Problem 19 - Precalculus

Question

Convert each point to exact polar coordinates. Assume that \(0 \leq \theta<2 \pi.\) $$(-\sqrt{3},-1)$$

Step-by-Step Solution

Answer

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

Step 1: Determine the radius

Calculate the radius by using the formula \( r = \sqrt{x^2 + y^2} \). Substitute \( x = -\sqrt{3} \) and \( y = -1 \) into the formula: \( r = \sqrt{(-\sqrt{3})^2 + (-1)^2} \). This simplifies to \( r = \sqrt{3 + 1} = \sqrt{4} = 2 \).

Step 2: Calculate the angle

Find the angle \( \theta \) using \( \tan\theta = \frac{y}{x} \). Substitute \( x = -\sqrt{3} \) and \( y = -1 \) into the formula: \( \tan\theta = \frac{-1}{-\sqrt{3}} = \frac{1}{\sqrt{3}} \). Thus, \( \theta \) should be \( \frac{\pi}{6} \), but note that the point is in the third quadrant, adjust \( \theta \) by adding \( \pi \) to get \( \theta = \frac{7\pi}{6} \).

Step 3: Verify the range of angle

Ensure \( \theta \) is within the required range \([0, 2\pi)\). Since \( \frac{7\pi}{6} \) is already between 0 and \( 2\pi \), no further adjustment is necessary.

Step 4: Write the polar coordinates

Combine the results into polar coordinates: \((r, \theta) = (2, \frac{7\pi}{6})\). This is the final result in exact polar coordinates.

Key Concepts

Radius Calculation

In the realm of polar coordinates, the radius serves as a fundamental measurement. It quantifies the distance from the origin to the point in question. To determine the radius, we utilize the formula: \[ r = \sqrt{x^2 + y^2} \]This formula comes from the Pythagorean theorem, which dictates how we calculate diagonal distances on a coordinate plane. In the given exercise, we're provided with the point \((-\sqrt{3}, -1)\). By substituting these values into our formula, we compute:\[ r = \sqrt{(-\sqrt{3})^2 + (-1)^2} \]This simplifies to:\[ r = \sqrt{3 + 1} = \sqrt{4} = 2 \]Thus, the radius of the point from the origin is 2. This tells us how far the point is from the center of the coordinate system.

Angle Determination

After finding the radius, the next step is to determine the angle \(\theta\) in polar coordinates. The angle helps us understand the direction in which the point is situated relative to the positive x-axis. We find \(\theta\) using:\[ \tan\theta = \frac{y}{x} \]For our example point \((-\sqrt{3}, -1)\), we substitute into the tangent formula:\[ \tan\theta = \frac{-1}{-\sqrt{3}} = \frac{1}{\sqrt{3}} \]Normally, this corresponds to \(\theta = \frac{\pi}{6}\). However, direct computation of \(\theta\) doesn't take into account which quadrant the point resides in. Therefore, one has to consider the correct angle by possibly adjusting \(\theta\) to reflect the actual quadrant.

Quadrant Analysis

Analyzing the quadrant in which the point lies is essential for determining the correct angle in polar coordinates. Basic knowledge of quadrants tells us that the sign of the x and y coordinates determines the quadrant:

• Quadrant I:
  • x > 0, y > 0
• Quadrant II:
  • x < 0, y > 0
• Quadrant III:
  • x < 0, y < 0
• Quadrant IV:
  • x > 0, y < 0

Considering our point \((-\sqrt{3}, -1)\), both coordinates are negative. This places our point in the third quadrant. In the third quadrant, we must add \(\pi\) to the initial angle \(\frac{\pi}{6}\) to account for its position:\[\theta = \frac{\pi}{6} + \pi = \frac{7\pi}{6} \]This adjustment ensures that the angle accurately reflects the point's location in the coordinate system.

Trigonometric Functions

Understanding trigonometric functions is crucial when working with polar coordinates. They link angles with ratios of sides in triangles and are key when converting between Cartesian and polar coordinates:

• \(\sin\theta = \frac{\text{opposite side}}{\text{hypotenuse}}\)
• \(\cos\theta = \frac{\text{adjacent side}}{\text{hypotenuse}}\)
• \(\tan\theta = \frac{\text{opposite side}}{\text{adjacent side}}\)

In the given exercise, the tangent function is used to find the angle associated with the coordinates. The tangent of an angle in a right triangle is the ratio of the y-coordinate to the x-coordinate, providing us a measure of its direction. By setting \(\tan\theta = \frac{1}{\sqrt{3}}\), we first find \(\theta = \frac{\pi}{6}\) ignoring the quadrant adjustment.Finally, understanding these functions better equips us to leverage them for calculations such as converting between coordinate systems, which requires a grasp of how angles and ratios interrelate.