Problem 52 - Calculus for Biology and Medicine

Question

Parameterize the equation of the line given in standard form. $$ x-2 y+5=0 $$

Step-by-Step Solution

Answer

The parametric equations are \( x = -5 + 5t \) and \( y = 2.5t \).

Step 1: Understand the Standard Form Equation

The given equation \( x - 2y + 5 = 0 \) is in standard form, which is generally written as \( Ax + By + C = 0 \) where \( A = 1 \), \( B = -2 \), and \( C = 5 \). Our goal is to express this line in parametric form.

Step 2: Find Points on the Line

To parameterize the line, we need a direction vector and a point on the line. Set \( y = 0 \) to find a point: \( x + 5 = 0 \), so \( x = -5 \). One point is \((-5, 0)\). Another point can be found by setting \( x = 0 \): \( -2y + 5 = 0 \), so \( y = 2.5 \). Another point is \((0, 2.5)\).

Step 3: Determine the Direction Vector

From two points on the line, we can find the direction vector. Taking points \((-5, 0)\) and \((0, 2.5)\), the direction vector \( \vec{d} \) is \( \langle 0 - (-5), 2.5 - 0 \rangle = \langle 5, 2.5 \rangle \).

Step 4: Write the Parametric Equations

Using the point \((-5,0)\) and the direction vector \(\langle 5, 2.5 \rangle\), the parametric equations are \( x = -5 + 5t \) and \( y = 0 + 2.5t \).

Step 5: Finalize and Simplify If Needed

The parametric form of the line is \( x = -5 + 5t \) and \( y = 2.5t \). You can simplify the expression by leaving it as is or factoring common terms if desired.

Key Concepts

Direction Vector

In order to understand the concept of a direction vector, imagine a line in a two-dimensional space. A direction vector gives you a sense of how this line extends infinitely in both directions. Simply put, it determines the 'slope' or 'tilt' of the line in space. The direction vector \( \vec{d} \) is essentially a way to 'travel' along the line. If you have two points on the line, say \((-5, 0)\) and \((0, 2.5)\), the direction vector can be derived from these points. The components of the vector are obtained by subtracting the coordinates of the starting point from the coordinates of the ending point:

• The x-component: \( 0 - (-5) = 5 \)
• The y-component: \( 2.5 - 0 = 2.5 \)

Hence, the direction vector is \( \langle 5, 2.5 \rangle \), pointing from one on-line point to another. This vector helps us move along the line by any scalar multiple, \( t \), hence parameterizing it.

Standard Form Equation

Standard Form is one of the ways to express a linear equation, typically written as \( Ax + By + C = 0 \). It is simple and concise, gathering all terms on one side with zero on the other. In this form, the coefficients \( A \), \( B \), and \( C \) play a specific role:

• \( A \): coefficient of \( x \)
• \( B \): coefficient of \( y \)
• \( C \): constant term

Analyzing the equation \( x - 2y + 5 = 0 \), you'll see:

• \( A = 1 \)
• \( B = -2 \)
• \( C = 5 \)

Understanding this format is crucial when transitioning to other forms, such as parametric equations, as it provides the necessary coefficients and constants needed to move forward.

Line Parameterization

Line parameterization is the process of expressing a line using one parameter, often \( t \), which allows you to find every possible point on the line. The parametric form is highly useful, especially in vector mathematics, because it extends the idea of a direction vector to describe how far and in what direction you travel from an initial point on the line. To parameterize a line, you begin with:

• A point on the line (e.g., \((-5, 0)\))
• A direction vector (e.g., \( \langle 5, 2.5 \rangle \))

Using these, the parametric equations are formed:

• For \( x \): \( x = -5 + 5t \)
• For \( y \): \( y = 0 + 2.5t \)

As \( t \) varies over all real numbers, \( (x, y) \) traces out every point on the line. This allows us to not only identify points but also understand the line's spatial orientation and extensions.