Problem 4 - Calculus

Question

Find the polar coordinates \(r \geqslant 0\) and \(0 \leqslant 0<2 \pi\) of these points. $$ (x, y)=(-1, \sqrt{3}) $$

Step-by-Step Solution

Answer

The polar coordinates are \((2, \frac{2\pi}{3})\).

Step 1: Calculate the radius

The polar coordinate radius \( r \) can be found using the formula \( r = \sqrt{x^2 + y^2} \). For the point \((-1, \sqrt{3})\), we have:\[r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2\]

Step 2: Calculate the angle

The angle \( \theta \) in polar coordinates can be found using the tangent function, \( \tan(\theta) = \frac{y}{x} \). For the given point, we calculate:\[\tan(\theta) = \frac{\sqrt{3}}{-1} = -\sqrt{3}\]The reference angle \( \theta_r \) that satisfies \( \tan(\theta_r) = \sqrt{3} \) is \( \frac{\pi}{3} \). Since \(x < 0\) and \(y > 0\), the angle \( \theta \) is in the second quadrant, so:\[\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\]

Step 3: Express in polar coordinates

Combine the calculated \( r \) and \( \theta \) to express the point in polar coordinates. The polar coordinates for \((-1, \sqrt{3})\) are:\[(r, \theta) = (2, \frac{2\pi}{3})\]

Key Concepts

Radius Calculation

Understanding how to calculate the radius in polar coordinates is crucial. The radius, represented as \( r \), describes the distance from the origin (0,0) to a given point in the plane. To find it, use the formula \( r = \sqrt{x^2 + y^2} \). This formula is based on the Pythagorean theorem, which relates the sides of a right triangle. Here, \( x \) and \( y \) are the Cartesian coordinates of the point.For example, given the point \((-1, \sqrt{3})\), plug in the values of \( x = -1 \) and \( y = \sqrt{3} \):

• Calculate \( x^2 \): \( (-1)^2 = 1 \)
• Calculate \( y^2 \): \( (\sqrt{3})^2 = 3 \)
• Add them up to find \( r^2 \): \( 1 + 3 = 4 \)
• Take the square root to find \( r \): \( \sqrt{4} = 2 \)

This gives us the radius \( r = 2 \). It's a simple method once you get the hang of it, providing a clear way to understand how far a point is from the origin.

Angle Determination

Finding the angle \( \theta \) is another critical part of converting to polar coordinates. The angle tells you the direction of the point relative to the positive x-axis. Use the tangent function: \( \tan(\theta) = \frac{y}{x} \). This works because tangent in trigonometry relates the opposite side \( y \) to the adjacent side \( x \) in a right triangle.Let's look at the point \((-1, \sqrt{3})\). You calculate the tangent:

• \( \tan(\theta) = \frac{\sqrt{3}}{-1} = -\sqrt{3} \)

Knowing \( \tan(\theta_r) = \sqrt{3} \), your reference angle \( \theta_r \) is \( \frac{\pi}{3} \). Yet, because our point's \( x \) is negative and \( y \) is positive, it lies in the second quadrant of the coordinate system.In the second quadrant:

• The formula becomes \( \theta = \pi - \theta_r \).
• Substitute your reference angle: \( \theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \)

This calculation aligns \( \theta \) correctly with the point's position on the plane, giving us \( \theta = \frac{2\pi}{3} \).

Coordinate Conversion

Converting between Cartesian and polar coordinates is a common task. The goal is to describe a 2D point using the radius and angle rather than \( x \) and \( y \). This new perspective can often simplify complex situations, making polar coordinates handy in fields like physics or engineering.Here's the process for the point \((-1, \sqrt{3})\):From previous calculations:

• Radius \( r = 2 \)
• Angle \( \theta = \frac{2\pi}{3} \)

The polar coordinates are thus expressed as \((r, \theta) = (2, \frac{2\pi}{3})\).This new format is sometimes visually more intuitive, as it directly relates to the distance and direction from the origin. To convert back, if needed:

• \( x = r \cos(\theta) = 2 \cos(\frac{2\pi}{3}) \)
• \( y = r \sin(\theta) = 2 \sin(\frac{2\pi}{3}) \)

Double-checking these calculated values helps guarantee a correct two-way conversion.