Question

For the following exercises, sketch the parametric curve and eliminate the parameter to find the Cartesian equation of the curve. $$ x=4 \cos \phi, y=1-\sin \phi, 0 \leq \phi \leq 2 \pi $$

Short answer

The Cartesian equation is \( \frac{x^2}{16} + (y-1)^2 = 1 \), representing an ellipse.

Step-by-step solution

Understand the Parametric Equations

The given equations are \( x = 4 \cos \phi \) and \( y = 1 - \sin \phi \). We need to understand how these describe points on a plane as \( \phi \) varies from 0 to \( 2\pi \).

Express \( \cos \phi \) and \( \sin \phi \) in terms of \( x \) and \( y \)

From the equation \( x = 4 \cos \phi \), solve for \( \cos \phi \): \( \cos \phi = \frac{x}{4} \). Similarly, from \( y = 1 - \sin \phi \), solve for \( \sin \phi \): \( \sin \phi = 1 - y \).

Use the Pythagorean Identity

Since \( \cos^2 \phi + \sin^2 \phi = 1 \), substitute \( \cos \phi \) and \( \sin \phi \) from the expressions found in Step 2: \[ \left(\frac{x}{4}\right)^2 + (1-y)^2 = 1 \].

Simplify the Equation

Expand and simplify the equation: \[ \frac{x^2}{16} + (1 - y)^2 = 1 \]. \( (1-y)^2 = (1 - 2y + y^2) \), so \[ \frac{x^2}{16} + 1 - 2y + y^2 = 1 \].

Solve for the Cartesian Equation

Rearranging gives: \( \frac{x^2}{16} + y^2 - 2y + 1 = 1 \). Simplify to find \( \frac{x^2}{16} + y^2 - 2y = 0 \).

Complete the Square in \( y \)

Complete the square for \( -2y \, \to\, (y-1)^2 \), giving: \( \frac{x^2}{16} + (y-1)^2 = 1 \). This simplifies to the equation of an ellipse centered at \((0, 1)\) with a horizontal radius (semi-major axis) of 4 and a vertical radius (semi-minor axis) of 1.

Key concepts

Cartesian Coordinates

Cartesian coordinates are a way of locating points on a plane. Think of it like a grid where each point has two numbers: one for how far to move right or left (x-coordinate), and one for how far to move up or down (y-coordinate). In this exercise, we use parametric equations, which include parameters (in this case \(\phi\)) to express a curve in terms of coordinates. By eliminating the parameter, we find a direct relationship between \(x\) and \(y\), resulting in a Cartesian equation.

The transformation of parametric equations \(x = 4 \cos \phi\) and \(y = 1 - \sin \phi\) into Cartesian coordinates involves combining these to find a single equation without \(\phi\). This highlights how parametric curves can be expressed in different forms, providing various perspectives on geometric shapes.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions like sine and cosine that hold true for all values of their variables. One of the most important identities is the Pythagorean identity: \(\cos^2 \phi + \sin^2 \phi = 1\). This identity becomes crucial when converting from parametric to Cartesian form.

In our exercise, we start with \(x = 4 \cos \phi\) and \(y = 1 - \sin \phi\). To eliminate \(\phi\), we solve for \(\cos \phi\) and \(\sin \phi\). Substituting these expressions into the Pythagorean identity provides a path to finding the Cartesian equation. By utilizing these relationships, we transform the parametric form into a familiar geometric shape description.

Ellipse

An ellipse is a geometric shape resembling an elongated circle. It is defined by two distances known as axes. The longer axis is the major axis, and the shorter is the minor axis. For an ellipse centered at the origin, the standard form equation is \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), where \((h, k)\) is the center, \(a\) is the semi-major axis, and \(b\) is the semi-minor axis.

In this exercise, after simplifying, we found the equation \(\frac{x^2}{16} + (y-1)^2 = 1\). This represents an ellipse centered at \((0, 1)\), with \(a = 4\) and \(b = 1\). Understanding the structure of this equation helps in visualizing the ellipse's orientation and scale relative to the Cartesian plane. Ellipses have properties that make them significant in physics and astronomy, such as the orbits of planets, which often resemble elliptical paths.