Problem 17 - Calculus

Question

At what \(\theta\) 's does the cardioid \(r=1+\cos \theta\) have infinite slope? Which points are furthest to the left (minimum \(x\) )?

Step-by-Step Solution

Answer

Infinite slope occurs at \(\theta = \frac{\pi}{2}, \frac{3\pi}{2}, 0, \pi\). Minimum \(x\) at \(\theta = \pi\).

Step 1: Polar to Cartesian Conversion

The equation of the cardioid is given in polar coordinates as \( r = 1 + \cos \theta \). To find the points where the slope is infinite, we first convert this into Cartesian coordinates. The relations are \( x = r \cos \theta \) and \( y = r \sin \theta \). Thus, \( x = (1+\cos \theta) \cos \theta \) and \( y = (1+\cos \theta) \sin \theta \).

Step 2: Derive the Slope Expression

To find the slope \( \frac{dy}{dx} \), we need \( \frac{dy}{d\theta} \) and \( \frac{dx}{d\theta} \). We have:\( dx/d\theta = \frac{d}{d\theta}((1+\cos\theta)\cos\theta) \) and \( dy/d\theta = \frac{d}{d\theta}((1+\cos\theta)\sin\theta) \).Calculate these derivatives. - \( dx/d\theta = -\sin\theta\cos\theta + \cos\theta - \cos^2\theta \).- \( dy/d\theta = \sin^2\theta + \cos\theta\sin\theta \).

Step 3: Set Up Infinite Slope Condition

The slope is infinite when \( \frac{dy}{dx} = \infty \), which occurs when \( dx/d\theta = 0 \) while \( dy/d\theta eq 0 \). Therefore, solve \(-\sin\theta\cos\theta + \cos\theta - \cos^2\theta = 0 \) for \(\theta\).

Step 4: Solve for \( \theta \)

Solving \(-\sin\theta \cos\theta + \cos\theta - \cos^2\theta = 0\) gives: Factor out \(\cos\theta\): \( \cos\theta(- \sin\theta + 1 - \cos\theta) = 0 \). So, either \(\cos\theta = 0 \) or \(- \sin\theta + 1 - \cos\theta = 0 \). For \(\cos\theta = 0\), \( \theta = \frac{\pi}{2} \) or \( \theta = \frac{3\pi}{2} \). For \(- \sin\theta + 1 - \cos\theta = 0\), re-arrange to \(\sin\theta = 1 - \cos\theta\), which holds for \(\theta = 0\) and \(\theta = \pi\).

Step 5: Determine Points Furthest Left

From the solutions obtained, evaluate for which \(\theta\) the x-coordinate \( x = (1 + \cos\theta) \cos\theta = 0 \). This occurs when \(\theta = \pi\) as previously \(\cos\theta = -1\) gives the minimum \(x = 0\).

Step 6: Conclusion

The \(\theta\)'s where the cardioid has infinite slope are \(\theta = \frac{\pi}{2} \), \(\theta = \frac{3\pi}{2}\), \(\theta = 0\), and \(\theta = \pi\). The point furthest to the left is at \(\theta = \pi\) where the x-coordinate is minimized.

Key Concepts

Polar to Cartesian conversion

Converting polar coordinates to Cartesian coordinates can be a crucial step in understanding curves better. Polar equations like the cardioid given by \(r = 1 + \cos \theta\) express relationships based on radius \(r\) and angle \(\theta\). To analyze such curves in a Cartesian plane, we use the conversion formulas:

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

For the cardioid \(r = 1 + \cos \theta\), substituting for \(x\) and \(y\) gives us \(x = (1 + \cos \theta) \cos \theta\) and \(y = (1 + \cos \theta) \sin \theta\).
This transformation simplifies the understanding of the curve's shape and features by plotting it on familiar Cartesian axes.

Infinite slope conditions

To find where a curve has an infinite slope, we look for points where the change in \(y\) with respect to the change in \(x\), i.e., \(\frac{dy}{dx}\), becomes infinitely large. Mathematically, this means finding where the denominator in the fraction \(\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\) is zero.

For the cardioid equation, we calculate:

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

The slope becomes infinite when \(\frac{dx}{d\theta} = 0\) but \(\frac{dy}{d\theta} eq 0\). Solving \(\frac{dx}{d\theta} = 0\) results in the angles: \(\theta = \frac{\pi}{2}, \frac{3\pi}{2}, 0,\) and \(\pi\).
These are the points where the graph of the cardioid is vertical and thus the slope is considered infinite.

Derivative of polar equations

Taking the derivative of polar equations is essential for analyzing curves, particularly when assessing slopes. To find the slope of a polar curve like the cardioid \(r = 1 + \cos \theta\), we must compute \(\frac{dy}{dx}\) by using derivatives with respect to \(\theta\).

This involves two steps:

• Finding \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\)
• Using these derivatives in \(\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\)

For our cardioid:

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

These derivatives allow us to determine the behavior of the curve and calculate when the slope reaches critical points, helping to identify features like infinite slope conditions or minimum and maximum locations on the curve.