Problem 6 - A Graphical Approach to Precalculus with Limits

Question

Graph each pair of parametric equations by hand, using values of tin \([-2,2] .\) Make a table of \(t_{*}, x_{*}\) and \(y\) -values, using \(t=-2,-1,0,1,\) and \(2 .\) Then plot the points and join them with a line or smooth curve for all values of \(t\) in \([-2,2] .\) Do not use a calculator. $$x=-t^{2}+2, \quad y=t+1$$

Step-by-Step Solution

Answer

The graph forms a downward-opening parabola through points (-2,-1), (1,0), (2,1), (1,2), (-2,3).

Step 1: Setting Up the Table

Create a table with columns for \(t\), \(x = -t^2 + 2\), and \(y = t + 1\). Fill in the first column with the values of \(t\): -2, -1, 0, 1, 2.

Step 2: Calculate \(x\)-values

Use the equation \(x = -t^2 + 2\) to calculate the \(x\)-values for each \(t\). For example:- For \(t = -2\), \(x = -(-2)^2 + 2 = -4 + 2 = -2\).- For \(t = -1\), \(x = -(-1)^2 + 2 = -1 + 2 = 1\).- For \(t = 0\), \(x = -(0)^2 + 2 = 2\).- For \(t = 1\), \(x = -(1)^2 + 2 = -1 + 2 = 1\).- For \(t = 2\), \(x = -(2)^2 + 2 = -4 + 2 = -2\).Fill these values in the second column of the table.

Step 3: Calculate \(y\)-values

Use the equation \(y = t + 1\) to calculate the \(y\)-values for each \(t\). For example:- For \(t = -2\), \(y = -2 + 1 = -1\).- For \(t = -1\), \(y = -1 + 1 = 0\).- For \(t = 0\), \(y = 0 + 1 = 1\).- For \(t = 1\), \(y = 1 + 1 = 2\).- For \(t = 2\), \(y = 2 + 1 = 3\).Fill these values in the third column of the table.

Step 4: Plot the Points

Using the table, plot the points \((-2, -1)\), \((1, 0)\), \((2, 1)\), \((1, 2)\), and \((-2, 3)\) on the coordinate plane. These points represent the parametric equations for the given \(t\) values.

Step 5: Connect and Interpret the Graph

Connect the plotted points with a smooth curve. The resulting figure should show the path that the points follow as \(t\) changes from -2 to 2. It forms a parabolic path that opens downward, indicating the curve behaves like a parabola in the \(xy\)-plane.

Key Concepts

Plotting Points

When dealing with parametric equations like \(x = -t^2 + 2\) and \(y = t + 1\), plotting points serves as a crucial step in visualizing the behavior of these equations on the coordinate plane. To start, create a table to calculate the x and y coordinates corresponding to various values of \(t\). This exercise breaks \(t\) down into simpler integer values, such as -2, -1, 0, 1, and 2, to make calculations more manageable.

• For instance, if \(t = -2\), substitute into each equation to find \(x = -2\) and \(y = -1\).
• Repeat for other \(t\) values, such as \(-1, 0, 1,\) and \(2\), to generate respective \((x, y)\) pairs.

After filling your table with \((t, x, y)\) values, plot these points on your graph. It's important to remember that these points will provide a snapshot of the parametric equations and offer insight into the curve's appearance.

Coordinate Plane

The coordinate plane acts as our canvas to graphically represent the parametric equations. It is divided into four quadrants by the x-axis and y-axis, where each point plotted from our parametric equations is shown as a specific coordinate based on its x and y values.When you plot these coordinates \((x, y)\) on the graph:

• Start by labeling each axis accordingly. Typically, the x-axis is horizontal while the y-axis is vertical.
• Each point represents a unique position on this plane. For example, the point \((-2, -1)\) lies 2 units left along the x-axis and 1 unit down along the y-axis from the origin \((0,0)\).

As you plot these points, the coordinate plane helps to clearly visualize the path formed as \(t\) shifts between -2 and 2. This visual representation is key to understanding the form and direction of the resulting path, a parabolic curve in this instance.

Parabolic Path

From plotting the points of the given parametric equations \(x = -t^2 + 2\) and \(y = t + 1\), the plotted points can be connected with a smooth, continuous curve. This curve represents a parabolic path, which is a specific type of curve that resembles the graph of a quadratic function.In our context:

• The path is shaped like a downward-opening parabola because the larger the absolute value of \(t\), the more negative the corresponding \(x\) value, due to the \(-t^2\) term.
• The \(y = t + 1\) simply adjusts the height of each point relative to the t-value, causing the curve to shift upwards as \(t\) increases.

Visualizing this parabolic path is essential for recognizing the behavior of parametric curves, aiding in the comprehension of more complex mathematical systems which rely on similar concepts. Understanding the shape and direction of the path provides context for what the equations represent in practical terms.