Problem 32 - A Graphical Approach to Precalculus with Limits

Question

Find a rectangular equation for each curve and graph the curve. $$x=1+2 \sin t, y=2+3 \cos t ; \text { for } t \text { in }[0,2 \pi]$$

Step-by-Step Solution

Answer

The rectangular equation for the curve is \( \frac{(x - 1)^2}{4} + \frac{(y - 2)^2}{9} = 1 \). This represents an ellipse centered at (1, 2).

Step 1: Identify the Parametric Equations

We are given the parametric equations: \( x = 1 + 2\sin(t) \) and \( y = 2 + 3\cos(t) \). Our goal is to eliminate the parameter \( t \) and find a rectangular equation that relates \( x \) and \( y \) directly.

Step 2: Solve for Sin and Cos Terms

To eliminate the parameter \( t \), we need to isolate \( \sin(t) \) and \( \cos(t) \) in each equation. From \( x = 1 + 2\sin(t) \), we get \( \sin(t) = \frac{x - 1}{2} \). From \( y = 2 + 3\cos(t) \), we get \( \cos(t) = \frac{y - 2}{3} \).

Step 3: Use the Pythagorean Identity

Recall the Pythagorean identity: \( \sin^2(t) + \cos^2(t) = 1 \). Substitute \( \sin(t) = \frac{x - 1}{2} \) and \( \cos(t) = \frac{y - 2}{3} \) into this identity. This gives: \( \left( \frac{x - 1}{2} \right)^2 + \left( \frac{y - 2}{3} \right)^2 = 1 \).

Step 4: Simplify the Equation

Simplify the equation \( \left( \frac{x - 1}{2} \right)^2 + \left( \frac{y - 2}{3} \right)^2 = 1 \). This gives us the rectangular equation of the ellipse: \( \frac{(x - 1)^2}{4} + \frac{(y - 2)^2}{9} = 1 \).

Step 5: Graph the Curve

The equation \( \frac{(x - 1)^2}{4} + \frac{(y - 2)^2}{9} = 1 \) represents an ellipse centered at \( (1, 2) \). The semi-axis along the \( x \)-axis has a length of 2 (horizontal direction) and the semi-axis along the \( y \)-axis has a length of 3 (vertical direction). Graph the curve accordingly.

Key Concepts

Parametric Equations

Parametric equations are a pair of equations that express a set of related quantities as explicit functions of an independent parameter, usually denoted as \( t \). In most cases, this parameter represents time. In our problem, we are given the parametric equations \( x = 1 + 2\sin(t) \) and \( y = 2 + 3\cos(t) \). These equations describe the relationship between \( x \) and \( y \) coordinates as functions of \( t \), effectively allowing us to trace a curve in the Cartesian plane as \( t \) varies over a particular interval.

• \( x \) and \( y \) are expressed separately in terms of \( t \).
• The main goal is to eliminate \( t \) to find a rectangular equation that relates \( x \) and \( y \) directly.

To convert these parametric equations into a rectangular form, we typically solve each equation for the sinusoidal component in terms of \( x \) and \( y \) and then use a known identity to find a single equation that links \( x \) and \( y \) without \( t \). This process is crucial for graphing purposes, as it allows us to see the full shape of the curve on a Cartesian plane without considering the parameter \( t \).

Pythagorean Identity

The Pythagorean identity is a fundamental equation in trigonometry that states \( \sin^2(t) + \cos^2(t) = 1 \). This identity plays a key role in converting parametric equations into a rectangular form.
In our problem, we have:

• \( \sin(t) = \frac{x - 1}{2} \) from the equation \( x = 1 + 2\sin(t) \).
• \( \cos(t) = \frac{y - 2}{3} \) from the equation \( y = 2 + 3\cos(t) \).

By substituting these expressions into the Pythagorean identity, we get \( \left( \frac{x - 1}{2} \right)^2 + \left( \frac{y - 2}{3} \right)^2 = 1 \).
This calculation allows us to transform the separate sin and cos parametric forms into a single rectangular equation. The use of the Pythagorean identity here is essential for eliminating the parameter \( t \), effectively translating the curve into a form easier for plotting and interpretation in the xy-plane.

Graphing Ellipses

Graphing an ellipse requires understanding its standard form equation: \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \). Here, \((h, k)\) is the center of the ellipse, \(a\) is the semi-major axis, and \(b\) is the semi-minor axis.
For our solution, we ended up with \( \frac{(x-1)^2}{4} + \frac{(y-2)^2}{9} = 1 \). This implies an ellipse centered at \((1, 2)\).

• The term \((x-1)^2/4\) indicates a horizontal semi-axis of length 2 (\(a = 2\)).
• \((y-2)^2/9\) indicates a vertical semi-axis of length 3 (\(b = 3\)).

When graphing:

• Start by plotting the center at \((1, 2)\).
• Draw the axes with the specified lengths: 2 units horizontally from the center in both directions, 3 units vertically from the center in both directions.

These steps help you accurately sketch the ellipse on a graph, allowing a clear visual representation of the solution.