Problem 11 - Precalculus Mathematics for Calculus

Question

Find the missing coordinate of \(P,\) using the fact that \(P\) lies on the unit circle in the given quadrant. Coordinates $$P\left(\quad, \frac{1}{3}\right)$$ Quadrant II

Step-by-Step Solution

Answer

\(x = -\frac{2\sqrt{2}}{3}\)

Step 1: Identify Characteristics of the Unit Circle

The unit circle is defined by the equation \(x^2 + y^2 = 1\), where \((x, y)\) are coordinates on the circle. Since point \(P\) is on the unit circle, it must satisfy these conditions.

Step 2: Plug Known Values into the Unit Circle Equation

Given that \(P\left(x, \frac{1}{3}\right)\), substitute \(y = \frac{1}{3}\) into the equation \(x^2 + y^2 = 1\). This gives us: \(x^2 + \left(\frac{1}{3}\right)^2 = 1\).

Step 3: Simplify the Equation

Simplify \(\left(\frac{1}{3}\right)^2\) to get \(\frac{1}{9}\). The equation becomes: \(x^2 + \frac{1}{9} = 1\).

Step 4: Solve for \(x^2\)

Subtract \(\frac{1}{9}\) from both sides to solve for \(x^2\): \(x^2 = 1 - \frac{1}{9} = \frac{9}{9} - \frac{1}{9} = \frac{8}{9}\).

Step 5: Solve for \(x\)

Take the square root of both sides to solve for \(x\): \(x = \pm\sqrt{\frac{8}{9}} = \pm\frac{\sqrt{8}}{3}\). Simplify \(\sqrt{8}\) to \(2\sqrt{2}\), so \(x = \pm\frac{2\sqrt{2}}{3}\).

Step 6: Determine the Sign of \(x\) in Quadrant II

In Quadrant II, \(x\) values are negative, so \(x = -\frac{2\sqrt{2}}{3}\).

Step 7: Conclusion: Write the Missing Coordinate

The missing coordinate \(x\) of point \(P\) is \(-\frac{2\sqrt{2}}{3}\). Thus, \(P\left(-\frac{2\sqrt{2}}{3}, \frac{1}{3}\right)\).

Key Concepts

Coordinates

When we talk about coordinates, we are referring to a pair of values that show an exact position on a plane. On the unit circle, each point is represented by an

• **x-coordinate** - the horizontal distance from the origin
• **y-coordinate** - the vertical distance from the origin

Both coordinates must satisfy the unit circle equation: \(x^2 + y^2 = 1\).This equation is derived from the Pythagorean Theorem, which guarantees that any point lying on the unit circle will always be a distance of one from the center (origin).
In our exercise, we are given a y-coordinate of \(\frac{1}{3}\), and we need to find the x-coordinate that complements it to satisfy the equation of the unit circle.

Quadrants

The coordinate plane is divided into four sections, known as quadrants. These quadrants are numbered I to IV and are generally represented counter-clockwise starting from the positive x-axis.Each quadrant help us understand the sign of the coordinates:

• Quadrant I: Both x and y are positive.
• Quadrant II: x is negative and y is positive.
• Quadrant III: Both x and y are negative.
• Quadrant IV: x is positive and y is negative.

In the current problem, point \(P\) is in Quadrant II where the x-coordinate is negative, and the y-coordinate is positive. So once we found the possible values of x, we must choose the negative value to ensure that the coordinates align with the properties of Quadrant II.

Pythagorean identity

The Pythagorean identity is key for understanding how coordinates work on the unit circle. It arises from a generalization of the Pythagorean Theorem to trigonometric functions. For the unit circle, this identity is written as:

• \(x^2 + y^2 = 1\)

This means that for any point \((x, y)\) on the unit circle, the sum of the squares of the x and y coordinates will always equal 1.
This identity is what ensures that our coordinates of \(P(x, \frac{1}{3})\) must be such that \(x^2 + \left(\frac{1}{3}\right)^2 = 1\). By solving this equation, step-by-step, we can determine that \(x = -\frac{2\sqrt{2}}{3}\), due to the negative x requirement of Quadrant II.