Problem 47 - Calculus for Biology and Medicine

Question

In Problems \(47-50\), find the parametric equation of the line in the \(x-y\) plane that goes through the given points. Then eliminate the parameter to find the equation of the line in standard form. $$ (-1,2) \text { and }(3,4) $$

Step-by-Step Solution

Answer

The line's parametric form is \(x = -1 + 4t\), \(y = 2 + 2t\). Standard form is \(x - 2y = -5\).

Step 1: Identify Direction Vector

To find the parametric equation of the line, we first need to determine the direction vector. Given points are \((-1, 2)\) and \((3, 4)\). The direction vector can be found by subtracting the coordinates of these points: \((3 - (-1), 4 - 2) = (4, 2)\).

Step 2: Write the Parametric Equations

Using one of the points, say \((-1, 2)\), and the direction vector \((4, 2)\), the parametric equations of the line are: \(x = -1 + 4t\) and \(y = 2 + 2t\), where \(t\) is the parameter.

Step 3: Eliminate the Parameter

To eliminate the parameter \(t\), solve the parametric equation for \(t\) in terms of \(x\): \(x = -1 + 4t \Rightarrow t = \frac{x + 1}{4}\). Substitute this expression for \(t\) into the equation for \(y\): \(y = 2 + 2t = 2 + 2\left(\frac{x + 1}{4}\right)\).

Step 4: Simplify and Rearrange

Substitute \(t = \frac{x + 1}{4}\): \(y = 2 + 2\left(\frac{x + 1}{4}\right)\). Simplify this to get \(y = 2 + \frac{x}{2} + \frac{1}{2}\). Combine terms: \(y = \frac{x}{2} + 2.5\). Rearrange it in standard form by multiplying through by 2: \(2y = x + 5\), or equivalently, \(x - 2y = -5\).

Key Concepts

Direction Vector

A direction vector is a fundamental concept when dealing with parametric equations. It indicates the direction in which a line or vector points in space. When calculating the direction vector from two given points, for instance, Point A \((-1, 2)\) and Point B \((3, 4)\), we subtract the components of Point A from Point B. This results in the vector \((3 - (-1), 4 - 2) = (4, 2)\).
The direction vector \(\overrightarrow{d} = (4, 2)\) tells us that for every unit increase in the parameter, the line moves 4 units in the x-direction and 2 units in the y-direction. This vector is crucial in forming the parametric equations of the line.
By representing where the line starts and how it progresses, the direction vector provides a complete picture of the line's trajectory in the plane.

Standard Form of a Line

The standard form of a linear equation is often expressed as \(Ax + By = C\), where \A, B,\ and \C\ are constants. This form is commonly used because it offers a clear, neat way to understand the relationship between x and y. After finding the parametric equations for a line, converting them to this standard form can be highly illuminating.
Using our previously determined parametric equations, \(x = -1 + 4t\) and \(y = 2 + 2t\), we eliminated the parameter \(t\) and derived the equation \(y = \frac{x}{2} + 2.5\). By multiplying through by 2 to clear the fraction, we eventually attain \(x - 2y = -5\).
This resulting equation presents the line in a format that's ready for comparison with other lines or for solving systems of equations, making standard form valuable in both academic and practical settings.

Parameter Elimination

Parameter elimination is the process that allows us to take parametric equations and convert them into a more traditional form of a line equation. This involves solving the parametric equations to remove the parameter, typically denoted by \(t\), which serves as a bridge between x and y values.
Let's illustrate this with our parametric equations \(x = -1 + 4t\) and \(y = 2 + 2t\). By solving the x-equation for \(t\), we get \(t = \frac{x+1}{4}\). Substituting this expression into the y-equation gives us \(y = 2 + 2\left(\frac{x+1}{4}\right)\).
Simplifying yields \(y = \frac{x}{2} + 2.5\), effectively removing \(t\) from equations, resulting in an equation that directly relates x and y. This not only simplifies computation but also reveals more about the geometric and algebraic properties of the line, offering deeper insights into its behavior.