91Ó°ÊÓ

Parametric equations are a way to define a set of equations that express the coordinates of the points on a curve or surface, typically in terms of a single parameter. In our given problem, the equations are:

• \( x = 1 + \cos t \)
• \( y = 3 - \sin t \)

These equations define the horizontal and vertical coordinate values \( x \) and \( y \) respectively, using the parameter \( t \). This parameter often represents an angle, making parametric equations especially useful for describing circular and elliptical shapes. By having two separate functions for \( x \) and \( y \), it allows us to trace the path of a curve as the parameter \( t \) changes.

The challenge in converting parametric equations into rectangular form lies in eliminating the parameter. This requires expressing the relation in terms of only \( x \) and \( y \), as we see in common Cartesian coordinates.

The Pythagorean identity is a fundamental relation in trigonometry, and it plays a key role in converting parametric equations to rectangular form. It states that for any angle \( t \), the sum of the squares of the sine and cosine of that angle equals 1:

• \( \sin^2 t + \cos^2 t = 1 \)

This identity allows us to relate the sine and cosine functions to each other, and by extension, relate \( x \) and \( y \) in our problem. By solving the parametric equations for \( \sin t \) and \( \cos t \) and substituting them into the Pythagorean identity, we get:

\((3 - y)^2 + (x - 1)^2 = 1.\)

This new equation is free of the parameter \( t \) and relates directly to our goal of finding a rectangular (Cartesian) form. It's an integral step in many problems involving trigonometric parameters.

Completing the square is a technique used to transform a quadratic equation into a more useful form, typically into something that resembles \((u-h)^2=k\). This approach is particularly helpful when seeking to convert equations into the standard form of a circle or ellipse.

In our example, we had a quadratic equation:

• (x^2 - 2x + y^2 - 6y + 9 = 1)

To complete the square, you manipulate the quadratic expression to add and subtract the necessary constants so each squared term becomes a perfect square trinomial. For \( x \), you add and subtract 1, and for \( y \), you add and subtract 9:

• \((x - 1)^2 = x^2 - 2x + 1\)
• \((y - 3)^2 = y^2 - 6y + 9\)

After completing the square, the equation simplifies to:
\[(x - 1)^2 + (y - 3)^2 = 1\]

This form visually represents a circle, showing a clear geometric interpretation with its center at \((1, 3)\) and a radius of 1, making it easier to understand and interpret the original problem.