Problem 64 - Calculus Early Transcendentals

Question

\(63-68\) Find the points on the given curve where the tangent line is horizontal or vertical. $$r=1-\sin \theta$$

Step-by-Step Solution

Answer

The points are \((0,0)\) for both horizontal and vertical tangents, and \((0,-2)\) for a horizontal tangent.

Step 1: Understand Horizontal and Vertical Tangents

For a polar curve, a tangent is horizontal where \(\frac{dy}{d\theta} = 0\) and vertical where \(\frac{dx}{d\theta} = 0\). For the curve given by \(r = 1 - \sin\theta\), we need to convert it to Cartesian coordinates and then find where these conditions hold.

Step 2: Convert Polar to Cartesian Coordinates

The relationship between polar and Cartesian coordinates is \(x = r \cos \theta\) and \(y = r \sin \theta\). Substituting \(r = 1 - \sin\theta\), we get: \(x = (1 - \sin\theta) \cos \theta\) and \(y = (1 - \sin\theta) \sin \theta\).

Step 3: Compute Derivatives

Compute \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\) using the product and chain rules. This gives us:\[\frac{dx}{d\theta} = -(1-\sin\theta) \sin\theta + \cos\theta \cos\theta\] \[\frac{dy}{d\theta} = (1-\sin\theta) \cos\theta + \sin\theta \sin\theta\].

Step 4: Set \(\frac{dy}{d\theta} = 0\) for Horizontal Tangents

Solving \((1-\sin\theta) \cos\theta + \sin\theta \sin\theta = 0\), simplify to \(\cos\theta = 0\) or \(\theta = \frac{\pi}{2}, \frac{3\pi}{2}\). Calculate \(r\) for these \(\theta\) values to find corresponding points.

Step 5: Set \(\frac{dx}{d\theta} = 0\) for Vertical Tangents

Solving \(-(1-\sin\theta) \sin\theta + \cos^2\theta = 0\), simplify to \(\sin\theta = 1\), then \(\theta = \frac{\pi}{2}\). Calculate \(r\) to find the corresponding point.

Step 6: Calculate Points from \(r = 1 - \sin \theta\)

For \(\theta = \frac{\pi}{2}\), \(r = 1 - 1 = 0\), thus the point is \((0,0)\). For \(\theta = \frac{3\pi}{2}\), \(r = 1 - (-1) = 2\), thus the point is \((0,-2)\).

Key Concepts

Horizontal Tangents

In polar coordinates, a tangent line to a curve is horizontal when the rate of change of the y-coordinate with respect to the angle \( \theta \) is zero. Mathematically, this means we're looking for where \( \frac{dy}{d\theta} = 0 \).

• Identifying Horizontal Tangents: For the curve \( r = 1 - \sin \theta \), the y-coordinate in Cartesian form is \( y = (1 - \sin \theta) \sin \theta \). Using the chain and product rules of derivatives, we find \( \frac{dy}{d\theta} \).
• Solving: Set \( (1-\sin\theta) \cos\theta + \sin^2\theta = 0 \). Simplifying, we find \( \cos\theta = 0 \), leading to the values \( \theta = \frac{\pi}{2} \text{ and } \theta = \frac{3\pi}{2} \).
• Corresponding Points: Calculate \( r \) using these \( \theta \) values:
  • For \( \theta = \frac{\pi}{2} \), \( r = 1 - 1 = 0 \), resulting in the point \((0,0)\).
  • For \( \theta = \frac{3\pi}{2} \), \( r = 1 - (-1) = 2 \), resulting in the point \((0,-2)\).

Horizontal tangents indicate flat spots on the curve, where the direction changes but the height remains constant momentarily.

Vertical Tangents

For a tangent line to be vertical, the sensitivity of the x-coordinate to changes in \( \theta \) should be zero. This corresponds to \( \frac{dx}{d\theta} = 0 \) for a vertical tangent.

• Identifying Vertical Tangents: The x-coordinate equation is \( x = (1 - \sin\theta) \cos\theta \). Using the derivatives, we find \( \frac{dx}{d\theta} \).
• Solving: We solve \( -(1-\sin\theta) \sin\theta + \cos^2\theta = 0 \). Simplifying, we set \( \sin\theta = 1 \), leading to \( \theta = \frac{\pi}{2} \).
• Corresponding Point: For \( \theta = \frac{\pi}{2} \), \( r = 1 - 1 = 0 \), giving the point \((0,0)\).

Vertical tangents represent points where the curve reaches an instantaneous vertical movement, without horizontal displacement around that point.

Cartesian Coordinates

Converting from polar coordinates to Cartesian coordinates is crucial for understanding and solving problems related to tangents. Polar coordinates \( (r, \theta) \) express a point in terms of the distance from the origin and the angle from the positive x-axis.

• Conversion Formulas: The relationships are given by:
  • \( x = r \cos \theta \)
  • \( y = r \sin \theta \)
• Application: For the specific curve \( r = 1 - \sin\theta \):
  • We substitute the value for \( r \) into the equations above to convert polar to Cartesian:
    • \( x = (1 - \sin \theta) \cos \theta \)
    • \( y = (1 - \sin \theta) \sin \theta \)

Switching to Cartesian coordinates can simplify some operations, especially when handling derivatives and finding tangents.

Derivatives

Derivatives are tools that help us determine how a function changes at any given point, which in turn helps us understand the geometry of curves.

• Role of Derivatives in Tangents:
  • For tangents, we determine the slopes along the x and y directions using derivatives \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \).
  • Horizontal tangents occur where \( \frac{dy}{d\theta} = 0 \).
  • Vertical tangents occur where \( \frac{dx}{d\theta} = 0 \).
• Calculating Derivatives:
  • For \( x = (1 - \sin \theta) \cos \theta \), apply the product and chain rules to find \( \frac{dx}{d\theta} \).
  • For \( y = (1 - \sin \theta) \sin \theta \), apply the same differentiation techniques to find \( \frac{dy}{d\theta} \).

Derivatives provide the precise slopes needed to identify critical points like horizontal and vertical tangents on the curve.