Problem 15 - 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. $$x=\cos 2 t, \quad y=\sin t, \quad-\pi \leq t \leq \pi$$

Step-by-Step Solution

Answer

The equation is \(x = 1 - 2y^2\). Orientation: starts at (1,0), leftward loop.

Step 1: Identifying Parametric Equations

We start with the parametric equations given: \(x = \cos 2t\) and \(y = \sin t\). These equations describe the curve \(C\).

Step 2: Relationship Between \(x\) and \(t\)

The equation \(x = \cos 2t\) can use trigonometric identities. We know that \(\cos 2t = 2\cos^2 t - 1\) and also \(\cos 2t = 1 - 2\sin^2 t\). This gives us a starting point to relate \(x\) with another trigonometric function.

Step 3: Relationship Between \(y\) and \(t\)

Given \(y = \sin t\), express \(\sin t\) in terms of \(y\): \(\sin t = y\).

Step 4: Expressing \(\cos 2t\) in terms of \(y\)

Using the double-angle identity for cosine, \(\cos 2t = 1 - 2\sin^2 t\), substitute \(\sin t = y\) to get \(\cos 2t = 1 - 2y^2\).

Step 5: Formulating the Equation in \(x\) and \(y\)

Substitute \(x = \cos 2t\) with the expression from the previous step to get: \(x = 1 - 2y^2\).

Step 6: Sketching the Graph

To sketch the graph of the equation \(x = 1 - 2y^2\), recognize this is a sideways parabola opening towards negative \(x\)-axis. The vertex is at \((1,0)\).

Step 7: Indicating the Orientation on Graph

As \(t ightarrow -\pi\) to \(t \rightarrow \pi\), \(y = \sin t \) traces from \(0\to 1 \to 0 \to -1 \to 0\). This means the graph begins at \((1,0)\), moves leftward through \((0,1)\), \((-3,0)\), \((0,-1)\), and back to \((1,0)\). Mark these points and draw arrows along this path.

Key Concepts

Trigonometric Identities

Trigonometric identities are essential tools in mathematics. They involve relationships between the trigonometric functions, such as sine and cosine. These identities can simplify expressions and solve equations. In this context, the double-angle identity for the cosine function is particularly useful.

The identity says that:

• \(\cos 2t = 2\cos^2 t - 1\)
• \(\cos 2t = 1 - 2\sin^2 t\)

These expressions allow us to convert and express cosine in terms of sine or vice versa.
In the problem, we use the identity \(\cos 2t = 1 - 2\sin^2 t\) to eliminate the parameter \(t\) and express everything in terms of \(y\). By substituting \(\sin t = y\), we replaced \(t\) and described the curve just with \(x\) and \(y\). Thus, we found the equation for the graph as \(x = 1 - 2y^2\).
This transformation is key to moving from a parametric form to a clearer algebraic equation.

Graph Orientation

Understanding the orientation of a graph is important in parametric equations. Orientation tells us the direction in which a curve is traced as the parameter changes. Here, this involves analyzing how the values of \(t\) affect \(x\) and \(y\).

For the equation \(x = 1 - 2y^2\), by setting the range \(-\pi \leq t \leq \pi\), we detail the journey of the point on the curve. As \(t\) progresses from \(-\pi\) to \(\pi\):

• The point starts at \((1,0)\)
• Moves to \((0,1)\)
• Progresses to \((-3,0)\)
• Descends to \((0,-1)\)
• Finally returns back to \((1,0)\)

The path describes a clockwise loop from its starting position. Arrows on a graph provide a visual cue for direction, helping in understanding this flow along a curve. Thus, orientation adds critical dynamic information, illustrating how a point travels through the curve over the given range of \(t\).

Parabolic Curve

In mathematics, a parabolic curve is one of the standard types of curves, shaped like a parabola. Parabolas can open in different directions: up, down, left, or right, depending on their equations. This exercise presents a unique case, a sideways or horizontal parabola.

The equation \(x = 1 - 2y^2\) reveals such a curve commonly called a 'horizontal parabola'. Let's understand its characteristics:

• It opens leftwards since its main axis aligns horizontally towards the negative x-axis.
• The vertex of this parabola is located at \((1,0)\).

Drawing this parabola involves determining key points that satisfy the equation:

• \(x = 1\), when \(y = 0\) (vertex point)
• \(x = 0\), when \(y = 1\) or \(y = -1\)
• \(x = -3\), when \(y = 0\)

the graph symmetry reflects these features, forming a smooth curve from right to left. Recognizing these properties enriches our comprehension of graphs and plotting complex curves efficiently in coordinate systems.