Problem 16 - 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 \(-\frac{1}{3},\) and the \(x\) -coordinate is positive.

Step-by-Step Solution

Answer

The point \(P\) is \(\left(\frac{2\sqrt{2}}{3}, -\frac{1}{3}\right)\)."

Step 1: Understanding Unit Circle Equation

The equation for the unit circle is given by \(x^2 + y^2 = 1\). Any point \(P(x, y)\) on the unit circle satisfies this equation since its radius is 1.

Step 2: Plug in Known Values

We know the \(y\)-coordinate is \(-\frac{1}{3}\). Substitute this into the unit circle equation: \(x^2 + \left(-\frac{1}{3}\right)^2 = 1\).

Step 3: Calculate \(y^2\) and Substitute

Calculate \(y^2\): \(\left(-\frac{1}{3}\right)^2 = \frac{1}{9}\). Substitute this back into the equation: \(x^2 + \frac{1}{9} = 1\).

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

Rearrange the equation to solve for \(x^2\): \(x^2 = 1 - \frac{1}{9}\).

Step 5: Simplify the Equation

Calculate \(1 - \frac{1}{9}\): \(1 - \frac{1}{9} = \frac{9}{9} - \frac{1}{9} = \frac{8}{9}\), so \(x^2 = \frac{8}{9}\).

Step 6: Solve for \(x\)

Since \(x\) is positive, take the positive square root: \(x = \sqrt{\frac{8}{9}} = \frac{\sqrt{8}}{3}\). Further simplify: \(x = \frac{2\sqrt{2}}{3}\).

Step 7: Write the Coordinates of \(P\)

Now that we know both coordinates, the point \(P\) is \(\left(\frac{2\sqrt{2}}{3}, -\frac{1}{3}\right)\).

Key Concepts

Understanding Coordinate Geometry and the Unit Circle

In coordinate geometry, the unit circle is a fundamental concept. It is a circle with a radius of 1, centered at the origin of the coordinate plane, which is at point \(0, 0\). All the points on this circle satisfy the equation \(x^2 + y^2 = 1\). This is because, according to the Pythagorean theorem, the sum of the squares of a right triangle's legs is equal to the square of the hypotenuse, which, in this case, is always 1. Understanding the coordinates of a point on the unit circle is crucial as they correspond to certain angles in trigonometry. For example, when you know one coordinate, you can always determine the other using the unit circle equation.

Exploring the Pythagorean Identity

The Pythagorean identity is derived directly from the unit circle equation \(x^2 + y^2 = 1\). It relates the sine and cosine functions to one another. In terms of a unit circle:

• The x-coordinate represents the cosine of an angle \(\theta\).
• The y-coordinate represents the sine of the same angle \(\theta\).

Thus, any point on the unit circle can be expressed using trigonometric functions: \((\cos(\theta), \sin(\theta))\). When given either the x or y coordinate of a point on the circle, this identity helps to find the other coordinate. It's crucial to remember that this principle aligns with the equation \(x^2 + y^2 = 1\), which represents a Pythagorean triple for the hypotenuse equal to 1.

Techniques for Solving Equations on the Unit Circle

Solving equations involving the unit circle typically starts with substituting the known values into the unit circle equation \(x^2 + y^2 = 1\). When a coordinate is given, like the \(y\)-coordinate \(-\frac{1}{3}\), you substitute it into the equation producing \(x^2 + \left(-\frac{1}{3}\right)^2 = 1\).By solving for \(x^2\), we isolate the unknown variable. First, calculate \(-\frac{1}{3})^2\) resulting in \(rac{1}{9}\), then reconfigure: \(x^2 + \frac{1}{9} = 1\). Simplifying this gives \(x^2 = \frac{8}{9}\).Finding the value of \(x\) requires taking the square root: \(x = \pm\sqrt{\frac{8}{9}}\). Given that the context specifies a positive x-coordinate, select the positive root, \(x = \frac{2\sqrt{2}}{3}\). This standard step utilizes algebraic manipulations frequently when engaging with unit circle problems.

Simplifying with Radical Expressions

When solving equations on the unit circle, encountering radical expressions is common. In our discussion, when determining \(x\), we arrived at the expression \(\sqrt{\frac{8}{9}}\). Simplification involves several steps:

• Starting with the expression \(\frac{\sqrt{8}}{\sqrt{9}}\) simplifies to \(\frac{\sqrt{8}}{3}\).
• So, \(\sqrt{8}\), which is equal to \(2\sqrt{2}\), further simplifies the expression to \(\frac{2\sqrt{2}}{3}\).

These steps define the process of rationalizing and simplifying radicals, a skill vital for various mathematical disciplines. Mastery of handling radical expressions can assist in solving complex geometry and trigonometry problems efficiently. This often involves transforming expressions into their simplest radical form, a reliable technique within the unit circle framework.