Problem 8 - 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. $$x=t^{2}, \quad y=t^{4}+1$$

Step-by-Step Solution

Answer

The rectangular equation is \(y = x^2 + 1\).

Step 1: Understand the Given Parametric Equations

We are given the equations \(x = t^2\) and \(y = t^4 + 1\). Our task is to sketch the curve represented by these equations and then find a rectangular equation by eliminating the parameter \(t\).

Step 2: Sketch the Curve

To sketch the curve, we can create a table of values for \(t\), \(x\), and \(y\). Try some values like \(t = -2, -1, 0, 1, 2\). Calculate \(x\) and \(y\) for those values: \((t, x, y)\) values are \((-2, 4, 17)\), \((-1, 1, 2)\), \((0, 0, 1)\), \((1, 1, 2)\), \((2, 4, 17)\). Plot these points, and note the symmetry about the x-axis since \(t\) values are squared.

Step 3: Eliminate the Parameter

We begin by expressing \(t\) in terms of \(x\) from \(x = t^2\). Solving for \(t\) gives \(t = \pm \sqrt{x}\). Substitute \(t\) into \(y = t^4 + 1\) to get: \(y = (\pm \sqrt{x})^4 + 1 = x^2 + 1\). This is the rectangular equation of the curve.

Step 4: Analyze the Rectangular Equation

The derived equation \(y = x^2 + 1\) represents a parabola opening upwards with its vertex at (0, 1). Since both expressions of \(t\) yield the same rectangular equation, and \(t^2\) results in non-negative values, the domain for \(x\) is \([0, \infty)\).

Key Concepts

Rectangular Coordinates

Rectangular coordinates, also known as Cartesian coordinates, are a way to define every point in a plane using two numbers, x and y. The x-value tells you how far to move horizontally, while the y-value tells you how far to move vertically from a starting point called the origin (0,0).
In our exercise, we derived the rectangular equation from the given parametric equations. The original equations were expressed in terms of a parameter \( t \). However, in rectangular coordinates, we want to express everything just using \( x \) and \( y \).
By eliminating the parameter \( t \), we found a direct relationship between \( x \) and \( y \): the equation \( y = x^2 + 1 \). This transformation allows us to sketch the curve directly using the familiar x and y axes, without needing to calculate intermediate values for \( t \).
This approach simplifies the curve analysis as it revolves entirely around the Cartesian plane, aligning fully with the usual geometric intuitions based on horizontal and vertical movements.

Parabola

A parabola is a specific type of curve defined by a quadratic equation, typically written in the form \( y = ax^2 + bx + c \). In our exercise, after eliminating the parameter \( t \), we derived the equation \( y = x^2 + 1 \).
This is a simple form of a parabola, where \( a = 1 \) and both \( b \) and \( c \) are absent, except for the constant \(+1\). What this means is our parabola opens upwards, without any tilts, due to the non-zero \( a \), and it is shifted one unit above the horizontal (x-axis).
Key characteristics of this parabola here include:

• The vertex at (0, 1), the point where the curve changes direction.
• The axis of symmetry, which is a vertical line through the vertex, \( x = 0 \).
• It extends indefinitely in the upward direction from the vertex.

Understanding these features helps in sketching the parabola simply and accurately in rectangular coordinates.

Curve Sketching

Curve sketching involves drawing a graph based on the properties of an equation. It helps visually represent the relationship between variables, here \( x \) and \( y \).
The first step in sketching the curve from our original parametric equations involved calculating points by substituting values for \( t \). This helped us understand the symmetry of those points. When plotted together, these points suggested the shape of the curve.
Converting the parametric equations into the rectangular coordinate equation \( y = x^2 + 1 \) allowed us to use the familiar process of sketching a parabola, as discussed earlier. The transformation confirmed the upward opening shape, its vertex, and symmetrical properties.
By knowing these critical features:

• The direction the parabola opens (upwards).
• The vertex and axis of symmetry location (\( x = 0 \) and (0, 1)).
• The domain, which here only includes non-negative \( x \) values (\( x \geq 0 \)).

Students can graph the curve with confidence, even without detailed calculations for each point.