Problem 23 - A Graphical Approach to Precalculus with Limits

Question

For each plane curve, (a) graph the curve, and (b) find a rectangular equation for the curve. $$x=t+2, y=t-4 ; \text { for } t \text { in }(-\infty, \infty)$$

Step-by-Step Solution

Answer

The rectangular equation is \( y = x - 6 \), and it's a line with a slope of 1 and a y-intercept of -6.

Step 1: Understand the Parametric Equations

The given parametric equations for the curve are \( x = t + 2 \) and \( y = t - 4 \). Here, \( t \) is the parameter that can take any real value.

Step 2: Express t in Terms of x

To eliminate the parameter \( t \), we first solve one of the parametric equations for \( t \). From \( x = t + 2 \), we get \( t = x - 2 \).

Step 3: Substitute t in the Equation for y

Substitute \( t = x - 2 \) into the equation for \( y \). This gives us \( y = (x - 2) - 4 \). Simplifying this, we get \( y = x - 6 \).

Step 4: Derive the Rectangular Equation

The rectangular equation of the curve is \( y = x - 6 \). This is a straight line where the slope (m) is 1 and the y-intercept (b) is -6.

Step 5: Graph the Rectangular Equation

To graph the line \( y = x - 6 \), we note the y-intercept is (0,-6) and the slope is 1. This means for a change of +1 in x, y also increases by +1. Plot the y-intercept and a couple more points by adjusting x and y accordingly, then draw the line through these points. The graph is a straight line extending infinitely in both directions.

Key Concepts

Rectangular Equation

The rectangular equation is derived by eliminating the parameter from the parametric equations, resulting in a single equation in terms of only two variables, typically x and y. In our exercise, we started with two parametric equations:

• \( x = t + 2 \)
• \( y = t - 4 \)

The goal is to eliminate the parameter \( t \) and express \( y \) as a function of \( x \). By solving for \( t \) in the first equation, \( t = x - 2 \), and substituting into the second equation, we obtain the rectangular equation \( y = x - 6 \).
This equation represents a straight line where the slope (m) is 1 and the y-intercept (b) is -6. This form, known as the slope-intercept form \( y = mx + b \), succinctly captures the relationship between \( x \) and \( y \) without involving the parameter \( t \).

Graphing

Graphing plays a critical role in visualizing the relationship described by the rectangular equation. The process begins by identifying key components like the y-intercept and the slope. For the equation \( y = x - 6 \):

• The y-intercept is the point where the line crosses the y-axis, in this case, (0, -6).
• The slope is 1, indicating that for each unit increase in \( x \), \( y \) also increases by one unit.

To graph this line, start at the y-intercept. Then use the slope to find subsequent points. Increase \( x \) by 1 and \( y \) by 1 to locate the next point. Repeat this to establish a few points on either side of the y-intercept. Once plotted, draw a continuous line through these points.
Since the slope is constant, the line will extend infinitely in both directions, illustrating the linear path dictated by the equation.

Eliminating Parameters

Eliminating parameters is an essential step in converting a set of parametric equations into a single rectangular equation. Parameters, often denoted as \( t \), serve as a linking variable that defines both \( x \) and \( y \) independently.
To eliminate the parameter, we focus on expressing one of the variables, say \( x \), in terms of \( t \), which was spearheaded by the equation \( x = t + 2 \). Solving this gives \( t = x - 2 \). Next, we replace \( t \) in the \( y \) equation: \( y = t - 4 \). By substitution, it becomes \( y = (x - 2) - 4 \). Simplifying this, we find \( y = x - 6 \).
Thus, by removing the parameter, we streamline the equations into a form that shows direct dependence of one variable on the other. The resulting equation, \( y = x - 6 \), is simpler to analyze and graph, providing a clearer picture of the relationship between \( x \) and \( y \).