Problem 15 - Precalculus Mathematics for Calculus

Question

The point \(P\) is on the unit circle. Find \(P(x, y)\) from the given information. The \(x\) -coordinate of \(P\) is \(\frac{4}{5},\) and the \(y\) -coordinate is positive.

Step-by-Step Solution

Answer

The coordinates of point \(P\) are \(\left(\frac{4}{5}, \frac{3}{5}\right)\).

Step 1: Understanding the Unit Circle

A point \(P\) on the unit circle means that it lies on the circle described by the equation \(x^2 + y^2 = 1\). The unit circle has a radius of 1 and is centered at the origin \((0, 0)\) on the coordinate plane.

Step 2: Use Given Information

The problem states that the \(x\)-coordinate of point \(P\) is \(\frac{4}{5}\). Insert this value into the unit circle equation to find the \(y\)-coordinate. Begin by substituting \(\frac{4}{5}\) into the equation: \(\left(\frac{4}{5}\right)^2 + y^2 = 1\).

Step 3: Solve for \(y^2\)

Calculate \(\left(\frac{4}{5}\right)^2 = \frac{16}{25}\). Substitute back into the equation: \[\frac{16}{25} + y^2 = 1\]. To find \(y^2\), subtract \(\frac{16}{25}\) from 1: \[y^2 = 1 - \frac{16}{25}\].

Step 4: Simplify the Equation

Convert 1 into a fraction with a denominator of 25, which is \(\frac{25}{25}\). Thus, \[y^2 = \frac{25}{25} - \frac{16}{25} = \frac{9}{25}\].

Step 5: Solve for \(y\)

To find \(y\), take the square root of both sides of the equation \(y^2 = \frac{9}{25}\). Thus, \[y = \pm \frac{3}{5}\]. Since the \(y\)-coordinate is positive, select \(y = \frac{3}{5}\).

Step 6: Find the Coordinates of P

Now that we have both the \(x\) and \(y\) coordinates, the point \(P\) on the unit circle is \(\left(\frac{4}{5}, \frac{3}{5}\right)\).

Key Concepts

Unit Circle

When we talk about the unit circle in trigonometry, we refer to a special circle with a radius of exactly one unit, centered at the origin of a coordinate plane at the point (0, 0). What makes the unit circle so significant in mathematics is its simplicity and its role in defining the trigonometric functions. Every point

• The equation of the unit circle is given by \( x^2 + y^2 = 1 \). This equation ensures that any point \((x, y)\) on the unit circle is exactly one unit away from the origin.
• As you move around the circle, you change not only the coordinates of the point \((x, y)\), but also the corresponding angle \(\theta\) in standard position with respect to the positive x-axis.
• The unit circle helps in understanding the properties and relationships of trigonometric functions, such as sine, cosine, and tangent, by defining them in terms of coordinates.

The coordinate values directly relate to the cosine and sine of the angle formed by the radius and the positive x-axis. Exploring and understanding the unit circle is a fundamental step in mastering trigonometry.

Coordinates

Coordinates are numerical values that denote a specific point in a plane or space. In the context of a unit circle, each point has an

• \(x\) coordinate and a \(y\) coordinate. These determine the point's position with respect to the center of the circle (0, 0).
• The coordinates on the unit circle are determined by the intersection of the circle and a line drawn from the origin that forms a specific angle with the positive x-axis.

In trigonometric terms, if you draw a radius from the origin, the \(x\) coordinate of any given point on the circle can be thought of as \(\cos(\theta)\), and the \(y\) coordinate as \(\sin(\theta)\), where \(\theta\) is the angle in question. For example,

• In the exercise, the given \(x\)-coordinate is \(\frac{4}{5}\).
• The task involves deriving the positive \(y\)-coordinate using the unit circle equation.

Thus, understanding coordinates and how they represent a point's position is very important when working with the unit circle.

Equation of a Circle

The equation of a circle is a powerful tool in geometry and algebra, expressing all the points that form a circle's circumference. For any circle in a plane, the standard equation takes the form: \[ (x - h)^2 + (y - k)^2 = r^2 \]

• Where \((h, k)\) are the coordinates of the circle's center.
• \(r\) is the radius of the circle.

For the unit circle, the center is at the origin, \((0, 0)\), which simplifies the equation to \(x^2 + y^2 = 1\). This equation states that the distance from any point on the circle to the origin must always be precisely 1 unit.

• In the given problem, with \(x = \frac{4}{5}\), substituting into the circle equation helps find \(y\).
• The equation illustrates the symmetrical properties of the circle, providing solutions for both positive and negative values for \(y\).

Learning to solve this equation is essential for calculating and plotting points around the circle, thus reinforcing the fundamental principles of trigonometry and analytic geometry.