Problem 20 - 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 \(-\frac{2}{5},\) and \(P\) lies above the \(x\) -axis.

Step-by-Step Solution

Answer

The coordinates of point \(P\) are \((-\frac{2}{5}, \frac{\sqrt{21}}{5})\).

Step 1: Understand Unit Circle Equation

The unit circle has the equation \(x^2 + y^2 = 1\). Since point \(P(x, y)\) is on the unit circle, it must satisfy this equation.

Step 2: Substitute Given x-value

We know \(x = -\frac{2}{5}\). Substitute this into the unit circle equation: \((-\frac{2}{5})^2 + y^2 = 1\).

Step 3: Simplify and Solve for y^2

Calculate \((-\frac{2}{5})^2 = \frac{4}{25}\). Substitute this value into the equation: \(\frac{4}{25} + y^2 = 1\). Then, subtract \(\frac{4}{25}\) from 1 to find \(y^2\): \(y^2 = 1 - \frac{4}{25} = \frac{21}{25}\).

Step 4: Solve for y

Since \(y^2 = \frac{21}{25}\), take the square root of both sides: \(y = \pm\sqrt{\frac{21}{25}}\). Simplify this to get \(y = \pm\frac{\sqrt{21}}{5}\).

Step 5: Determine Correct y-value Based on Location

Point \(P\) is said to be above the x-axis, which means \(y > 0\). Therefore, choose the positive value for \(y\). So \(y = \frac{\sqrt{21}}{5}\).

Step 6: Final Result

Therefore, the coordinates of point \(P\) are \(\left(-\frac{2}{5}, \frac{\sqrt{21}}{5}\right)\).

Key Concepts

Equation of a Circle

Imagine a perfect circle drawn in a coordinate plane. The equation that describes this circle is fundamental in understanding its characteristics and points on its edge. Particularly, a **unit circle** has a radius of 1 and is centered at the origin, point (0, 0).The equation of any circle in the coordinate plane can be written in the general form: \[ (x - h)^2 + (y - k)^2 = r^2 \] Where (h, k) is the center of the circle, and **r** is the radius. For a unit circle specifically, the equation simplifies to: \[ x^2 + y^2 = 1 \] This special equation tells us that any point (x, y) lying on the unit circle will satisfy this equation. If you place any point on that circle, plugging your coordinates into this equation should equal 1. This understanding forms the basis of much of analytical geometry when discussing circles.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, merges algebra and geometry using a coordinate system. It allows us to find the relationship between algebraic equations and geometric figures. This approach is particularly useful for solving various geometry problems using algebraic methods. In the case of circles, the coordinate system helps us comprehensively understand points and lines, intersecting figures, and distances between points:- **Points and Coordinates:** Every point in coordinate geometry is represented as (x, y). For problems involving circles, knowing the x or y coordinate helps us find the other missing coordinate.- **Equivalent transformations:** You can substitute known values into equations, transforming them to solve for unknowns. This is how we found y in our exercise by manipulating the equation \(x^2 + y^2 = 1\).Coordinate geometry's beauty lies in its ability to visualize algebraic solutions on a graph, making complex problems simpler to comprehend.

Solving Equations

At the heart of solving any problem in mathematics lies the ability to manipulate and solve equations. When a point like **P(x, y)** is given on a circle, and you're tasked to find its coordinates, this involves several steps of problem-solving.Here's a quick breakdown of solving such equations:- **Substitute Known Values:** Begin by substituting known values (like x in our exercise) into the circle's equation. This replacement helps reduce the number of unknowns in the equation.- **Simplify and Rearrange the Equation:** Break down terms and rearrange them to isolate the unknown variable (in this case, y). Remember, breaking down complex terms can make the solution clear. To find the value of y, we simplify the equation \(\frac{4}{25} + y^2 = 1\) by solving for \(y^2\), and then take the square root to find y. This particular example shows: - Using squares and roots carefully, as they can both reduce complexities and introduce new solutions (like positive and negative roots).Understanding these steps will boost your ability to tackle similar problems, primarily when coordinates and geometries decide the fate of an equation.