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Chapter 5: Problem 4

The point \(P\) is on the unit circle. If the \(x\) -coordinate of \(P\) is \(\frac{1}{5},\) and \(P\) is in quadrant IV, find the \(y\) coordinate.

Short Answer

Expert verified
The y-coordinate of point P is \(-\frac{2\sqrt{6}}{5}\).

Step by step solution

01

Understanding the Unit Circle Equation

The equation of the unit circle is \(x^2 + y^2 = 1\). Any point \((x, y)\) on the unit circle will satisfy this equation.
02

Substitute for the x-coordinate

Since the point \(P\) has an \(x\)-coordinate of \(\frac{1}{5}\), substitute \(x = \frac{1}{5}\) into the equation \(x^2 + y^2 = 1\). This gives \(\left(\frac{1}{5}\right)^2 + y^2 = 1\).
03

Simplify the Equation

Calculate \(\left(\frac{1}{5}\right)^2\) which is \(\frac{1}{25}\). Substitute this back into the equation to get \(\frac{1}{25} + y^2 = 1\).
04

Isolate y^2

Subtract \(\frac{1}{25}\) from 1 to solve for \(y^2\). This gives \(y^2 = 1 - \frac{1}{25} = \frac{24}{25}\).
05

Solving for y

Take the square root of both sides to solve for \(y\). This gives \(y = \pm \frac{\sqrt{24}}{5}\). Simplify \(\sqrt{24}\) to \(2\sqrt{6}\), resulting in \(y = \pm \frac{2\sqrt{6}}{5}\).
06

Determine the Correct Quadrant IV y-coordinate

Since point \(P\) is located in quadrant IV, the \(y\)-coordinate must be negative. Therefore, \(y = -\frac{2\sqrt{6}}{5}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadrant IV
A unit circle is split into four different sections, known as quadrants. Each quadrant is characterized by specific signs for the x-coordinate and the y-coordinate of the points located within them. In Quadrant IV:
  • The x-coordinate is positive.
  • The y-coordinate is negative.
This quadrant is located at the bottom-right of the unit circle, covering angles from 270掳 to 360掳 or equivalently from \(-\frac{\pi}{2}\) to 0 in radians. Points in Quadrant IV have their y-coordinate below the x-axis, signifying a negative y-value. Given the problem situation, since point \(P\) is located in Quadrant IV, it makes sense for \(y\) to be \(-\frac{2\sqrt{6}}{5}\), a negative value, verifying the position with the quadrant's typical properties.
Equation of the Circle
The concept of the unit circle equation is central to understanding circular behavior in trigonometry. For a standard unit circle centered at the origin on the coordinate plane:
  • The equation is \(x^2 + y^2 = 1\).
  • This equation ensures that every point \((x, y)\) along the circle鈥檚 circumference is equal to a radius of 1.
By plugging in a specific x-coordinate, you can find the corresponding y-coordinate that maintains the equality, representing position along this circular path. In the instance of the original problem, substituting \(x = \frac{1}{5}\) into the equation, \(x^2 + y^2 = 1\), and solving allows us to find the specific y-coordinate for point \(P\). This demonstrates how you can use the equation of the circle to navigate the possible points on the circle.
x-coordinate and y-coordinate
In a coordinate system, every point is represented by an \(x\) and \(y\) coordinate. These coordinates help to determine the precise location of the point in a geometric plane.
  • The \(x\)-coordinate refers to the horizontal position, while the \(y\)-coordinate refers to the vertical position.
Given the equation \(x^2 + y^2 = 1\), one can see how the coordinates interrelate to maintain the circle's properties. For point \(P\) in the unit circle, knowing \(x = \frac{1}{5}\) allows us to determine that \(y\) has to be \(-\frac{2\sqrt{6}}{5}\) to satisfy the circle's equation. Understanding this interaction between the coordinates is fundamental in graphically displaying relations and interactions within the unit circle.
Trigonometric Functions
Trigonometric functions rely heavily on the measurements and properties of angles and distances in the unit circle. Here are some basic insights:
  • The sine function corresponds to the y-coordinate of a point on the unit circle.
  • The cosine function corresponds to the x-coordinate.
  • The tangent function is the ratio of sine to cosine.
For example, in the unit circle, each trigonometric function has specific values determined by the angle made with the positive x-axis. While tackling Quadrant IV, observe the signs as these rules help solve and simplify equations regarding points like \(P\). Knowing that \(y\) is negative influences how we interpret the sine function's resulting value in this particular quadrant.

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Most popular questions from this chapter

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