Problem 51 - Calculus for Biology and Medicine

Question

Parameterize the equation of the line given in standard form. $$3 x+2 y-1=0$$

Step-by-Step Solution

Answer

The parametric equations are \(x(t) = 2t\) and \(y(t) = \frac{1}{2} - 3t\).

Step 1: Identify the Standard Form

The given equation is in the standard form, which is \(Ax + By + C = 0\). In this case, \(A = 3\), \(B = 2\), and \(C = -1\).

Step 2: Convert to Slope-Intercept Form

To find the parameterization, it's helpful to express the equation in slope-intercept form, \(y = mx + b\). Solve for \(y\) by isolating it: \[3x + 2y - 1 = 0 \Rightarrow 2y = -3x + 1 \Rightarrow y = -\frac{3}{2}x + \frac{1}{2}\]So, the slope \(m\) is \(-\frac{3}{2}\) and the y-intercept \(b\) is \(\frac{1}{2}\).

Step 3: Use Parameterization Formulas

Line parameterization can use a parameter \(t\) where \(x\) and \(y\) are functions of \(t\). Choose a point on the line (for convenience, use the y-intercept point \((0, \frac{1}{2})\)) and the direction vector derived from the slope. The direction vector from slope \(m = -\frac{3}{2}\) for change in \(x\) and \(y\) is \( \langle 2, -3 \rangle \).

Step 4: Parameterization of the Line

Using parameter \(t\), parameterize the equations for \(x\) and \(y\): \[\begin{align*}x(t) &= 0 + 2t \y(t) &= \frac{1}{2} - 3t\end{align*}\]Thus, the parametric equations are \(x(t) = 2t\) and \(y(t) = \frac{1}{2} - 3t\).

Key Concepts

Standard Form Equation

The standard form of an equation for a line is expressed as \(Ax + By + C = 0\). This is a versatile form that highlights the coefficients directly. Here, \(A\), \(B\), and \(C\) represent real numbers. In our equation, \(3x + 2y - 1 = 0\), we specifically identify:

• \(A = 3\)
• \(B = 2\)
• \(C = -1\)

Standard form is particularly useful for performing algebraic operations, like finding intercepts quickly or converting to other forms. When we set \(x = 0\), we can immediately solve for \(y\) to find the y-intercept, and vice versa for the x-intercept.
Additionally, equations in this form can work effectively in integer arithmetic, avoiding fractions. However, for graphing purposes, converting to other forms may provide better insights into the line's characteristics.

Slope-Intercept Form

The slope-intercept form of a linear equation is \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. It's highly useful for graphing and understanding how a line behaves visually on a coordinate plane.
From our example, converting \(3x + 2y - 1 = 0\) into slope-intercept form involves solving for \(y\). This yields \(y = -\frac{3}{2}x + \frac{1}{2}\). Here:

• The slope \(m\) is \(-\frac{3}{2}\), indicating the line falls as it moves from left to right.
• The y-intercept \(b\) is \(\frac{1}{2}\), showing where the line crosses the y-axis.

Knowledge of the slope is crucial for understanding line steepness and direction. Positive slopes rise, negative slopes fall. Slope-intercept form gives immediate visual cues helpful for sketching and analyzing linear behavior.

Parametric Equations

Parametric equations describe a line by expressing its coordinates as functions of a parameter \(t\). This method offers a dynamic view of a line or curve. For instance, we use the equations \(x(t) = 2t\) and \(y(t) = \frac{1}{2} - 3t\) from the standard form.
Here's how it works:

• Select a point on the line – we use \((0, \frac{1}{2})\), the y-intercept.
• Find a direction vector from the slope \(-\frac{3}{2}\). The vector \(\langle 2, -3 \rangle\) is derived from the change in \(x\) and \(y\).
• Parameter \(t\) represents the movement along the line. While \(t\) changes, \(x(t)\) and \(y(t)\) define points on the line.

• Setting \(t = 0\) gives the base point, while changes in \(t\) map the line's direction and extension.

Parametric equations are powerful for describing motion and curves, connecting geometry with algebraic representation in versatile ways.