91Ó°ÊÓ

To translate parametric equations into a Cartesian equation, we should aim to remove the parameter, typically denoted as \( t \). In this exercise, we started with the parametric equations \( x(t) = 3 \cos^2 t \) and \( y(t) = -3 \sin^2 t \). The goal is to express the relationship between \( x \) and \( y \) without referencing \( t \).

One useful trigonometric identity here is \( \cos^2 t + \sin^2 t = 1 \). By rearranging the equation \( \cos^2 t = 1 - \sin^2 t \), we can relate \( x \) and \( y \):

• Start with the expressions: \( x = 3 \cos^2 t \) and \( y = -3 \sin^2 t \).
• Substitute \( \cos^2 t = 1 + \frac{y}{3} \).
• Substitute this into the \( x \) equation to yield \( x + y = 0 \).

This Cartesian equation, \( x + y = 0 \), describes a straight line where each point \( (x, y) \) satisfies the relation between these variables without dependency on \( t \). Understanding this conversion is crucial when sketching curves as it clarifies the shape shown by the parametric equations.

Curve sketching is the art of converting an equation into a visual representation on a coordinate plane. With parametric equations like the ones in the exercise, sketching may seem complex due to their dependence on \( t \). However, after translating to the Cartesian equation \( x + y = 0 \), sketching becomes more straightforward.

For the equation \( x + y = 0 \), all points that fit this equation create a line with a negative slope of -1, passing through the origin. To graph it:

• Identify a few points where the equation holds true, like (3, -3) or (-3, 3).
• Draw a straight line through these points, ensuring it passes through the origin (0,0).

Visualizing these steps simplifies the process and provides clarity about how equations form shapes on a graph. Sketching parametric curves often requires understanding their Cartesian form first.

Orientation determines the direction in which a parametric curve is traced as the parameter \( t \) progresses. Even though a line itself has no beginning or end, the parametric representation grants this directionality.

In the exercise, we start with \( x(t) = 3 \cos^2 t \) and \( y(t) = -3 \sin^2 t \). To find out how the line is traced:

• At \( t = 0 \), \( x(0) = 3 \) and \( y(0) = 0 \), marking point (3, 0).
• As \( t \rightarrow \frac{\pi}{2} \), \( x(t) \rightarrow 0 \) and \( y(t) \rightarrow -3 \), tracing towards (0, -3).

This pathway showcases the orientation from left to right across the graph, beginning at (3, 0) and moving to (0, -3) when \( t \) runs from 0 to \( \frac{\pi}{2} \). Understanding orientation helps in correctly portraying the movement across the curve.

Trigonometric identities play a pivotal role in simplifying and solving parametric equations. These identities are essential tools for connecting trigonometric functions under one algebraic umbrella.

In this exercise, the identity \( \cos^2 t + \sin^2 t = 1 \) simplified our expressions to find the Cartesian form. Here are a few pointers on how it helps:

• With \( x(t) = 3 \cos^2 t \) and \( y(t) = -3 \sin^2 t \), using the identity reintroduced a connection between \( x \) and \( y \) without \( t \).
• It allowed us to form the equation \( x + y = 0 \), wrapping the parametric curve into a single straightforward relationship.

Comprehending and applying such identities can vastly reduce complexity, turning intricate parametric relations into familiar Cartesian equations while aiding in further analysis like curve sketching or determining orientation.