Problem 17 - Precalculus Mathematics for Calculus

Question

The point \(P\) is on the unit circle. Find \(P(x, y)\) from the given information. The \(y\) -coordinate of \(P\) is \(\frac{2}{3},\) and the \(x\) -coordinate is negative.

Step-by-Step Solution

Answer

The point is \(P = \left(-\frac{\sqrt{5}}{3}, \frac{2}{3}\right)\).

Step 1: Understanding the Unit Circle

The unit circle is a circle with a radius of 1 centered at the origin (0, 0) in the coordinate plane. Any point \((x, y)\) on the unit circle satisfies the equation \(x^2 + y^2 = 1\). This is because the radius is 1, and according to the Pythagorean theorem, the sum of the squares of the coordinates equals the square of the radius.

Step 2: Substitute the Known y-Coordinate

We are given that \(y = \frac{2}{3}\). We will substitute \(y = \frac{2}{3}\) into the unit circle equation:\[x^2 + \left(\frac{2}{3}\right)^2 = 1\]Simplifying, we get:\[x^2 + \frac{4}{9} = 1\]

Step 3: Solve for the x-Coordinate

Rearrange the equation to find \(x^2\):\[x^2 = 1 - \frac{4}{9}\]Combine fractions:\[x^2 = \frac{9}{9} - \frac{4}{9} = \frac{5}{9}\]

Step 4: Calculate the Values of x

Take the square root of both sides to solve for \(x\):\[x = \pm \sqrt{\frac{5}{9}} = \pm \frac{\sqrt{5}}{3}\]Since the problem states that the \(x\)-coordinate is negative, we choose:\[x = -\frac{\sqrt{5}}{3}\]

Step 5: Combine Coordinates for the Point P

Now that we have both coordinates for \(P\), we can write the point as:\[P = \left(-\frac{\sqrt{5}}{3}, \frac{2}{3}\right)\]

Key Concepts

Pythagorean Theorem

The Pythagorean Theorem is a fundamental principle in mathematics that relates the sides of a right triangle. It's usually written as \(a^2 + b^2 = c^2\), where \(c\) is the hypotenuse—the longest side of the triangle. In the context of a unit circle, this theorem is employed to specify the relationship between the \(x\) and \(y\) coordinates of any point on the circle.
If you imagine a triangle transposed onto a coordinate plane, with one vertex at the origin, the x-coordinate forms the base and the y-coordinate forms the height of the triangle. The radius of the circle acts as the hypotenuse, here being 1 unit in length.

• Each point \((x, y)\) on the unit circle satisfies \(x^2 + y^2 = 1\).
• This equation is derived directly from the Pythagorean theorem set against the unit circle's parameters.
• It magnificently delineates the unity between geometry and algebra through this simple yet robust equation.

This concept is pivotal as it forms the foundation for trigonometry and helps in finding unknown values when given partial coordinates.

Coordinate Geometry

Coordinate Geometry, or analytic geometry, brings algebra and geometry together through graphs and coordinates. In a coordinate plane, every point is denoted with an \(x\) and \(y\) coordinate that signifies its position.
This system helps visualize geometric problems and derive equations that express relationships among different entities within a plane.

• The unit circle is a classic example used in coordinate geometry.
• Positioned at the origin, it allows easy visualization of trigonometric functions, with every angle corresponding to a unique point on the circle.
• This facilitates calculating trigonometric identities as sine and cosine relate directly to the \(y\) and \(x\) coordinates, respectively.

Through coordinate geometry, abstract algebraic equations find real-world application, such as the fundamental equation \(x^2 + y^2 = 1\) of the unit circle.

Square Roots

Square roots represent one way to simplify or solve equations, especially those involving exponents. The square root of a number \(x\) is a value that, when multiplied by itself, gives \(x\).
Symbolically, it is expressed as \(\sqrt{x}\). In the context of this exercise, square roots are crucial for finding the coordinates.

• To find the \(x\) value, we took the square root of \(\frac{5}{9}\): \(x = \pm \frac{\sqrt{5}}{3}\).
• The square root operation typically yields both a positive and negative result, as both \(n\) and \(-n\) squared yield the same product \(n^2\).
• However, the condition given is that \(x\) is negative, which directs us to choose the negative root.

Understanding square roots is essential, as it often comes into play when solving coordinate problems involving the unit circle and discovering certain unknown values.