Problem 15 - College Trigonometry

Question

In Exercises \(1-20\), plot the set of parametric equations by hand. Be sure to indicate the orientation imparted on the curve by the parametrization. $$ \left\\{\begin{array}{l} x=\sec (t) \\ y=\tan (t) \end{array} \text { for }-\frac{\pi}{2}

Step-by-Step Solution

Answer

Plot a hyperbola using \(x = \sec(t)\) and \(y = \tan(t)\), indicating orientation from left to right as \(t\) increases.

Step 1: Understand Parametric Equations

We are given the parametric equations \(x = \sec(t)\) and \(y = \tan(t)\). These need to be plotted for the interval \(-\frac{\pi}{2} < t < \frac{\pi}{2}\). Understanding the trigonometric functions, we know that \(\sec(t) = \frac{1}{\cos(t)}\) and \(\tan(t) = \frac{\sin(t)}{\cos(t)}\).

Step 2: Determine Range for \(t\)

The range of \(t\) is restricted to \(-\frac{\pi}{2} < t < \frac{\pi}{2}\). In this range, \(\cos(t)\) never equals zero, but approaches it as \(t\) approaches \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\), which means the functions are defined within the open interval. The values of \(x\) (which are \(\sec(t)\)) can range from negative to positive infinity as \(t\) nears those limits.

Step 3: Plot Points for Different Values of \(t\)

Choose specific values of \(t\) such as \(-\frac{\pi}{3}, -\frac{\pi}{4}, 0, \frac{\pi}{4}, \frac{\pi}{3}\), and calculate corresponding \(x\) and \(y\) values. For example, for \(t = 0\), \(x = \sec(0) = 1\), \(y = \tan(0) = 0\). This provides points \((1, 0)\), \(\left( \frac{2}{\sqrt{3}}, \frac{-1}{\sqrt{3}} \right)\) for \(t = \frac{-\pi}{6}\), and so forth.

Step 4: Understand the Curve Shape

As \(t\) approaches \(-\frac{\pi}{2}\) or \(\frac{\pi}{2}\), the values of \(x = \sec(t)\) become larger in absolute terms. The parametric equations describe a hyperbola, as the expression \(x^2 - y^2 = 1\) can be derived from replacing \(x = \sec(t)\) and \(y = \tan(t)\).

Step 5: Indicate the Orientation

With increasing \(t\) from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), we can evaluate that the orientation of the curve will go upwards on the right branch and downwards on the left branch because \(y\) (or \(\tan(t)\)) increases as \(t\) moves from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).

Step 6: Draw the Curve Based on Analysis

Using the points calculated and understanding of the behavior of \(x\) and \(y\), plot a hyperbola with its branches open horizontally and plotted from left to right following the orientation. Indicate the increasing \(t\) direction along the curve.

Key Concepts

Hyperbola

When working with parametric equations like those given in the exercise, you should be aware that they can describe various types of curves, including hyperbolas. A hyperbola consists of two separate curves or branches, appearing as mirror images. Specifically, for this exercise, we define the hyperbola using the expressions for secant and tangent functions: \[\begin{array}{l} x=\sec (t) \ y=\tan (t) \end{array}\]This transforms into the standard hyperbola form:\[ x^2 - y^2 = 1 \]A hyperbola in this form opens horizontally. Its defining characteristics include:

• Two branches of the curve open outwards in opposite horizontal directions.
• The curve has a center at the origin (0,0) in the coordinate plane.
• Each side of the hyperbola approaches the asymptotes but never touches them.

The asymptotic behavior is crucial for sketching the hyperbola accurately. Knowing that it stretches indefinitely along these lines helps understand the shape and placement of the graph.

Trigonometric Functions

Trigonometric functions are key to analyzing and plotting parametric equations, particularly those dealing with circular and hyperbolic functions. In the exercise, we use two such important functions:

• Secant (
  72ec(t)): Defined as the reciprocal of the cosine function, or 1 divided by
  72os(t). The secant function can be undefined at points where the cosine is zero due to division by zero risks, which is why
  72os(t) must not be zero for defined outputs.
• Tangent (
  72an(t)): This function is the ratio of the sine function over the cosine function, expressed as
  72an(t) =
  72in(t)/
  72os(t). It measures the angle's slope of the line intersecting a circle, increasing or decreasing based on the input values.

These trigonometric functions have specific ranges and intervals where they are valid, especially in this context, where the secant and tangent are considered over the interval
72iaa{-
72i/2,
72i/2}. Understanding where these functions are undefined is key to successfully plotting them, as it determines the hyperbola's structure and where the curve breaks or continues smoothly.

Plotting Graphs

Plotting graphs of parametric equations can initially seem daunting, but by breaking the task into steps and using the properties of trigonometric functions, you can plot them successfully. Here's a high-level approach:

• Select Key Points: Choose values of
  72
  72 from the interval
  72iaa{-
  72i/2,
  72i/2} that are easy to calculate, like 0,
  72i/4, and
  72i/3. Calculate corresponding x and y values using the given parametric equations.
• Calculate and Plot: Determine x and y for each selected value of t. For example, for
  72r(0), x =
  72ec(0) = 1, and y =
  72an(0) = 0, giving a point at (1,0).
• Indicate Orientation: This refers to the direction the sketch should grow as t increases. As
  72 increases from -
  72i/2 to
  72i/2, the graph's orientation or direction also shifts, curving up and down along its branches.

By understanding these steps, mapping key points, and visualizing how the curve behaves across the positive and negative directions of both the x and y axes, you can accurately plot the curve's trajectory and define its precise structure.