^2\theta \).

Step 1: Understand the Problem

The problem asks to find where the polar curve \( r = 1 + \sin\theta \) has horizontal or vertical tangents. In polar coordinates, this involves differentiating \( r \) with respect to \( \theta \).

Step 2: Differentiate the Polar Equation

To find tangent lines in polar coordinates, first express \( x \) and \( y \) in terms of \( r \) and \( \theta \): \( x = r \cos \theta \) and \( y = r \sin \theta \). Therefore, differentiate both expressions with respect to \( \theta \).

Step 3: Find the Derivatives

The derivatives are found using the product rule: \( \frac{dx}{d\theta} = \frac{dr}{d\theta} \cos \theta - r \sin \theta \) and \( \frac{dy}{d\theta} = \frac{dr}{d\theta} \sin \theta + r \cos \theta \). Here, \( \frac{dr}{d\theta} = \cos\theta \) because \( r = 1 + \sin \theta \).

Step 4: Determine Critical Condition for Tangent Lines

A horizontal tangent occurs when \( \frac{dy}{d\theta} = 0 \) and a vertical tangent occurs when \( \frac{dx}{d\theta} = 0 \).

Step 5: Solve for Horizontal Tangents

\( \frac{dy}{d\theta} = \cos\theta \sin\theta + (1 + \sin\theta)\cos \theta = \cos \theta (\sin\theta + 1 + \sin\theta) = \cos\theta (1 + 2\sin\theta) \). Set this equal to zero: \( \cos\theta (1 + 2\sin\theta) = 0 \) which gives \( \cos\theta = 0 \) or \( 1 + 2\sin\theta = 0 \).

Step 6: Solve for Vertical Tangents

\( \frac{dx}{d\theta} = \cos\theta \cos\theta - (1 + \sin\theta)\sin \theta = \cos^2\theta - \sin\theta (1+ \sin\theta) \). Set equal to zero: \( \cos^2\theta = \sin\theta (1+ \sin\theta) \).

Step 7: Find Specific Theta Values

For horizontal tangents: \( \cos\theta = 0 \) gives \( \theta = \frac{\pi}{2}, \frac{3\pi}{2} \). For \( 1 + 2\sin\theta = 0 \), solve \( \sin\theta = -\frac{1}{2} \) giving \( \theta = \frac{7\pi}{6}, \frac{11\pi}{6} \).For vertical tangents: Solve \( \cos^2\theta = \sin\theta + \sin^2\theta \). This equation can be more simply solved by substituting possible angles to meet symmetry and the known values from the cosine and sine identities.

Step 8: Result Interpretation

The values from Step 7 are used to determine the exact points in polar coordinates by substituting \( \theta \) values back into the equation \( r = 1 + \sin \theta \), ensuring symmetry across axes.

Key Concepts

Horizontal Tangent Lines

In polar coordinates, finding horizontal tangent lines involves considering the changes in the vertical direction, which is linked to the derivative \( \frac{dy}{d\theta} \). For the cardioid given by the equation \( r = 1 + \sin \theta \), we look for the points where \( \frac{dy}{d\theta} = 0 \). By expressing the position in terms of both \( r \) and \( \theta \) as \( x = r \cos \theta \) and \( y = r \sin \theta \), we use the product rule to compute the derivative.

• The expression for \( \frac{dy}{d\theta} \) becomes \( \cos\theta (1 + 2\sin\theta) \).
• Horizontal tangent lines occur when \( \cos\theta = 0 \) which is true at \( \theta = \frac{\pi}{2} \) and \( \theta = \frac{3\pi}{2} \).
• Another possibility for a horizontal tangent is \( 1 + 2\sin\theta = 0 \), leading to \( \sin\theta = -\frac{1}{2} \), which further yields \( \theta = \frac{7\pi}{6}, \frac{11\pi}{6} \).

These \( \theta \) values give rise to certain radial points on the cardioid representing horizontal tangency.

Vertical Tangent Lines

Vertical tangent lines arise when the change in the horizontal direction vanishes, signified by \( \frac{dx}{d\theta} = 0 \). We similarly analyze this condition in the context of our cardioid \( r = 1 + \sin \theta \), adopting the polar positions for \( x \) and \( y \). Through differentiation, we get:

• The derivative \( \frac{dx}{d\theta} \) simplifies to \( \cos^2\theta - \sin\theta (1 + \sin\theta) \).
• This zeroing condition, \( \cos^2\theta = \sin\theta (1 + \sin\theta) \), requires substituting possible values of \( \theta \) matching known cosine and sine values.

Using trigonometric identities and symmetry, we identify potential angles where vertical tangents are achieved. These computations, checked against polar symmetry, ensure correct angles are picked for vertical tangent lines.

Cardioid

A cardioid is a distinctive heart-shaped polar curve produced when a circle rolls around another circle of the same radius. In mathematics, it often appears in the equation form \( r = 1 + \sin \theta \) or \( r = 1 + \cos \theta \). In exploring this specific curve:

• The cardioid has one loop, distinguished by its cusp at the origin, giving it its "heart" shape.
• This polar curve exhibits symmetry across the axis which simplifies finding points of tangency.
• The study of tangent lines enriches understanding of the curve's properties, such as continuity and smoothness except at the cusp.

Recognizing a cardioid's attributes allows students to anticipate its behavior; the same work will apply to similar polar curves, revealing broader applications.

Differentiation in Polar Coordinates

Differentiating in polar coordinates involves converting Cartesian functions \( x \) and \( y \) into radial forms \( r \) and angular forms \( \theta \). This is necessary for curves represented by polar equations like our cardioid \( r = 1 + \sin \theta \). Here’s what you need to know:

• Express \( x \) as \( r \cos \theta \) and \( y \) as \( r \sin \theta \) to reformulate the curve.
• Use the product rule to derive \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \), which are paramount for identifying tangents.
• Setting these derivatives to zero reveals conditions for horizontal and vertical tangent lines.

This process, though intricate, offers a robust method for analyzing polar curve characteristics and understanding the geometric meaning behind their derivatives.