Problem 18 - 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 positive, and the \(y\) -coordinate of \(P\) is \(-\sqrt{5} / 5\).

Step-by-Step Solution

Answer

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

Step 1: Write the Equation of the Unit Circle

Recall that the unit circle is defined by the equation \(x^2 + y^2 = 1\). This equation represents all points \((x, y)\) that are exactly one unit away from the origin \((0, 0)\). We will use this equation to find the \(x\)-coordinate.

Step 2: Substitute the Given y-coordinate

We are given that the \(y\)-coordinate of the point is \(-\frac{\sqrt{5}}{5}\). Substitute this value into the unit circle equation: \[x^2 + \left(-\frac{\sqrt{5}}{5}\right)^2 = 1\].

Step 3: Simplify the Equation

Simplify the equation by calculating \(\left(-\frac{\sqrt{5}}{5}\right)^2\): \[\left(-\frac{\sqrt{5}}{5}\right)^2 = \frac{5}{25} = \frac{1}{5}.\]Substitute this into the unit circle equation: \[x^2 + \frac{1}{5} = 1.\]

Step 4: Solve for x²

To find \(x^2\), subtract \(\frac{1}{5}\) from both sides of the equation:\[x^2 = 1 - \frac{1}{5} = \frac{5}{5} - \frac{1}{5} = \frac{4}{5}.\]

Step 5: Find the x-coordinates

Take the square root of both sides to solve for \(x\):\[x = \pm \sqrt{\frac{4}{5}} = \pm \frac{2}{\sqrt{5}}.\]Rationalize the denominator by multiplying the numerator and the denominator by \(\sqrt{5}\):\[x = \pm \frac{2}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \pm \frac{2\sqrt{5}}{5}.\]

Step 6: Determine the Correct Sign for x

Since we are told that the \(x\)-coordinate is positive, choose the positive solution:\[x = \frac{2\sqrt{5}}{5}.\]

Step 7: Write the Coordinates of Point P

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

Key Concepts

Equation of the Unit Circle

The equation of a unit circle is a fundamental concept in trigonometry. It describes a perfect circle with a radius of one unit, centered at the origin of a coordinate system, \(0, 0\). The equation is represented as \(x^2 + y^2 = 1\). This set equation signifies that any point \(x, y\) lying on this circle is exactly one unit away from the origin.

• The unit circle helps to visualize and solve trigonometric problems.
• Its equation is derived from the Pythagorean Theorem, which states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

It primarily serves as a tool for understanding the relationship between the angles and coordinates on the plane, which are crucial when dealing with sine, cosine, and other trigonometric functions. It also plays a significant role when trying to find the sine and cosine of angles, where the x-coordinate represents cosine and the y-coordinate mirrors sine.

X-Coordinate

In the given problem, we want to find the x-coordinate of the point on the unit circle. We were given that the y-coordinate is \(-\frac{\sqrt{5}}{5}\). This information needs to be plugged into the equation of the unit circle to solve for the x-coordinate. Here's how you manage that:

• Substitute the y-coordinate into the unit circle equation: \(x^2 + \left(-\frac{\sqrt{5}}{5}\right)^2 = 1\).
• Simplify \(y^2\) to get \(\frac{1}{5}\).
• Then solve the equation \(x^2 + \frac{1}{5} = 1\) to find \(x^2 = \frac{4}{5}\).
• Finally, take the square root to solve for x, noting that \(x\) could be positive or negative.

In this problem, it is stated that the x-coordinate is positive, so you select the positive value, ending with x = \(+\frac{2\sqrt{5}}{5}\). Ensuring the right sign is crucial based on the quadrant or region of interest.

Y-Coordinate

The y-coordinate in the context of the unit circle can provide insight into the specific position of a point on the circle. Given the problem, the y-coordinate is specified as \(-\frac{\sqrt{5}}{5}\), which influences where the point lies: the lower half of the circle.

• Knowing the y-coordinate allows you to find compatible x-coordinates that fulfill the unit circle's equation.
• The position of the y-coordinate on the negative axis indicates that, for a unit circle, the point is below the x-axis.

Understanding y's value helps in determining x because it gives an anchor point for the calculations in terms of placement on the circle. This is crucial for accurately solving problems and understanding rotational dynamics and symmetry in trigonometric contexts.