Problem 9 - Algebra and Trigonometry

Question

\(3-24=A\) pair of parametric equations is given. (a) Sketch the curve represented by the parametric equations. (b) Find a rectangular coordinate equation for the curve by eliminating the parameter. $$ x=\frac{1}{t}, \quad y=t+1 $$

Step-by-Step Solution

Answer

(a) Sketch a hyperbola with asymptotes at \( x=0 \) and \( y=1 \). (b) The rectangular equation is \( y = 1 + \frac{1}{x} \).

Step 1: Assign the parametric equations

The parametric equations given are \( x = \frac{1}{t} \) and \( y = t + 1 \). Here, \( t \) is the parameter that links both equations to describe the curve.

Step 2: Express the parameter t in terms of x

From the equation \( x = \frac{1}{t} \), solve for \( t \). Multiply both sides by \( t \) to get \( xt = 1 \). Then solve for \( t \) to get \( t = \frac{1}{x} \).

Step 3: Substitute t in the y equation

Substitute \( t = \frac{1}{x} \) into the equation \( y = t + 1 \). This gives \( y = \frac{1}{x} + 1 \).

Step 4: Simplify the rectangular equation

The rectangular coordinate equation derived from the parametric equations is \( y = \frac{1}{x} + 1 \). This can be written as \( y = 1 + \frac{1}{x} \).

Step 5: Sketch the curve

The equation \( y = 1 + \frac{1}{x} \) describes a curve that is a hyperbola. Sketch this curve by noting that it has a vertical asymptote at \( x = 0 \) and a horizontal asymptote at \( y = 1 \). The curve will be in the first and third quadrants due to the restrictions of the parameter \( t \) being positive and negative.

Key Concepts

Rectangular Coordinate Equation

Parametric equations are a fun tool in mathematics that allow you to describe curves using a third variable, often denoted as a parameter like \( t \). They provide a powerful way to represent complex geometries. However, the equations involving a parameter might not always be the most convenient to work with in every context. So, what do we do? We convert them into a familiar rectangular coordinate equation using \( x \) and \( y \).

For instance, given the parametric equations \( x = \frac{1}{t} \) and \( y = t + 1 \), we eliminate the parameter \( t \) to find a more recognizable equation. By solving for \( t \) in terms of \( x \), we find \( t = \frac{1}{x} \), which allows us to express \( y \) solely in terms of \( x \), resulting in the rectangular equation \( y = \frac{1}{x} + 1 \).

This translation gives us a continuous curve on the traditional \( xy \)-coordinate plane, making it simpler to analyze and visualize.

Hyperbola

A hyperbola is an interesting type of curve that appears frequently in mathematics and various applications. It's one of the conic sections - curves formed by the intersection of a plane and a double-napped cone. A hyperbola has two main branches, each tending towards asymptotes, and they open away from each other.

In our example, the rectangular coordinate equation \( y = 1 + \frac{1}{x} \) represents a hyperbola. Here's how:

• The term \( \frac{1}{x} \) introduces a vertical asymptote at \( x = 0 \), which means the curve will never cross this line.
• Meanwhile, the \( +1 \) part of the equation creates a horizontal asymptote at \( y = 1 \).

The curve forms in the first and third quadrants. This structure results because \( \frac{1}{x} \) changes sign depending on whether \( x \) is positive or negative. Thus, the hyperbola's two distinct branches reflect this sign change behavior.

Eliminating the Parameter

Eliminating the parameter is a crucial step when we want to convert a parametric equation into a regular, or rectangular, coordinate equation. This process removes the extra variable, allowing us to express the relationship between \( x \) and \( y \) directly.

The process goes like this:

• Identify the parameter you're using; in many cases, this is \( t \).
• Express \( t \) in terms of \( x \) or \( y \).
• Substitute this expression back into the equation for the other variable (in our case, \( y \)).

For the problem with equations \( x = \frac{1}{t} \) and \( y = t + 1 \), we follow these steps:

First, solve \( x = \frac{1}{t} \) which gives \( t = \frac{1}{x} \). Then, replace \( t \) in the \( y \) equation. The result: \( y = \frac{1}{x} + 1 \).

By doing this, you've effectively stripped away the dependency on the parameter and yielded a straightforward expression connecting \( x \) and \( y \). This is often simpler to analyze and plot on a standard Cartesian coordinate system.