Problem 29 - Contemporary Precalculus

Question

The given curve is part of the graph of an equation in \(x\) and \(y .\) Find the equation by eliminating the parameter. $$x=3 \cos t, \quad y=3 \sin t, \quad 0 \leq t \leq 2 \pi$$

Step-by-Step Solution

Answer

Question: Find the equation of the curve represented by the parametric equations \(x = 3\cos t\) and \(y = 3\sin t\) for \(0 \leq t \leq 2\pi\). Answer: The equation of the curve in terms of \(x\) and \(y\) is \(x^2 + y^2 = 9\).

Step 1: Solve for \(\cos t\) and \(\sin t\) from the given functions

From the given functions, we can solve for \(\cos t\) and \(\sin t\) in terms of \(x\) and \(y\) as follows: $$\cos t = \frac{x}{3}, \quad \sin t = \frac{y}{3}$$

Step 2: Use the Pythagorean identity to eliminate \(t\)

We know that for any angle \(t\), the Pythagorean identity holds: $$\sin^2 t + \cos^2 t = 1$$ Now substitute the expressions we found in Step 1 for \(\sin t\) and \(\cos t\): $$\left(\frac{y}{3}\right)^2 + \left(\frac{x}{3}\right)^2 = 1$$

Step 3: Simplify the equation

Now, let's simplify the equation by eliminating the fractions and combining terms: $$\frac{y^2}{9} + \frac{x^2}{9} = 1$$ Multiplying both sides of the equation by \(9\) gives us the final equation: $$x^2 + y^2 = 9$$ So, the equation representing the curve in terms of \(x\) and \(y\), after eliminating the parameter \(t\), is \(x^2 + y^2 = 9\).

Key Concepts

Pythagorean Identity

The Pythagorean Identity is an essential concept in trigonometry, which reflects the relationship between the sine and cosine of an angle. Simply put, for any angle \(t\), the identity states that

• \( \sin^2 t + \cos^2 t = 1 \)

This identity is named after the Pythagorean Theorem because it completes the analogy of forming a right triangle using a unit circle. Here, the sine and cosine values essentially represent the legs of a right triangle with a hypotenuse of 1. Therefore, just like the triangle’s side lengths fulfill the equation \(a^2 + b^2 = c^2\), sine and cosine of angle \(t\) fulfill \(\sin^2 t + \cos^2 t = 1\).

In the exercise, the identity is used to connect the parametric equations back to a standard form. By substituting \(\cos t = \frac{x}{3}\) and \(\sin t = \frac{y}{3}\) into the Pythagorean identity, we can remove the parameter \(t\) and describe the curve by a singular equation in terms of \(x\) and \(y\). This transition is crucial for simplifying parametric equations into familiar shapes.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are fundamental in expressing relationships between angles and sides in right triangles, as well as in modeling periodic phenomena. In this exercise, these functions are defined parametrically. For a given parameter \(t\):

• \(x = 3 \cos t\)
• \(y = 3 \sin t\)

Each of these functions captures the coordinate position of a point moving along a circle of radius 3, as \(t\) varies. This illustrates how trigonometric functions can describe circular motion.

Furthermore, understanding these functions allows us to convert parametric equations into Cartesian equations (expressed as \(x\) and \(y\)). By identifying the parametric forms of sine and cosine and manipulating these forms, students can master the transition between formats. Recognizing these forms as part of a circular trajectory helps ensure clarity when visualizing solutions graphically.

Eliminating Parameters

Eliminating parameters is a process by which we remove the parameter \(t\) from parametric equations to express the relationship in a more conventional form, like a Cartesian equation. In our exercise, we begin with two parametric equations denoted by trigonometric functions:

• \(x = 3 \cos t\)
• \(y = 3 \sin t\)

To eliminate \(t\), we first expressed \(\cos t\) and \(\sin t\) in terms of \(x\) and \(y\):

• \(\cos t = \frac{x}{3}\)
• \(\sin t = \frac{y}{3}\)

By substituting these expressions into the Pythagorean identity, \( \sin^2 t + \cos^2 t = 1 \), we eliminated the parameter and ended up with \( \left(\frac{y}{3}\right)^2 + \left(\frac{x}{3}\right)^2 = 1 \). After simplifying, the equation \(x^2 + y^2 = 9\) was obtained.

This final equation represents the original parametric curve in Cartesian form, describing a circle with a radius of 3 centered at the origin.