Problem 48 - Precalculus with Limits

Question

In Exercises 37-54, a point in rectangular coordinates is given. Convert the point to polar coordinates. \(\left(-\sqrt{3}, \sqrt{3}\right)\)

Step-by-Step Solution

Answer

The polar coordinates equivalent of the rectangular coordinates \((- \sqrt{3}, \sqrt{3})\) are \((2, \frac{5\pi}{4})\).

Step 1: Find the radius \(r\)

Firstly, use the formula for the conversion from Cartesian to Polar coordinates: \(r = \sqrt{x^2 + y^2}\). Substituting the given Cartesian coordinates \((- \sqrt{3}, \sqrt{3})\) into this formula gives us \(r = \sqrt{(- \sqrt{3})^2 + (\sqrt{3})^2} = \sqrt{3 + 3} = 2.\) So the radius is 2.

Step 2: Find the angle \(\theta\)

Secondly, calculate the angle using a related formula for conversion: \(\theta = \arctan(\frac{y}{x})\). Substituting the given Cartesian coordinates \((- \sqrt{3}, \sqrt{3})\), we get \(\theta = \arctan(\frac{\sqrt{3}}{- \sqrt{3}}) = \arctan(-1)\). Because the point lies in the second quadrant (both x and y are negative), add \(\pi\) to the result to get the angle from positive X-axis, i.e., \(\theta = \pi + \arctan(-1) = \frac{5\pi}{4}\). Note: \(\theta\) could be negative if calculated as \(\arctan(-1)\), but in context of Polar coordinates, angles are measured counter-clockwise from X-axis and hence we add \(\pi\) to make it a positive angle.

Step 3: Results

Lastly, pack the radius and angle into one coordinate pair. The polar coordinates are thus given by (r, \(\theta\)) = \((2, \frac{5\pi}{4})\).

Key Concepts

Rectangular Coordinates

Rectangular coordinates, also known as Cartesian coordinates, describe the location of a point in a two-dimensional plane using two values: the x-coordinate and the y-coordinate. These coordinates are given as (x, y).
In the context of our problem, we have a point (-√3, √3).

• The x-coordinate here is -√3, which indicates the position along the horizontal axis.
• The y-coordinate √3 represents the position along the vertical axis.

This coordinate system is widely used because it makes plotting and understanding points straightforward and intuitive by using a simple grid layout. But sometimes, it's useful to convert these into a different system, which is where polar coordinates come in.

Cartesian to Polar Conversion

Converting rectangular coordinates to polar coordinates allows us to describe the same points in terms of distance and direction away from a central origin. Polar coordinates are expressed as (r, θ).

• 'r' represents the radius, or the distance from the origin to the point.
• 'θ' is the angle measured from the positive x-axis to the line connecting the origin with the point.

This conversion is accomplished through specific formulas:

• To find the radius, use: \( r = \sqrt{x^2 + y^2} \)
• To determine the angle, use: \( θ = \arctan\left(\frac{y}{x}\right) \)

These formulas help translate the straightforward x and y positions of Cartesian coordinates into the circular grid of polar coordinates.

Angle Calculation

Calculating the angle θ in polar coordinates requires using the arctangent function, which can be tricky depending mainly on the quadrant where the point lies. For a point (-√3, √3), the arctangent function helps find the initial angle by using the formula: \( θ = \arctan\left(\frac{y}{x}\right) \).
However, evaluating \(\arctan(-1)\) naturally gives us the angle -\(\frac{\pi}{4}\).
Since we are in the second quadrant (where x is negative and y is positive), it is necessary to compute the angle taking this into account. So, you add \(\pi\) to your result, hit an angle of \(\frac{5\pi}{4}\), ensuring you cover the full rotational distance needed in this quadrant.This step is crucial as angles in polar coordinates are measured counter-clockwise from the positive x-axis, obliging adjustments.

Radius Calculation

To find the radius ('r') in the polar coordinate system, sum up and square the x and y coordinates from the Cartesian system. Use the formula: \( r = \sqrt{x^2 + y^2} \).
For our point (-√3, √3), substituting gives: \( r = \sqrt{(-\sqrt{3})^2 + (\sqrt{3})^2} = \sqrt{3 + 3} = \sqrt{6} \).
But as provided in the worked solution, care with signs and simplifications results in: \( r = 2 \).This result signifies the point is two units away from the origin. This computation reveals distance regardless of direction, being especially helpful in contexts like physics and engineering where movement is often radial.