Problem 6 - Precalculus: Functions and Graphs

Question

Find an equation in \(x\) and \(y\) whose graph contains the points on the curve \(C\). Sketch the graph of \(C\), and indicate the orientation. $$x=1-9 t, \quad y=3 t+1\quad t \text { in } \mathbb{R}$$

Step-by-Step Solution

Answer

The equation is \(x + 3y = 4\); the curve is oriented from right to left, upwards.

Step 1: Understanding the Parametric Equations

The given parametric equations are \(x = 1 - 9t\) and \(y = 3t + 1\). These equations describe the coordinates \((x, y)\) of a point on the curve \(C\) as the parameter \(t\) varies over all real numbers.

Step 2: Eliminate Parameter \(t\)

To find an equation in terms of \(x\) and \(y\), we need to eliminate the parameter \(t\). Start by solving the equation for \(x\), \(x = 1 - 9t\), to express \(t\) in terms of \(x\):\[ t = \frac{1 - x}{9} \]

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

Substitute \(t = \frac{1 - x}{9}\) into the equation for \(y\):\[ y = 3\left(\frac{1 - x}{9}\right) + 1 \]Simplify it:\[ y = \frac{3(1 - x)}{9} + 1 = \frac{3 - 3x}{9} + 1 \]\[ y = \frac{1 - x}{3} + 1 \]

Step 4: Simplify to Find the Equation in \(x\) and \(y\)

Simplify the expression to find the equation of the curve in terms of \(x\) and \(y\):Combine fractions:\[ y = \frac{1 - x}{3} + \frac{3}{3} \]\[ y = \frac{4 - x}{3} \]Multiply through by 3 to clear the fraction:\[ 3y = 4 - x \]Rearrange to standard form:\[ x + 3y = 4 \]

Step 5: Sketch the Graph and Indicate the Orientation

The equation \(x + 3y = 4\) is a linear equation, which represents a line.- To sketch, find the intercepts: - For \(x\)-intercept, set \(y = 0\) ➔ \(x = 4\). - For \(y\)-intercept, set \(x = 0\) ➔ \(y = \frac{4}{3}\).- Plot the intercepts and draw a line through them. To determine orientation, follow the change in \(t\):- As \(t\) increases, \(x = 1 - 9t\), so \(x\) decreases.- As \(t\) increases, \(y = 3t + 1\), so \(y\) increases.- Therefore, the curve is oriented from right to left, upwards in the plane.

Key Concepts

Graphing Parametric Curves

Graphing parametric curves allows us to represent paths traced out by moving points with the help of two separate equations, one for each coordinate. In the provided problem, the equations given are:

• \(x = 1 - 9t\)
• \(y = 3t + 1\)

These equations describe how a point moves along the curve as the parameter \(t\) changes.
By altering \(t\), the point traces out a curve. Here, \(t\) is any real number, meaning the curve can extend infinitely in the directions determined by the functions. Thus, plotting these equations involves calculating specific points by substituting different values of \(t\) into the equations. This generates a series of \((x, y)\) coordinates that, when plotted, trace out the path of the curve.
Additionally, understanding the orientation is crucial—this tells us the direction the point moves along the curve as \(t\) increases. For the given example, as \(t\) rises, \(x\) decreases while \(y\) increases, indicating the curve is oriented from right to left and upwards in the plane.

Eliminating the Parameter

Eliminating the parameter involves finding a direct relationship between \(x\) and \(y\), removing the dependency on \(t\). This simplification helps in visualizing and sketching the graph easily.
Starting with the given:

• \(x = 1 - 9t\)
• \(y = 3t + 1\)

The first step is expressing \(t\) in terms of \(x\):\[ t = \frac{1 - x}{9} \]Next, substitute this \(t\) into the equation for \(y\):\[ y = 3\left(\frac{1 - x}{9}\right) + 1 \]Simplifying gives:\[ y = \frac{4 - x}{3} \]Finally, converting to standard form yields the linear equation: \[ x + 3y = 4 \]This equation now represents the same curve as the parametric equations but in a more straightforward form that relates \(x\) and \(y\) directly.

Equation of a Line

The final equation we derived—\(x + 3y = 4\)—is the equation of a straight line. This configuration can be identified via its standard linear form, \(Ax + By = C\).
To plot the line, it’s useful to find its intercepts:

• For the \(x\)-intercept, set \(y = 0\), resulting in \(x = 4\).
• For the \(y\)-intercept, set \(x = 0\), resulting in \(y = \frac{4}{3}\).

These intercepts are key points that help in sketching the line on a graph. Once plotted, drawing a line through these points reveals the path along which the correlation between \(x\) and \(y\) holds true.
Lines are a foundational element in mathematics because they depict consistent relationships in a simple, linear format. Here, this connection showcases the direct proportional relationship gathered from the elimination of the parametric variable \(t\), thus simplifying our original parametric form into straightforward Cartesian coordinates.