Problem 18 - 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=\cot t, \quad y=\csc t, \quad 0

Step-by-Step Solution

Answer

Rectangular equation: \( x^2 = y^2 - 1 \) (a horizontal hyperbola).

Step 1: Express the Parametric Equations

We are given the parametric equations \( x = \cot t \) and \( y = \csc t \). These equations determine the position of points on the curve as the parameter \( t \) changes from \( 0 \) to \( \pi \). We will sketch this curve by plotting several points for values of \( t \) in this interval.

Step 2: Calculate Sample Points

To plot the curve, let's calculate the values of \( x \) and \( y \) for several values of \( t \). For example: - If \( t = \frac{\pi}{4} \), then \( x = \cot \frac{\pi}{4} = 1 \) and \( y = \csc \frac{\pi}{4} = \sqrt{2} \).- If \( t = \frac{\pi}{2} \), then \( x = \cot \frac{\pi}{2} = 0 \) and \( y = \csc \frac{\pi}{2} = 1 \).- If \( t = \frac{3\pi}{4} \), then \( x = \cot \frac{3\pi}{4} = -1 \) and \( y = \csc \frac{3\pi}{4} = \sqrt{2} \).Plot these points on the Cartesian plane to get a sense of the curve.

Step 3: Sketch the Curve

Plot the calculated points in the previous step on the Cartesian plane. Notice that as \( t \to 0^+ \), \( x \to +\infty \) and \( y \to +\infty \), and as \( t \to \pi^- \), \( x \to -\infty \) and \( y \to -\infty \). This results in a hyperbola shape opening along the x-axis. Draw the curve based on these points and observations.

Step 4: Eliminate the Parameter

To find a rectangular-coordinate equation, we eliminate the parameter \( t \). We know:\[ x = \cot t = \frac{\cos t}{\sin t}, \quad y = \csc t = \frac{1}{\sin t}\]This implies \( y = \frac{1}{\sin t} \Rightarrow \sin t = \frac{1}{y} \). Substitute \( \sin t \) in the expression for \( x \):\[ x = \frac{\cos t}{\sin t} = \cos t \cdot y\]To find \( \cos t \), use the Pythagorean identity \( \sin^2 t + \cos^2 t = 1 \):\[ \left(\frac{1}{y}\right)^2 + \cos^2 t = 1 \Rightarrow \cos^2 t = 1 - \frac{1}{y^2}\]Thus, \( \cos t = \pm \sqrt{1 - \frac{1}{y^2}} \). Therefore:\[ x = \pm y \sqrt{1 - \frac{1}{y^2}}\]Squaring both sides gives us the rectangular equation:\[ x^2 = y^2 - 1\]

Step 5: Resulting Rectangular Equation

The rectangular equation for the given parametric equations is \( x^2 = y^2 - 1 \), which represents a horizontal hyperbola centered at the origin.

Key Concepts

Rectangular Equation

A rectangular equation is a mathematical expression that relates two variables, typically using Cartesian coordinates. When you have parametric equations, they describe a curve or line based on a parameter, often denoted as \( t \). To convert these parametric equations into a rectangular format, one must eliminate the parameter to express the relationship solely in terms of \( x \) and \( y \).

For example, consider the parametric equations given as \( x = \cot t \) and \( y = \csc t \). Our objective is to transform these into a rectangular-coordinate equation. This involves manipulating the equations to remove \( t \) and establish a direct relationship between \( x \) and \( y \).

By doing this, you end up with a clearer understanding of the curve's shape, as this format allows you to graph the curve on a regular coordinate plane. The final expression often lends itself more straightforwardly to analysis and graphing, especially if it represents classic shapes like lines, circles, ellipses, or hyperbolas.

Eliminating the Parameter

Eliminating the parameter is a crucial technique in converting parametric equations to their rectangular form. The fundamental idea is to express one of the parametric variables in terms of the other without involving the parameter \( t \).

For the equations \( x = \cot t \) and \( y = \csc t \), you can start by expressing \( \sin t \) from \( y = \csc t \):

• \( \sin t = \frac{1}{y} \)

Next, substitute \( \sin t \) into the equation for \( x \):

• \( x = \cot t = \frac{\cos t}{\sin t} = \cos t \cdot y \)

To find \( \cos t \), use the Pythagorean identity \( \sin^2 t + \cos^2 t = 1 \):

• \( (\frac{1}{y})^2 + \cos^2 t = 1 \)
• \( \cos^2 t = 1 - \frac{1}{y^2} \)

The expressions thus obtained can be simplified to derive the rectangular equation, as shown. This process of parameter elimination transforms equations into a more interpretable and graphable form.

Hyperbola

A hyperbola is a type of conic section that is defined as the set of points where the difference in distances from two fixed points (foci) is constant. In rectangular equations, hyperbolas often appear in standard forms like:\[ x^2/a^2 - y^2/b^2 = 1 \]

In our problem, after eliminating the parameter from \( x = \cot t \) and \( y = \csc t \), we derived the rectangular equation \( x^2 = y^2 - 1 \). This equation represents a horizontal hyperbola centered at the origin. Here are some key features of this hyperbola:

• Its branches open along the x-axis, which is indicated by the \( x^2 \) term.
• The center is at the origin \((0,0)\).
• The equation suggests that as \( y^2 \) becomes larger than 1, \( x^2 \) grows larger, further indicating its hyperbolic nature.

Hyperbolas have distinct asymptotes that the curve approaches as \( x \) or \( y \) extends towards infinity, characterized by the graph's open branches. Understanding these properties helps in sketching and analyzing the hyperbola efficiently.