Problem 15 - Algebra and Trigonometry

Question

\(3-24=A\) 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. $$ x=\sin ^{2} t, \quad y=\sin ^{4} t $$

Step-by-Step Solution

Answer

The rectangular equation is \( y = x^2 \), and the curve is a parabola in the first quadrant.

Step 1: Understand the Parametric Equations

In the given parametric equations, we have \( x = \sin^2 t \) and \( y = \sin^4 t \). These describe the relationship between the angle \( t \) and the coordinates \( x \) and \( y \). Our goal is to sketch the curve and find the rectangular equation by eliminating \( t \).

Step 2: Sketch the Curve

To sketch the curve, consider different values of \( t \). As \( t \) varies between 0 and \( \pi \), \( \sin t \) varies between 0 and 1. Therefore, \( x = \sin^2 t \) varies from 0 to 1, and \( y = \sin^4 t \) also varies from 0 to 1. Note that as \( x \) increases, \( y \) increases, indicating the curve is always in the first quadrant.

Step 3: Eliminate the Parameter

To find the rectangular equation, we need to express \( y \) in terms of \( x \). We know \( x = \sin^2 t \), so \( \sin t = \sqrt{x} \). Therefore, \( y = \sin^4 t = (\sin^2 t)^2 = x^2 \). Thus, the rectangular equation of the curve is \( y = x^2 \).

Step 4: Verify and Finalize the Solutions

Let's verify if our rectangular equation is consistent with the parametric form. Since \( x = \sin^2 t \) and \( y = \sin^4 t \) can be written as \( y = (\sin^2 t)^2 = x^2 \), it aligns with the derived equation of \( y = x^2 \). Hence, every solution for \( x \) from the parametric form is consistent with this rectangular equation, confirming the accuracy of the transformation.

Key Concepts

Rectangular Coordinate Equation

When you work with parametric equations, each coordinate in your equation is described using a parameter which, in this scenario, is denoted by the variable \( t \). Sometimes, it's useful to convert these into a rectangular coordinate equation. This means that you express one variable, typically \( y \), directly in terms of another, such as \( x \), without involving the parameter \( t \).

In the given exercise, we have two parametric equations: \( x = \sin^2 t \) and \( y = \sin^4 t \). These describe a relationship where both \( x \) and \( y \) depend on \( t \). To eliminate \( t \), notice that

• \( x \) can be expressed as \( \sin^2 t \)
• \( y = \sin^4 t = (\sin^2 t)^2 \)

Substitute \( \sin^2 t \) with \( x \), and you get \( y = x^2 \). Congratulations, you now have the rectangular coordinate equation: \( y = x^2 \). It's a simple but crucial step for analyzing curves in a more traditional \( x-y \) graph.

Curve Sketching

To sketch a curve given by parametric equations, you need to evaluate the equations for a range of parameter values and plot the points. For the current set of equations, \( x = \sin^2 t \) and \( y = \sin^4 t \), you'd typically consider \( t \) values from 0 to \( \pi \).

As \( t \) changes:

• \( \sin t \) changes from 0 up to 1 and back to 0.
• Consequently, \( \sin^2 t \) (or \( x \)) varies from 0 to 1 and back to 0.
• Similarly, \( y = \sin^4 t \) starts at 0, peaks at 1, and returns to 0.

This path lies within the first quadrant because \( \sin t \) is always non-negative between 0 and \( \pi \). This exercise illustrates how curve sketching helps you visualize the relationship between \( x \) and \( y \) without solving for all points explicitly. The curve described is \( y = x^2 \), which is a parabola opening upwards.

Eliminating the Parameter

Eliminating the parameter in parametric equations involves removing the dependency on the parameter to get a direct relationship between \( x \) and \( y \). In our example, we started with \( x = \sin^2 t \) and \( y = \sin^4 t \). Here is how to remove \( t \):

• First, express \( t \) in terms of \( x \): since \( x = \sin^2 t \), we find \( \sin t = \sqrt{x} \).
• Next, substitute this into the equation for \( y \): \( y = \sin^4 t = (\sin^2 t)^2 = x^2 \).

By doing this, we've transitioned from a parameterized representation to a straightforward algebraic relationship between \( x \) and \( y \). This process makes it easier to evaluate the curve's geometry and align it with familiar Cartesian graphs.