Problem 11 - A Graphical Approach to College Algebra

Question

For each plane curve, use a graphing calculator to generate the curve over the interval for the parameter \(t\), in the window specified. Then, find a rectangular equation for the curve. $$\begin{aligned}&x=t^{3}+1, y=t^{3}-1, \text { for } t \text { in }[-3,3]\\\&\text { window: }[-30,30] \text { by }[-30,30]\end{aligned}$$

Step-by-Step Solution

Answer

The rectangular equation is \(x - y = 2\).

Step 1: Analyze Parametric Equations

The given parametric equations are \(x = t^3 + 1\) and \(y = t^3 - 1\). Our objective is to find a relationship between \(x\) and \(y\) by eliminating the parameter \(t\).

Step 2: Express t in Terms of x and y

First, we solve for \( t^3 \) from each equation separately. For \(x = t^3 + 1\), we get \(t^3 = x - 1\). Similarly, for \(y = t^3 - 1\), we get \(t^3 = y + 1\).

Step 3: Find the Rectangular Equation

Since both \(t^3 = x - 1\) and \(t^3 = y + 1\), we can equate them: \(x - 1 = y + 1\). Simplify to obtain the rectangular equation of the curve: \(x - y = 2\).

Step 4: Confirm Rectangular Equation

The rectangular equation \(x - y = 2\) represents a line. Verifying by substituting back, if \(x = t^3 + 1\) and \(y = t^3 - 1\), then \(x - y = (t^3 + 1) - (t^3 - 1) = 2\), confirming the derivation.

Key Concepts

Rectangular Equation

In the context of parametric equations, a rectangular equation is an expression that relates the variables directly without involving a parameter like \(t\).
Here, our task was to convert the parametric form into a direct relationship between \(x\) and \(y\).
The original parametric equations were \(x = t^3 + 1\) and \(y = t^3 - 1\). Through algebraic manipulation, we isolated the parameter \(t\) to arrive at a rectangular equation.
This process often involves solving each parametric equation for the parameter, and then equating them to eliminate \(t\).

• For instance, from \(x = t^3 + 1\), we found \(t^3 = x - 1\).
• Similarly, from \(y = t^3 - 1\), we got \(t^3 = y + 1\).

Upon equating \(x - 1\) and \(y + 1\), we derived the rectangular equation \(x - y = 2\). This gives us a clear linear relationship without the parameter.

Graphing Calculator

A graphing calculator is a powerful tool to visualize mathematical equations and functions, including parametric equations.
For our exercise, using a graphing calculator helped us generate the plane curve described by the parametric equations over the interval \([-3, 3]\).

• First, input the equations \(x = t^3 + 1\) and \(y = t^3 - 1\) into the calculator.
• Then, set the interval for \(t\) from \(-3\) to \(3\).
• Adjust the window to \([-30, 30]\) for both \(x\) and \(y\) axes.

This allows you to graph the equations and visually confirm their form, which in this case appears as a straight line due to the linear relationship found in our rectangular equation.
Graphing calculators are especially useful in verifying the correctness of algebraic manipulations and aiding the understanding of concepts through real-time graphical displays.

Plane Curve

A plane curve is essentially a curve that lies on a two-dimensional plane, representing the graphical interpretation of parametric or rectangular equations.
In parametric equations, a plane curve is generated by varying the parameter \(t\), where each value of \(t\) computes a point \((x, y)\) on the plane.

• The curve represented by our parameters \(x = t^3 + 1\) and \(y = t^3 - 1\) was plotted over the range of \(t\) from \(-3\) to \(3\).
• By graphing these, we recognized that the resulting plane curve formed a straight line.

Plane curves are crucial in depicting complex relationships visually, making them accessible and easier to analyze.
Understanding the notion of plane curves enables the transition from abstract formulae to tangible geometrical interpretations.

Algebra

Algebra, in this context, is the branch of mathematics used to manipulate the equations and eliminate parameters.
We used algebraic techniques to convert the parametric equations into a rectangular equation in this exercise.

• By expressing both equations in terms of \(t^3\), we derived expressions \(t^3 = x - 1\) and \(t^3 = y + 1\).
• Equating these expressions allowed us to simplify and obtain the relationship \(x - y = 2\).

Algebra was crucial in transforming the equations from dependent on the parameter \(t\) to a direct \(x\) and \(y\) relationship.
The ability to manipulate equations using algebraic principles is fundamental in enabling not just the solving of equations but also understanding deeper mathematical relationships.