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Horizontal and Vertical Tangency In Exercises 33-42, 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 horizontal and vertical tangency to the curve \(x = \cos^2\theta\), \(y = \cos\theta\) can be found by setting the derivatives of the \(x\) and \(y\) functions to zero and solving for \(\theta\). The corresponding \(x\) and \(y\) values are the points of tangency. Use a graphing utility to confirm these points.

Step by step solution

01

Determine the Derivatives of \(x\) and \(y\)

Start by finding the derivatives of the \(x\) and \(y\) functions with respect to \(\theta\). These are given by \(dx/d\theta = -2\cos\theta\sin\theta\) \(dy/d\theta = -\sin\theta\).
02

Set the Derivatives to Zero

Set each derivative to zero and solve for \(\theta\). This gives: \(dx/d\theta = -2\cos\theta\sin\theta = 0 \Rightarrow \cos\theta = 0 \) \(\sin\theta = 0 \Rightarrow \sin\theta = 0 \)
03

Compute the Points of Tangency

Now, substitute the solutions for \(\theta\) from Step 2 into the original \(x\) and \(y\) functions to find the corresponding points on the curve. These are the points of tangency.
04

Validate with Graphing Utility

Lastly, use a graphing utility to plot the functions \(x = \cos^2\theta\) and \(y = \cos\theta\) and the points of tangency calculated in Step 3. Check that these points are indeed the points of horizontal or vertical tangency on the curve.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Horizontal Tangency
When we talk about horizontal tangency on a curve, we're essentially looking for points where the tangent to the curve is perfectly horizontal. This occurs where the slope of the curve, or the derivative of the y-component with respect to the parameter \( \theta \), is zero. Mathematically, this is when \( \frac{dy}{d\theta} = 0 \). For the given curve, the y-component is \( y = \cos \theta \), thus its derivative is \( \frac{dy}{d\theta} = -\sin \theta \).
  • A horizontal tangent happens when \( -\sin \theta = 0 \), which simplifies to \( \sin \theta = 0 \).
  • Since \( \sin \theta \) equals zero at integer multiples of \( \pi \), such as \( \theta = 0, \pi, 2\pi, \ldots \), substitute these values into the parametric equations to find the actual points.
Substitute these into the original equations to find:
  • If \( \theta = 0 \), then \( x = \cos^2(0) = 1 \) and \( y = \cos(0) = 1 \).
  • If \( \theta = \pi \), then \( x = \cos^2(\pi) = 1 \) and \( y = \cos(\pi) = -1 \).
Vertical Tangency
Points of vertical tangency occur where the tangent line is vertical. Such points occur when the derivative of the x-component with respect to \( \theta \) is zero, which implies that \( \frac{dx}{d\theta} = 0 \). The x-component given is \( x = \cos^2 \theta \), and by taking the derivative, we have \( \frac{dx}{d\theta} = -2\cos\theta\sin\theta \).
  • Setting \( -2\cos\theta\sin\theta = 0 \), we solve for \( \cos \theta = 0 \) or \( \sin \theta = 0 \).
  • A vertical tangent happens when \( \cos \theta = 0 \), occurring at odd multiples of \( \frac{\pi}{2} \) such as \( \theta = \frac{\pi}{2}, \frac{3\pi}{2} \).
Substitute these values into the parametric equations:
  • If \( \theta = \frac{\pi}{2} \), then \( x = \cos^2(\frac{\pi}{2}) = 0 \) and \( y = \cos(\frac{\pi}{2}) = 0 \).
  • If \( \theta = \frac{3\pi}{2} \), then \( x = \cos^2(\frac{3\pi}{2}) = 0 \) and \( y = \cos(\frac{3\pi}{2}) = 0 \).
Thus, the curve has points of vertical tangency when it crosses the y-axis at these points.
Parametric Derivatives
Parametric derivatives allow us to study curves that are defined by separate equations for x and y in terms of a third parameter, often \( \theta \). By using parametric derivatives, we can find the slope of the tangent to the curve by comparing the rates of change for x and y with respect to the parameter. This involves:
  • Finding \( \frac{dx}{d\theta} \) using the derivative of the x-function. In our case, \( \frac{dx}{d\theta} = -2\cos\theta\sin\theta \).
  • Finding \( \frac{dy}{d\theta} \) using the derivative of the y-function, where \( \frac{dy}{d\theta} = -\sin\theta \).
The slope of the tangent to the curve at any point is then found using the formula \( \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \). Simplifying this gives us the ratio of changes and tells us how steeply the curve rises or falls at a given point. This process helps in identifying points of horizontal or vertical tangency by setting \( \frac{dy}{dx} \) to zero or undefined respectively.
Graphing Utility
A graphing utility is a helpful tool that allows visualization of mathematical equations and concepts. When studying parametric curves like the one given, it's beneficial to use a graphing utility because it helps:
  • Visualize the curve and confirm points of horizontal or vertical tangency.
  • Ensures our analytic solutions correspond to the actual shape of the curve.
With the graphing utility, you can input the parametric equations \( x = \cos^2\theta \) and \( y = \cos\theta \) to see the curve. Compare the points of tangency found analytically by checking if they match visually with the flat sections (horizontal tangents) or the places where the curve changes direction sharply (vertical tangents). The graphing utility reinforces our understanding by providing a visual representation of abstract mathematical concepts.

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