Problem 17 - Algebra and Trigonometry

Question

\begin{array}{l}{1-22 \text { a pair of parametric equations is given. }} \\\ {\text { (a) Sketch the curve represented by the parametric equations. }} \\\ {\text { (b) Find a rectangular-coordinate equation for the curve by }} \\\ {\text { eliminating the parameter. }}\end{array} $$ x=\sec t, \quad y=\tan t, \quad 0 \leq t<\pi / 2 $$

Step-by-Step Solution

Answer

The rectangular-coordinate equation is \((1 + y^2) = x^2\), for \(x \geq 1\) and \(y \geq 0\).

Step 1: Understand the Parametric Equations

We are given the parametric equations \(x = \sec t\) and \(y = \tan t\). These equations describe a curve as the parameter \(t\) varies from \(0\) to \(\pi/2\). The goal is to sketch this curve and find its rectangular-coordinate equation by eliminating the parameter \(t\).

Step 2: Sketch the Curve

To sketch the curve, examine how \(x = \sec t\) and \(y = \tan t\) behave over the interval \(0 \leq t < \pi/2\). As \(t\) approaches \(\pi/2\), both \(\sec t\) and \(\tan t\) approach infinity. Start plotting points by selecting values for \(t\) within the range, for instance \(t = 0, \frac{\pi}{4}, \frac{\pi}{3}, \)near\( \frac{\pi}{2}\). The sketch will appear as a portion of the curve moving upward and to the right from a point along the horizontal (positive x) axis, looking like part of a hyperbola.

Step 3: Find Relationship Between x and y

Use trigonometric identities to eliminate \(t\) from the equations. We know \(x = \sec t = \frac{1}{\cos t}\) and \(y = \tan t = \frac{\sin t}{\cos t}\). Using the identity \(1 + \tan^2 t = \sec^2 t\), substitute \(x\) for \(\sec t\) and \(y\) for \(\tan t\) to get: \[(1 + y^2) = x^2.\] This is the rectangular-coordinate form of the equation for the curve.

Step 4: Verify Domain for x and y

Given that \(0 \leq t < \pi/2\), \(\sec t\) is always greater than or equal to 1, aligning with \(x \geq 1\). Since \(\tan t\) ranges from 0 to infinity as \(t\) approaches \(\pi/2\), \(y\) (representing \(\tan t\)) ranges from 0 to infinity. Thus, in the coordinate plane, the portion of the curve lies where \(x \ge 1\) and \(y \ge 0\).

Key Concepts

Rectangular-Coordinate Equation

When working with parametric equations such as \[ x = \sec t \text{ and } y = \tan t, \] we aim to transform them into a rectangular-coordinate equation. This involves eliminating the parameter \( t \) to express \( y \) directly in terms of \( x \). This process often utilizes trigonometric identities to relate the two. Here, one key identity is \( 1 + \tan^2 t = \sec^2 t \). Using this identity, we substitute \( \sec t \) with \( x \) and \( \tan t \) with \( y \), resulting in the equation \[ 1 + y^2 = x^2. \]

• Start by identifying trigonometric relationships.
• Use identities to replace trigonometric functions.
• Rearrange resulting expressions into a familiar form.

By simplifying parametric equations into rectangular form, we can interpret and analyze curves using traditional coordinate planes.

Trigonometric Identities

Trigonometric identities are powerful tools in transforming parametric equations. They allow us to rewrite expressions involving trigonometric functions in different forms, aiding the elimination of parameters. Some common identities that are particularly useful include:

• \( \sin^2 t + \cos^2 t = 1 \)
• \( \tan^2 t + 1 = \sec^2 t \)
• \( \sec t = \frac{1}{\cos t} \)

For the given parametric equations, the identity \( 1 + \tan^2 t = \sec^2 t \) is indispensable. It seamlessly relates \( y \) and \( x \) by allowing substitution, leading to the simplified equation \( 1 + y^2 = x^2 \). Trigonometric identities not only simplify equations but also reveal deeper connections between geometric and algebraic forms.

Hyperbola Sketch

To sketch the curve from parametric equations, one plots points at selected values of the parameter \( t \). For \( x = \sec t \) and \( y = \tan t \), as \( t \) ranges from \( 0 \) to \( \pi/2 \), both \( \sec t \) and \( \tan t \) increase. Start by plotting:

• At \( t = 0 \): \( x = 1 \), \( y = 0 \)
• At \( t = \frac{\pi}{4} \): \( x = \sqrt{2} \), \( y = 1 \)
• At \( t = \frac{\pi}{3} \): \( x = 2 \), \( y = \sqrt{3} \)

These points show a pattern moving upward and outward, resembling a hyperbola, specifically one branch. The curve naturally extends to infinity as \( t \) approaches \( \pi/2 \). The rectangular-coordinate equation further confirms the sketch by identifying its hyperbolic nature, bounded by \( x \geq 1 \) and \( y \geq 0 \), which is a clear subset of the right-hand side of a hyperbola.