Problem 38 - Algebra and Trigonometry

Question

Convert the rectangular coordinates to polar coordinates with \(r>0\) and \(0 \leq \theta<2 \pi .\) $$ (-\sqrt{6},-\sqrt{2}) $$

Step-by-Step Solution

Answer

Polar coordinates: \((2\sqrt{2}, \frac{7\pi}{6})\).

Step 1: Identify the rectangular coordinates

The given rectangular coordinates are \[x = -\sqrt{6}, \, y = -\sqrt{2}.\]

Step 2: Find the radius \( r \)

Use the formula for the radius in polar coordinates: \[r = \sqrt{x^2 + y^2}.\]Substitute the values:\[r = \sqrt{(-\sqrt{6})^2 + (-\sqrt{2})^2} = \sqrt{6 + 2} = \sqrt{8} = 2\sqrt{2}.\]

Step 3: Find the angle \( \theta \)

Use the formula \( \tan \theta = \frac{y}{x} \):\[\tan \theta = \frac{-\sqrt{2}}{-\sqrt{6}} = \frac{\sqrt{2}}{\sqrt{6}} = \frac{1}{\sqrt{3}}.\]Thus, \(\theta\) is \(\frac{\pi}{6}\) as \(\tan^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}\).Since both \(x\) and \(y\) are negative, the point is in the third quadrant. Adjust accordingly:\[\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}.\]

Step 4: Write the polar coordinates

The polar coordinates are given by \[r = 2\sqrt{2}, \, \theta = \frac{7\pi}{6}.\]

Key Concepts

Rectangular Coordinates

Rectangular coordinates, commonly known as Cartesian coordinates, use an x-coordinate and a y-coordinate to specify the position of a point in a plane. This system is highly intuitive because it draws from horizontal and vertical lines. Think of it as a grid with two axes: the x-axis that runs horizontally and the y-axis that runs vertically.
In the problem at hand, the point given as (-\sqrt{6}, -\sqrt{2}) provides these x and y values explicitly. These coordinates tell you precisely how far to move from the origin:\[- \sqrt{6}\] to the left along the x-axis and \[- \sqrt{2}\] down along the y-axis. By plotting this point, you can visually see its location before converting it to another system. This helps in understanding why certain adjustments are needed later on in the process.

Radius

The radius in polar coordinates is equivalent to the hypotenuse of a right-angled triangle formed by the point and the axes. It is always positive, showing how far away the point is from the origin, regardless of direction.
To find it, we use the formula: \[r = \sqrt{x^2 + y^2}.\]
This formula comes from the Pythagorean theorem, where x and y represent the two shorter sides of a right triangle, and the radius is the hypotenuse.
In our exercise, substituting the given values results in \[r = \sqrt{(-\sqrt{6})^2 + (-\sqrt{2})^2} = \sqrt{6 + 2} = \sqrt{8} = 2\sqrt{2}.\]
This radius is crucial for further calculations, specifically in determining the polar angle.

Angle Calculation

The angle \( \theta \) in polar coordinates tells us the direction of the point from the origin. We calculate it using \(\tan \theta = \frac{y}{x}\). This formula arises from basic trigonometry, where tangent is the ratio of the opposite side to the adjacent side of a right triangle.
For our coordinates \((-\sqrt{6}, -\sqrt{2})\), we find \(\tan \theta = \frac{-\sqrt{2}}{-\sqrt{6}} = \frac{1}{\sqrt{3}}.\) The resulting angle before quadrant adjustment is \(\frac{\pi}{6}.\)
However, since both x and y are negative, the point resides in the third quadrant, requiring adjustment by adding \(\pi\) radians to \(\theta\). Therefore, the correct angle becomes \(\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}.\)

Third Quadrant

In polar coordinate conversion, understanding the quadrant where a point is located is essential for calculating the correct angle \(\theta\). The xy-plane is divided into four quadrants, and in our case, both x and y are negative, placing the point in the third quadrant.
The third quadrant is a space where both x and y values are negative. Here are a few important points about it:

• The angle \( \theta \), typically between \(\pi\) and \(\frac{3\pi}{2},\) adjusts depending on its position within that region.
• Being in this quadrant shifts our visually reckoned angles. This means we add \(\pi\) to \(\frac{\pi}{6}\) to reflect correctly within the intended space.

This distinction ensures the computed polar angle accurately represents the point's location on the plane.