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\(3-24=A\) pair of parametric equations is given. (a) Sketch the curve represented by the parametric equations. (b) Find a rectangular coordinate equation for the curve by eliminating the parameter. $$ x=6 t-4, \quad y=3 t, \quad t \geq 0 $$

Short Answer

Expert verified
The rectangular equation is \( y = \frac{x + 4}{2} \), a line starting from (-4, 0) with a slope of 0.5.

Step by step solution

01

Parameterize x and y

We have the parametric equations given by:\[ x = 6t - 4 \]\[ y = 3t \]These equations show how both x and y depend on the parameter \( t \).
02

Express t in terms of y

To eliminate the parameter \( t \), we need to express \( t \) in terms of one of the rectangular coordinates. Let's solve for \( t \) in terms of \( y \):\[ y = 3t \]\[ t = \frac{y}{3} \]
03

Substitute t in the x equation

Now, substitute the expression for \( t \) into the equation for \( x \):\[ x = 6\left(\frac{y}{3}\right) - 4 \]This simplifies to:\[ x = 2y - 4 \]
04

Rewrite as a rectangular equation

The rectangular equation of the curve is given by:\[ x = 2y - 4 \]Rearrange it to the familiar form by solving for \( y \):\[ x + 4 = 2y \]\[ y = \frac{x + 4}{2} \]
05

State the range for the parameter

Since \( t \geq 0 \), when \( t = 0 \), \( y = 0 \) and as \( t \to \infty \), \( y \to \infty \). This means the line starts from \( y = 0 \) and extends upwards. For \( x \), starting from \( x = -4 \) (when \( t = 0 \)) and increasing as \( y \) increases.
06

Sketch the graph

Plot a line starting at the point \((-4, 0)\) with a slope of \(\frac{1}{2}\). This line extends upwards to the right, indicating how \( x \) and \( y \) increase as the parameter \( t \) increases.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rectangular Equation
In mathematics, a **rectangular equation** is an equation that uses the rectangular coordinate system (usually coordinates x and y) to describe a curve or a line on a plane. In the given problem, the goal is to express the parametric equations in a single, unified form that uses the standard Cartesian coordinates.
  • Parametric equations are provided as \( x = 6t - 4 \) and \( y = 3t \). These parametric equations describe the relationship between x, y, and the parameter \( t \).
  • To derive a rectangular equation, one would normally eliminate the parameter to find a relationship solely between x and y.
In solving the exercise, the parameter \( t \) was removed through a process called parameter elimination. As a result, the final rectangular equation is formulated as \( y = \frac{x + 4}{2} \), describing a line in the xy-plane.
Coordinate Conversion
**Coordinate conversion** in this context refers to transitioning from parametric equations to a rectangular equation. This process involves replacing the parameter variable within the parametric equations to involve only the standard Cartesian coordinates.
  • Begin by solving one of the parametric equations for the parameter \( t \). In the solutions provided, the equation for \( y \) was used: \( y = 3t \), leading to \( t = \frac{y}{3} \).
  • Next, substitute this expression for \( t \) into the other parametric equation \( x = 6t - 4 \).
  • From here, simplify the expression to establish a compact equation in terms of x and y. The substitution yields \( x = 2y - 4 \), which can then be rearranged to a more familiar line equation \( y = \frac{x + 4}{2} \).
This conversion allows one to view the path traced by the parameterized curve in the regular Cartesian plane without relying on the parameter.
Parameter Elimination
**Parameter elimination** is a central technique in moving from parametric to rectangular equations. This involves removing the parameter (in this case, \( t \)) from the equations that represent x and y, leaving a direct relationship between these two variables.
  • First, isolate the parameter. In this example, start with the equation \( y = 3t \) to express \( t \) in terms of the Cartesian coordinate \( y \), resulting in \( t = \frac{y}{3} \).
  • Then, substitute this expression into the parametric equation for \( x \) which is \( x = 6t - 4 \). Substituting \( t = \frac{y}{3} \) into this equation results in \( x = 2y - 4 \).
  • This expression is further simplified and rearranged to provide the final rectangular equation, \( y = \frac{x + 4}{2} \).
The parameter elimination process simplifies the understanding and graphing of the curve by using the more conventional Cartesian form.

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