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Chapter 8: Problem 34

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

Expert verified
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

01

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 \(\theta\).
02

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 \(dy/dx = 1 / (2\cos\theta)\).
03

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).
04

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.

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Most popular questions from this chapter

Find the area of the circle given by \(r=\sin \theta+\cos \theta\). Check your result by converting the polar equation to rectangular form, then using the formula for the area of a circle.

Determine any differences between the curves of the parametric equations. Are the graphs the same? Are the orientations the same? Are the curves smooth? $$ \text { (a) } \begin{aligned} x &=t \\ y &=2 t+1 \end{aligned} $$ $$ \text { (b) } \begin{aligned} x &=\cos \theta \\ y &=2 \cos \theta+1 \end{aligned} $$ $$ \text { (c) } \begin{aligned} x &=e^{-t} \\ y &=2 e^{-t}+1 \end{aligned} $$ (d) \(x=e^{t}\) $$ y=2 e^{t}+1 $$

Use a graphing utility to graph the polar equation over the given interval. Use the integration capabilities of the graphing utility to approximate the length of the curve accurate to two decimal places. $$ r=2 \sin (2 \cos \theta), \quad 0 \leq \theta \leq \pi $$

Explain how the graph of each conic differs from the graph of \(r=\frac{4}{1+\sin \theta} .\) (a) \(r=\frac{4}{1-\cos \theta}\) (b) \(r=\frac{4}{1-\sin \theta}\) (c) \(r=\frac{4}{1+\cos \theta}\) (d) \(r=\frac{4}{1-\sin (\theta-\pi / 4)}\)

$$ \text { State the definition of a smooth curve } $$
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