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Chapter 11: Problem 103

Explain and carry out a method for graphing the curve \(x=1+\cos ^{2} y-\sin ^{2} y\) using parametric equations and a graphing utility.

Short Answer

Expert verified
Answer: The parametric equations for the curve \(x=1+\cos ^{2} y-\sin^{2}y\) are \(x(t) = 1 + \cos^2(t) - \sin^2(t)\) and \(y(t) = t\).

Step by step solution

01

Express x and y in terms of a parameter t

First, let's assign a parameter t to y, i.e., \(y=t\). Now we can rewrite the given equation as: \(x = 1 + \cos^2(t) - \sin^2(t)\). Now x and y are expressed in terms of parameter t as: \(x(t) = 1 + \cos^2(t) - \sin^2(t)\) \(y(t) = t\)
02

Identify the range of t for graphing the parametric equations

We need to determine the appropriate range of t to capture the complete curve. Since \(\cos^2(t) - \sin^2(t)\) take values between -1 and 1, the given curve will essentially be within the horizontal interval (-1,3). The range for t (or y) should include the entire periodicity of the function, so we should consider t within the range \([0, 2\pi]\) since all possible values of \(\cos^2(t) - \sin^2(t)\) can occur within one period of the trigonometric functions.
03

Use a graphing utility to plot the parametric equations

Now we can plot the parametric equations using a graphing utility like Desmos, GeoGebra, or Wolfram Alpha. Input the parametric equations \(x(t) = 1 + \cos^2(t) - \sin^2(t)\) and \(y(t) = t\) with t ranging from \(0\) to \(2\pi\). This will provide a visualization of the curve for the given equation. In summary, the parametric equations for the curve \(x=1+\cos ^{2} y-\sin^{2}y\) are \(x(t) = 1 + \cos^2(t) - \sin^2(t)\) and \(y(t) = t\). By plotting these equations using a graphing utility with the range of t from \(0\) to \(2\pi\), we can obtain an accurate visual representation of the curve.

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