Problem 11 - Precalculus Mathematics for Calculus

Question

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. $$x=4 t^{2}, \quad y=8 t^{3}$$

Step-by-Step Solution

Answer

The rectangular equation is \( y = x^{3/2} \). Sketched curve shows it crossing through quadrants with changing \( t \).

Step 1: Understanding the Parametric Equations

We are given the parametric equations: \( x = 4t^2 \) and \( y = 8t^3 \). These equations define the path of a curve in terms of the parameter \( t \). We will first sketch the curve based on these parametric equations.

Step 2: Sketching the Curve

To sketch the curve, calculate several points by substituting different \( t \) values into the parametric equations:\- For \( t = 0 \), \( x = 0 \), \( y = 0 \).\- For \( t = 1 \), \( x = 4 \), \( y = 8 \).\- For \( t = -1 \), \( x = 4 \), \( y = -8 \).\- For \( t = 2 \), \( x = 16 \), \( y = 64 \).\Plot these points to form a sketch which shows a curve that moves through quadrants the curve is upwards or downwards depending on the sign of \( t \).

Step 3: Eliminating the Parameter

We need to eliminate the parameter \( t \) to find a rectangular-coordinate equation. Start by expressing \( t \) in terms of \( x \):\\[ x = 4t^2 \Rightarrow t^2 = \frac{x}{4} \Rightarrow t = \pm \sqrt{\frac{x}{4}} \].\For \( y \), substitute \( t = \sqrt{\frac{x}{4}} \) leading to \( y = 8t^3 \).

Step 4: Substituting 't' into 'y' equation

Substitute \( t = \sqrt{\frac{x}{4}} \) into \( y = 8t^3 \): \\[ y = 8 \left( \sqrt{\frac{x}{4}} \right)^3 \].\This simplifies to: \\[ y = 8 \left( \frac{x}{4} \right)^{3/2} \].

Step 5: Rectangular Equation

Simplify the expression: \[ y = 8 \left( \frac{x}{4} \right)^{3/2} = 8 \times \frac{x^{3/2}}{8} = x^{3/2} \].\Thus, the rectangular-coordinate equation for the curve is \( y = x^{3/2} \).

Key Concepts

Rectangular-coordinate equation

A rectangular-coordinate equation is an equation written in terms of the standard Cartesian coordinates, typically labeled as \( x \) and \( y \). This type of equation describes the relationship between \( x \) and \( y \) without involving any parameters. In our exercise, we originally have two parametric equations: \( x = 4t^2 \) and \( y = 8t^3 \). By eliminating the parameter \( t \), the goal is to convert these into a single rectangular-coordinate equation.Understanding how to form a rectangular-coordinate equation is crucial for analyzing curves and their behaviors since it translates complex parameter-dependent movements into a simple \( y = f(x) \) format. This can facilitate easier graphing and comprehension of geometric properties.

Eliminating the parameter

Eliminating the parameter from parametric equations involves solving one of the equations for the parameter and substituting it into the other equation. This process is regardless of whether the parameter is time (like \( t \)) or another variable used to describe parts of the curve. In our given equations, \( x = 4t^2 \) is solved for \( t \):\[ t = \pm \sqrt{\frac{x}{4}} \]Notice the \( \pm \) sign introduces choices:

• \( t = \sqrt{\frac{x}{4}} \)
• \( t = -\sqrt{\frac{x}{4}} \)

Both options will reflect whether the path is traced in a positive or negative direction. Substituting \( t \) into \( y = 8t^3 \), you simplify to find:\[ y = 8 \left( \frac{x}{4} \right)^{3/2} = x^{3/2} \]This results in the rectangular-coordinate equation. The \( \pm \) disappears because raising a positive number or a negative number to an odd power results in the same absolute structure.

Curve sketching

Curve sketching with parametric equations requires you to first evaluate various points by choosing suitable values for the parameter. For example, evaluating \( t = 0, 1, -1, 2 \) inside the original parametric equations \( x = 4t^2 \) and \( y = 8t^3 \) gives points like:

• (0, 0) when \( t = 0 \)
• (4, 8) when \( t = 1 \)
• (4, -8) when \( t = -1 \)
• (16, 64) when \( t = 2 \)

Plotting these points on a coordinate system where \( x \) is the horizontal axis and \( y \) the vertical helps in sketching the curve. This approach shows how the curve positions in different quadrants based on the sign and value of \( t \). It's a graphical representation of how parametric equations describe a path over time or as the parameter varies.

Precalculus

Precalculus is a mathematical course that acts as a bridge between algebra, trigonometry, and calculus. It provides tools and concepts that are essential for understanding the deeper complexities of calculus. Parametric equations and their conversion to rectangular-coordinate equations form a key topic in precalculus. Understanding these topics prepares students to tackle complex problems, appreciate the origin and behavior of curves, and build intuition for calculus involving rates of change and areas under curves. It's about learning to visualize and manipulate equations in different forms to see the bigger picture in the overall function or geometric shape.