Question

Find all points (if any) of horizontal and vertical tangency to the curve. Use a graphing utility to confirm your results. $$ x=\cos ^{2} \theta, \quad y=\cos \theta $$

Short answer

The points of tangency to the curve described by the parametric equations \(x=\cos^{2}\theta, y=\cos\theta\) are vertical tangents at (0,0) and (0,-1). There are no horizontal tangents.

Step-by-step solution

Find the Derivatives

Using the equations \(x=\cos^{2}\theta\) and \(y=\cos\theta\), we'll first compute the derivative of \(y\) with respect to \(x\). This is also known as the slope of the curve and can indicate points of tangency. It is obtained by the derivative of \(y\) with respect to \(\theta\) divided by the derivative of \(x\) with respect to \(\theta\).

Compute \(dy/d\theta\) and \(dx/d\theta\)

The derivative of \(y = \cos\theta\) with respect to \(\theta\) is \(dy/d\theta = -\sin\theta\). For \(x = \cos^{2}\theta\), by applying the chain rule, the derivative \(dx/d\theta = 2\cos\theta(-\sin\theta) = -2\cos\theta\sin\theta\). Consequently, \(dy/dx = (dy/d\theta) / (dx/d\theta) = \frac{-\sin\theta}{-2\cos\theta\sin\theta}\). Simplifying this fraction yields \(dy/dx = 1 / (2\cos\theta)\).

Find Points of Tangency

Horizontal tangents occur when the derivative is 0. Since the denominator can never be 0, there are no horizontal tangents. Vertical tangents occur when the derivative is undefined, which happens when the denominator in \(dy/dx\) is zero i.e., when \(\cos\theta = 0\). Solving for \(\theta\), we get \(\theta = \pi/2, 3\pi/2\). For \(\theta = \pi/2\), \(x = \cos^{2}(\pi/2) = 0\) and \(y = \cos(\pi/2) = 0\). For \(\theta =3\pi/2\), \(x = \cos^{2}(3\pi/2) = 0\) and \(y = \cos(3\pi/2) = -1\). Thus, the points of vertical tangency are at (0,0) and (0,-1).

Confirm Results

Use a graphing tool to graph the functions \(x=\cos^{2}\theta\) and \(y=\cos\theta\). The graph shows vertical tangents at the points we calculated, confirming our results.