Problem 16 - 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. $$x=\sin ^{2} t, \quad y=\cos t$$

Step-by-Step Solution

Answer

The sketch shows a vertical line segment oscillating between \( y = -1 \) and \( y = 1 \) at \( x = 0 \). The rectangular equation is \( y^2 = 1 - x \).

Step 1: Understanding the Parametric Equations

The given parametric equations are \( x = \sin^2 t \) and \( y = \cos t \). Here, \( t \) is the parameter, and the equations describe a curve in the \(xy\)-plane as \( t \) varies.

Step 2: Analyze the Range of Parameters

For \( x = \sin^2 t \), the range of \( x \) is between 0 and 1, as \( \sin^2 t \) is always non-negative and less than or equal to 1. For \( y = \cos t \), the range of \( y \) is between -1 and 1.

Step 3: Sketch the Curve

To sketch the curve, observe the behavior of \( y = \cos t \) and \( x = \sin^2 t \). As \( t \) varies from 0 to \( 2\pi \), \( \cos t \) completes a full cycle from 1 to -1 back to 1, and \( \sin^2 t \) completes two cycles from 0 to 1 to 0. The curve starts at the point (0, 1), moves to (1, 0), (0, -1), and back to (0, 1). This describes a vertical line segment oscillating between \( y = -1 \) and \( y = 1 \) at \( x = 0 \).

Step 4: Eliminate Parameter to Find Rectangular Equation

Start by expressing \( \sin t \) in terms of \( x \): since \( x = \sin^2 t \), \( \sin t = \pm \sqrt{x} \). We also have \( y = \cos t \). Using the Pythagorean identity \( \sin^2 t + \cos^2 t = 1 \), substitute for \( \sin^2 t \) as \( x \), which gives \( \cos^2 t = 1 - x \). Therefore, \( y^2 = 1 - x \) is the rectangular-coordinate equation.

Key Concepts

Rectangular Coordinates

Rectangular coordinates are a way to describe a point's position in a two-dimensional plane using two values: typically called \( x \) and \( y \). These values respectively denote distances along the horizontal and vertical axes from a fixed origin point. When dealing with parametric equations like \( x = \sin^2 t \) and \( y = \cos t \), these equations convert a third variable, often referred to as a parameter (here, \( t \)), into a pair of coordinates. This results in tracing a path or curve on the \( xy \)-plane as \( t \) changes.
The aim of converting parametric equations into rectangular equations is to eliminate the parameter, thereby expressing the relation directly in terms of \( x \) and \( y \). By understanding and manipulating rectangular coordinates, we can effectively visualize and analyze the geometric properties of curves.

Eliminating the Parameter

Eliminating the parameter means removing the third variable (usually \( t \)) from parametric equations to derive a relationship solely between \( x \) and \( y \). This process transforms the parametric form into a simpler, more familiar equation, aiding in graphing and understanding the curve's behavior.
In our example, we start with \( x = \sin^2 t \) and \( y = \cos t \). To eliminate \( t \), we use trigonometric identities. Since \( x = \sin^2 t \), we have \( \sin t = \pm \sqrt{x} \). Also, using the Pythagorean identity \( \sin^2 t + \cos^2 t = 1 \), we find \( \cos^2 t = 1 - x \). This yields \( y^2 = 1 - x \), a rectangular equation expressing \( y \) only in terms of \( x \).
Eliminating the parameter simplifies the mathematical model, allowing for a more straightforward analysis of the curve's properties and making sketching easier.

Sketching Curves

Sketching curves from parametric equations involves understanding how \( x \) and \( y \) vary with the parameter \( t \). To sketch the curve represented by \( x = \sin^2 t \) and \( y = \cos t \), we trace how these values change as \( t \) progresses through its range. Observing the ranges, we note that \( x \) oscillates between 0 and 1, while \( y \) oscillates between -1 and 1 as \( t \) goes from 0 to \( 2\pi \).
The initial point is \( (0, 1) \) when \( t = 0 \). As \( t \) increases, \( y \) decreases to 0 when \( x = 1 \) and further to -1 as \( x \) returns to 0, completing the cycle and returning to the starting point at \( (0, 1) \). The resultant graph is a vertical line segment oscillating from \((0, 1)\) to \((0, -1)\). Both the symmetry and the periodic nature of trigonometric functions are key to the sketch, revealing the curve's vertical oscillation between \( y = 1 \) and \( y = -1 \) at \( x = 0 \). This visual representation helps in understanding the underlying behavior and shape of the parametric curve.