Problem 17 - Thomas Calculus

Question

Find the slopes of the curves in Exercises \(17-20\) at the given points. Sketch the curves along with their tangents at these points. Cardioid \(\quad r=-1+\cos \theta ; \quad \theta=\pm \pi / 2\)

Step-by-Step Solution

Answer

The slopes at \( \theta = \pm \pi/2 \) are vertical, indicating vertical tangents at these points.

Step 1: Parametrize the Cardioid

The cardioid is given in polar coordinates as \( r = -1 + \cos \theta \). To work with this in Cartesian coordinates, let's express it where \( x = r \cos \theta \) and \( y = r \sin \theta \). Begin by finding these in terms of \( \theta \).

Step 2: Convert to Cartesian Coordinates

Substitute \( r = -1 + \cos \theta \) into the equations for \( x \) and \( y \): \( x = (-1 + \cos \theta) \cos \theta \) \( y = (-1 + \cos \theta) \sin \theta \).

Step 3: Differentiate to Find Slopes

To find the slopes at given points, we need \( \frac{dy}{dx} \). Use the derivative chain rules: \( \frac{dx}{d\theta} = \frac{d}{d\theta}((-1 + \cos \theta) \cos \theta) \)\( \frac{dy}{d\theta} = \frac{d}{d\theta}((-1 + \cos \theta) \sin \theta) \).

Step 4: Evaluate Derivatives

Calculate the derivatives:\( \frac{dx}{d\theta} = \sin(2\theta) - \cos^2\theta \) \( \frac{dy}{d\theta} = \cos\theta - \cos(2\theta) \).

Step 5: Compute \( \frac{dy}{dx} \)

Using \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \), calculate the slope at \( \theta = \frac{\pi}{2} \) and \( \theta = -\frac{\pi}{2} \). For both values of \( \theta \), find:At \( \theta = \pm \frac{\pi}{2} \), \( \frac{dy}{dx} = \pm \infty \) (vertical tangent).

Step 6: Sketch the Cardioid

Draw the cardioid curve using its polar equation. Mark the points corresponding to \( \theta = \pm \frac{\pi}{2} \) and draw vertical tangent lines as these points have vertical tangents.

Key Concepts

Understanding the Cardioid

A cardioid is a heart-shaped curve that you may encounter in calculus. It typically appears in polar coordinates with an equation like \( r = -1 + \cos \theta \). Knowing how to express it in different coordinate systems is crucial.
The cardioid is an example of an epicycloid with one cusp, and its shape stems from altering traditional polar equations. In polar coordinates, \( r \) denotes the radial distance from the origin, and \( \theta \) is the angle from the positive x-axis.
Cardioids are valuable in different fields like physics and engineering due to their unique properties, including their symmetry and reflective properties.

Converting Polar Coordinates to Cartesian Coordinates

Once you have a polar function like \( r = -1 + \cos \theta \), converting it to Cartesian coordinates \((x, y)\) simplifies calculating derivatives. The conversion requires using:

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

By substituting \( r \), the equations become:

• \( x = (-1 + \cos \theta) \cos \theta \)
• \( y = (-1 + \cos \theta) \sin \theta \)

This transformation allows for easier manipulation using standard Cartesian calculus techniques.

Derivative Calculation to Determine Slopes

To find the slope of a curve at a specific point, derivative calculations are essential. When converting geometry from polar to Cartesian, knowing what \( \frac{dy}{dx} \) represents is key to understanding how the curve moves.
Instead of differentiating \( y \) concerning \( x \) directly, use the chain rule through their \( \theta \) parameter:

• \( \frac{dx}{d\theta} = \sin(2\theta) - \cos^2\theta \)
• \( \frac{dy}{d\theta} = \cos\theta - \cos(2\theta) \)

The slope of the tangent line \( \frac{dy}{dx} \) is then the ratio of these derivatives:
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \]This gives a practical method to find the slope, circumventing direct Cartesian differentiation issues.

Finding the Slope of the Tangent Line

The slope of a tangent line is critical in understanding how the curve behaves at specific points, such as where it intersects axes or reaches a cusp.
For a cardioid given by \( r = -1 + \cos \theta \), solving \( \frac{dy}{dx} \) at \( \theta = \pm \frac{\pi}{2} \) demonstrates vertical tangents. Here, both points correspond to:\
\[ \frac{dy}{dx} = \pm \infty \]
This signifies that the tangent lines are vertical at these points. Recognizing vertical tangents is crucial for sketching the curvatures accurately and understanding its geometrical properties.