Problem 80 - Calculus Single Variable

Question

In Exercises \(80-83,\) describe and sketch the curve that has the given parametric equations. \(x=2 t^{2}+t-1, y=t+1\)

Step-by-Step Solution

Answer

The curve is a rightward-opening parabola described by \(x = 2y^2 - 3y - 1\).

Step 1: Analyze the Parametric Equations

The parametric equations given are \(x = 2t^2 + t - 1\) and \(y = t + 1\). Our goal is to express \(x\) and \(y\) in a single Cartesian equation, eliminating the parameter \(t\).

Step 2: Solve for Parameter \(t\) in terms of \(y\)

From the equation \(y = t + 1\), solve for \(t\): \(t = y - 1\). This expression will be used to substitute for \(t\) in the equation for \(x\).

Step 3: Substitute \(t\) in the Equation for \(x\)

Substitute \(t = y - 1\) into \(x = 2t^2 + t - 1\):\[x = 2(y - 1)^2 + (y - 1) - 1.\]

Step 4: Simplify the Expression

Expand and simplify the equation:\[(y - 1)^2 = y^2 - 2y + 1\]\[x = 2(y^2 - 2y + 1) + y - 1 - 1 = 2y^2 - 4y + 2 + y - 2 - 1\]\[x = 2y^2 - 3y - 1\]

Step 5: Describe and Sketch the Curve

The equation \(x = 2y^2 - 3y - 1\) represents a parabola that opens sideways (along the x-axis). Because the coefficient of \(y^2\) is positive, the parabola opens to the right. Plotting key points from the parametric equations obtained by varying \(t\) will give the curve's orientation.

Key Concepts

Cartesian Equation

To understand the concept of a Cartesian equation, it's essential to start with parametric equations. These are equations where each coordinate depends on a third variable, called a parameter.
In our exercise, we have two parametric equations:

• \( x = 2t^2 + t - 1 \)
• \( y = t + 1 \)

Here, \( t \) is the parameter. Our goal is to eliminate this parameter and express the relationship directly between \( x \) and \( y \). This result is called a Cartesian equation and is free of parameters. By eliminating the parameter, we allow the equation to be analyzed and graphed using traditional \( xy \) coordinates which are easier to conceptualize and sketch.

Parabola

A parabola is a particular type of curve on a graph that forms a symmetric shape. They can open up, down, left, or right depending on their equations.
In this exercise, we derived the Cartesian equation:\( x = 2y^2 - 3y - 1 \)
Parabolas defined as functions of \( y \) typically open sideways:

• If \( y^2 \) has a positive coefficient, the parabola opens to the right.
• If \( y^2 \) has a negative coefficient, the parabola opens to the left.

In our case, because the coefficient of the \( y^2 \) term is positive (2), the parabola opens to the right. Recognizing this directionality aids in plotting and visualizing where the parabola lies in relation to the axes.

Eliminating Parameter

Eliminating the parameter involves solving one of the parametric equations for the parameter and substituting that expression into the other equation. This step aligns with steps where our \( t \) was isolated.
Here's how it works:

• Start with \( y = t + 1 \)
• Isolate \( t \): \( t = y - 1 \)
• Substitute back into the \( x \) equation: \( x = 2(y - 1)^2 + (y - 1) - 1 \)

This substitution helps merge our two parametric expressions into a single equation with \( x \) and \( y \), effectively eliminating \( t \). This is vital for sketching and understanding the behavior of the curve on an \( xy \) plane.

Curve Sketching

Once you have the Cartesian equation, you're equipped to sketch the curve. Understanding the form and direction of a parabola enables efficient sketching.
Steps to sketching:

• Identify the type of conic (a parabola in our case).
• Determine the orientation (opens to the right due to the positive \( y^2 \) coefficient).
• Plot key points by choosing values for \( y \) and solving for \( x \).
• Use symmetry about a vertical axis to ensure accuracy for left-to-right opening parabolas.

By following these steps, you gain a visual representation of the parabola, which provides critical insight into its behavior and allows for precise graphing on paper or software.