Problem 29 - A Graphical Approach to Precalculus with Limits

Question

Find a rectangular equation for each curve and graph the curve. $$x=\sin t, y=\csc t ; \text { for } t \text { in }(0, \pi)$$

Step-by-Step Solution

Answer

The rectangular equation is \(y = \frac{1}{x}\). The graph is a hyperbola symmetric about the origin.

Step 1: Understand the given parametric equations

The exercise gives parametric equations: \(x = \sin t\) and \(y = \csc t\) for \(t\) in the interval \((0, \pi)\). Our goal is to find the relationship between \(x\) and \(y\) that does not involve \(t\).

Step 2: Express \( \csc t \) in terms of \( \sin t \)

Recall the identity that the cosecant function is the reciprocal of the sine function, \(\csc t = \frac{1}{\sin t}\). This identity will be used to eliminate the parameter \(t\).

Step 3: Substitute \(\sin t\) with \(x\)

Since \(x = \sin t\), we can substitute \(\sin t = x\) into the expression for \(y\): \(y = \csc t = \frac{1}{\sin t} = \frac{1}{x}\).

Step 4: Write the rectangular equation

Now, we have \(y = \frac{1}{x}\), which is the rectangular equation in terms of \(x\) and \(y\), eliminating the parameter \(t\).

Step 5: Graph the rectangular equation

The graph of the equation \(y = \frac{1}{x}\) is a hyperbola that is symmetric with respect to the origin. It does not include any points on the x-axis or y-axis because \(x\) and \(y\) cannot be zero (since \(t\) is in \((0, \pi)\)). Let's graph this hyperbola considering these conditions.

Key Concepts

Rectangular Equation

In the world of mathematics, a **rectangular equation** is one that relates two variables, typically expressed in terms of coordinates. Unlike parametric equations which use a parameter to define relationships, a rectangular equation directly establishes the bond between the variables. In our exercise, the goal was to transition from parametric to rectangular form. Given the relationship:

• Parametric equations are given as:
  • \( x = \sin t \)
  • \( y = \csc t \)
• By using the identity \( \csc t = \dfrac{1}{\sin t}\) and knowing \( x = \sin t \), we substitute to find \( y = \dfrac{1}{x} \).

This provides us a clear relationship between \(x\) and \(y\), without referencing \(t\). The rectangular form \(y = \dfrac{1}{x}\) captures the essence of the equation's behavior as a hyperbolic shape.

Graphing Hyperbolas

**Graphing hyperbolas** involves understanding their unique curve properties. The standard hyperbola in rectangular form that we derived, \(y = \frac{1}{x}\), creates two separate curves or branches visible in the Cartesian plane. These curves stretch infinitely, approaching the axes but never actually touching them. When graphing the hyperbola \(y = \frac{1}{x}\), it's important to note:

• The graph is symmetric with respect to the origin, creating a sort of rotational mirror effect.
• The curves are in two negative-positive quadrants (top-right and bottom-left).
• Since \(x\) and \(y\) cannot be zero, the graph never intersects the x-axis or y-axis, forming asymptotes that define its boundaries.

Envision these properties to graph the hyperbola accurately and understand its asymptotic behavior as it gets closer to the axes.

Trigonometric Identities

Understanding **trigonometric identities** is essential when working with parametric equations. They allow for the simplification or transformation of expressions, facilitating the conversion between different forms of equations. Among these valuable identities, the reciprocal relationships are vital.For example, in this exercise, we employed:

• \( \csc t = \dfrac{1}{\sin t} \), which highlights that the cosecant is the reciprocal of sine.

By applying this identity, we efficiently moved from a parameter-based equation to a clear rectangular relationship. These identities simplify the manipulation of trigonometric equations, allowing mathematicians to express and solve problems more easily.