Problem 44
Question
The rectangular coordinates of a point are given. Find polar coordinates of each point. Express \(\theta\) in radians. $$ (-2 \sqrt{3}, 2) $$
Step-by-Step Solution
Verified Answer
The polar coordinates of the point with rectangular coordinates (-2\(\sqrt{3}\), 2) are (4, \(\frac{5\pi}{6}\)) radians.
1Step 1: Calculate the distance r from the point to the origin
We are given the rectangular coordinates of the point as (-2\(\sqrt{3}\), 2). We need to calculate the distance \(r\) from this point to the origin. We can use the equation \( r = \sqrt{x^2 + y^2} \). Substituting the values for \(x\) and \(y\) we get \( r = \sqrt{(-2\sqrt{3})^2 + 2^2} = \sqrt{12 + 4} = \sqrt{16} = 4 \). Therefore, \(r = 4\).
2Step 2: Calculate the counter clockwise angle \(\theta\) from the positive x-axis
We need to calculate the angle \(\theta\) measured counter clockwise from the positive x-axis. We can use the equation \(\theta= \arctan{\frac{y}{x}}\). As tan is undefined for \(-2\sqrt{3}\), we can consider the sign of \(\theta\) separately. Substituting the values for \(x\) and \(y\) we get \(\theta= \arctan{\frac{2}{-2\sqrt{3}}} = \frac{\pi}{6}\). As it lies in the second quadrant \(\theta\) = \(\pi - \frac{\pi}{6} = \frac{5\pi}{6}\). Therefore, \(\theta= \frac{5\pi}{6}\) radians.
Key Concepts
Rectangular CoordinatesDistance to the OriginAngle from Positive x-AxisQuadrant Identification
Rectangular Coordinates
Rectangular coordinates, also known as Cartesian coordinates, are a way to describe the location of a point in a plane using two values - the x-coordinate and y-coordinate. You can think about these coordinates as directions: how far a point is to the left or right (x-coordinate) and how far it is up or down (y-coordinate).
These coordinates provide a straightforward way to locate points on a two-dimensional plane. When we translate these to polar coordinates, we focus on the distance from the origin and the angle from the positive x-axis instead.
- x-coordinate: This tells us the horizontal position. For the point (-2√3, 2), the value -2√3 signifies the point is 2√3 units to the left of the origin.
- y-coordinate: This indicates the vertical position. Here, 2 means the point is 2 units upwards from the origin.
These coordinates provide a straightforward way to locate points on a two-dimensional plane. When we translate these to polar coordinates, we focus on the distance from the origin and the angle from the positive x-axis instead.
Distance to the Origin
Calculating the distance from a point to the origin is a crucial step when converting rectangular coordinates to polar form. This distance is usually represented by the letter \(r\), which stands for the radial distance in polar coordinates. To find \(r\), we apply the Pythagorean theorem: \[ r = \sqrt{x^2 + y^2} \]
For the point \((-2\sqrt{3}, 2)\), we substitute the values for \(x\) and \(y\) into the formula: \[ r = \sqrt{(-2\sqrt{3})^2 + 2^2} = \sqrt{12 + 4} = \sqrt{16} = 4\]
Therefore, the distance \(r\) from the point to the origin is 4 units. Understanding this distance helps visualize where the point lies in relation to the center of the graph, which is called the origin.
For the point \((-2\sqrt{3}, 2)\), we substitute the values for \(x\) and \(y\) into the formula: \[ r = \sqrt{(-2\sqrt{3})^2 + 2^2} = \sqrt{12 + 4} = \sqrt{16} = 4\]
Therefore, the distance \(r\) from the point to the origin is 4 units. Understanding this distance helps visualize where the point lies in relation to the center of the graph, which is called the origin.
Angle from Positive x-Axis
The angle in polar coordinates, denoted as \(\theta\), tells us how far a point is emerged from the positive x-axis, measured in radians. This angle is a critical element for converting to polar coordinates.
Thus, the angle \(\theta\) is \(\frac{5\pi}{6}\) radians. Correctly determining this angle is vital because it tells us where the point is oriented in the plane.
- To find \(\theta\), we use the equation given by the tangent function: \(\theta = \arctan{\frac{y}{x}}\).
- For our point \((-2\sqrt{3}, 2)\), this calculation becomes \(\theta = \arctan{\frac{2}{-2\sqrt{3}}}\).
- In mathematical calculations, we also need to account for which quadrant the point lies in to get the correct \(\theta\). Here, \(\theta = \frac{\pi}{6}\) is in the second quadrant, so we adjust by adding to \(\pi\): \(\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\).
Thus, the angle \(\theta\) is \(\frac{5\pi}{6}\) radians. Correctly determining this angle is vital because it tells us where the point is oriented in the plane.
Quadrant Identification
In the Cartesian plane, quadrants help us pinpoint the location of points by dividing the plane into four sections. This is essential, especially when determining polar angles, as it helps in adjusting \(\theta\) based on the point's position. Here’s a simple guide to remember the quadrant arrangement:
For the point \((-2\sqrt{3}, 2)\), both coordinates tell us it is in the second quadrant since \(x\) is negative and \(y\) is positive.
Identifying the correct quadrant ensures that we calculate the angle \(\theta\) accurately. This quadrant also influences the adjustment of \(\theta\) to get the correct direction of the point relative to the positive x-axis. Knowing how to identify the quadrants is a crucial step in many mathematical applications, especially when switching between coordinate systems.
- First Quadrant: Both x and y are positive.
- Second Quadrant: x is negative, y is positive.
- Third Quadrant: Both x and y are negative.
- Fourth Quadrant: x is positive, y is negative.
For the point \((-2\sqrt{3}, 2)\), both coordinates tell us it is in the second quadrant since \(x\) is negative and \(y\) is positive.
Identifying the correct quadrant ensures that we calculate the angle \(\theta\) accurately. This quadrant also influences the adjustment of \(\theta\) to get the correct direction of the point relative to the positive x-axis. Knowing how to identify the quadrants is a crucial step in many mathematical applications, especially when switching between coordinate systems.
Other exercises in this chapter
Problem 44
In Exercises \(39-46\), find the unit vector that has the same direction as the vector \(\mathbf{v}\) $$v=4 i-2 j$$
View solution Problem 44
In Exercises \(37-44,\) find the product of the complex numbers. Leave answers in polar form. $$ \begin{aligned} &z_{1}=1+i\\\ &z_{2}=2+2 i \end{aligned} $$
View solution Problem 45
Determine whether \(\mathbf{v}\) and \(\mathbf{w}\) are parallel, orthogonal, or neither. $$ \mathbf{v}=3 \mathbf{i}-5 \mathbf{j}, \quad \mathbf{w}=6 \mathbf{i}
View solution Problem 45
You are on a fishing boat that leaves its pier and heads east. After traveling for 25 miles, there is a report warning of rough seas directly south. The captain
View solution