Problem 16 - A Graphical Approach to Precalculus with Limits

Question

For each plane curve, (a) graph the curve, and (b) find a rectangular equation for the curve. $$x=2 t, y=t+1 ; \text { for } t \text { in }[-2,3]$$

Step-by-Step Solution

Answer

Rectangular equation: \(y = \frac{x}{2} + 1\); graph is a line from (-4,-1) to (6,4).

Step 1: Parametric Equations

We are given the parametric equations for the plane curve: \(x = 2t\) and \(y = t + 1\), with \(t\) ranging from -2 to 3. Our goal is to find the rectangular equation of the curve and sketch it.

Step 2: Express \(t\) in terms of \(x\)

We start by manipulating the equation for \(x\): \(x = 2t\). Solving for \(t\), we get \(t = \frac{x}{2}\).

Step 3: Substitute \(t\) in the \(y\) equation

Next, we'll substitute \(t = \frac{x}{2}\) into the equation for \(y\): \(y = t + 1\). Replace \(t\) to get \(y = \frac{x}{2} + 1\). This gives us the rectangular equation of the curve.

Step 4: Sketch the Graph of the Rectangular Equation

The rectangular equation \(y = \frac{x}{2} + 1\) is a linear equation with a slope of \(\frac{1}{2}\) and a y-intercept of 1. Plot this line and mark the endpoints corresponding to \(t = -2\) and \(t = 3\). For \(t = -2\), \(x = 2(-2) = -4\) and \(y = -2 + 1 = -1\); for \(t = 3\), \(x = 2(3) = 6\) and \(y = 3 + 1 = 4\). Draw the line segment between these points.

Key Concepts

Rectangular Equation

In mathematics, a rectangular equation, also known as a Cartesian equation, relates two variables, x and y, in one expression using Cartesian coordinates. This concept shines in its ability to describe geometric shapes, specifically curves, without needing to reference an additional parameter. In the context of the given problem, our parametric equations are:

• \( x = 2t \)
• \( y = t + 1 \)

 To derive the rectangular equation, we first isolate the parameter \( t \) from one of the parametric equations. From \( x = 2t \), solving for \( t \) gives us:

• \( t = \frac{x}{2} \)

 This expression for \( t \) is substituted into the second equation, \( y = t + 1 \). Replacing \( t \) yields:

• \( y = \frac{x}{2} + 1 \)

 This formula represents the rectangular equation for the curve, now devoid of the parameter \( t \). It's a linear relationship between \( x \) and \( y \), making it straightforward to graph or further analyze.

Plane Curve

A plane curve is a curve that lies entirely in a single plane. These curves can have various shapes and can be described in numerous ways, like parametric or rectangular equations. In this scenario, the plane curve is initially expressed through the parametric form using the equations \( x = 2t \) and \( y = t + 1 \).

What makes a plane curve interesting is how it can transition between different mathematical representations. Here, we've translated it from its parametric form to a simpler rectangular form, \( y = \frac{x}{2} + 1 \). This transformation indicates that our curve is indeed a straight line, as demonstrated by its linear equation.

It's important to visualize the path taken by the curve within its specified parameter range, from \( t = -2 \) to \( t = 3 \). While this may sound technical, it simply means we determine the starting and ending points of the curve that is plotted on a coordinate plane:

• For \( t = -2 \), the point is \((-4, -1)\).
• For \( t = 3 \), the point is \((6, 4)\).

This aids in fully grasping the span and reach of the plane curve on the graph.

Graphing Parametric Equations

Graphing parametric equations involves creating a visual representation of a set of equations that express coordinates \((x, y)\) in terms of a third variable, usually \( t \). In the given problem, our task was to graph the curve defined by \( x = 2t \) and \( y = t + 1 \) over a given range. Let's break down how to approach this task effectively.

To begin, identify the parameter’s range from the problem — here, \( t \) spans from -2 to 3. This pattern is essential because it confines our graph to a specific segment or path. For specific \( t \) values within this range, compute the corresponding \( x \) and \( y \) values:

• When \( t = -2 \), \( x = -4 \) and \( y = -1 \).
• When \( t = 3 \), \( x = 6 \) and \( y = 4 \).

These points identify where on the Cartesian plane the curve starts and ends.

Now, use the rectangular equation derived earlier, \( y = \frac{x}{2} + 1 \), to draw the line connecting these points. The graph is linear with a slope of \( \frac{1}{2} \) and a y-intercept at \( 1 \), indicating a steady, gentle incline from left to right. Parametric equations often graph intricate curves, but in this instance, our graph showcases a straightforward line, making visualization easier for this set of equations.