Problem 41 - Advanced Engineering Mathematics

Question

In Problems \(39-42\), convert the point given in rectangular coardinates to cylindrical coordinates. $$ (-\sqrt{2}, \sqrt{6}, 2) $$

Step-by-Step Solution

Answer

Cylindrical coordinates: \((2\sqrt{2}, \frac{5\pi}{3}, 2)\)."

Step 1: Understand the problem

We are given a point in rectangular coordinates \((-\sqrt{2}, \sqrt{6}, 2)\) and we need to convert it to cylindrical coordinates \((r, \theta, z)\). In cylindrical coordinates, \(r = \sqrt{x^2 + y^2}\), \(\theta = \tan^{-1}(\frac{y}{x})\), and \(z\) is the same in both systems.

Step 2: Find radius \(r\)

Use the formula \(r = \sqrt{x^2 + y^2}\). Substitute \(x = -\sqrt{2}\) and \(y = \sqrt{6}\).\[r = \sqrt{(-\sqrt{2})^2 + (\sqrt{6})^2} = \sqrt{2 + 6} = \sqrt{8} = 2\sqrt{2}\]

Step 3: Find angle \(\theta\)

Use the formula \(\theta = \tan^{-1}(\frac{y}{x})\). Substitute \(x = -\sqrt{2}\) and \(y = \sqrt{6}\).\[\theta = \tan^{-1}\left(\frac{\sqrt{6}}{-\sqrt{2}}\right) = \tan^{-1}\left(-\sqrt{3}\right)\]\(\theta\) is in the fourth quadrant.

Step 4: Find \(\theta\) in standard position

Since \(\tan(\theta) = -\sqrt{3}\) and we know it points in the fourth quadrant, \(\theta\) in radians is \(-\frac{\pi}{3}\). For standard position, add \(2\pi\) if necessary for a positive angle.\(\theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}\).

Step 5: Verify and format the cylindrical coordinates

Now we have the cylindrical coordinates \(r = 2\sqrt{2}\), \(\theta = \frac{5\pi}{3}\), and \(z = 2\). The final conversion is \((2\sqrt{2}, \frac{5\pi}{3}, 2)\).

Key Concepts

Rectangular Coordinates

Rectangular coordinates, also known as Cartesian coordinates, represent points in a plane using a pair of numerical values. Typically denoted as \(x, y, z\), this system finds the position of a point by measuring along orthogonal axes. Here is a basic understanding:

• **x-axis**: The horizontal line running left (negative) to right (positive).
• **y-axis**: The vertical line running down (negative) to up (positive). On planes, this measures how far up or down the point is from the origin.
• **z-axis**: In space, this adds a third dimension of depth.

In the given problem, the coordinates are presented as \((-\sqrt{2}, \sqrt{6}, 2)\). Each of these values indicates a specific distance from the origin along the respective axes.
Rectangular coordinates are particularly useful for plotting and visualizing points in a clear and straightforward manner, providing the framework for various mathematical explorations and applications.

Coordinate Conversion

Coordinate conversion refers to the process of translating coordinates from one system to another. This is crucial for numerous practical applications, such as switching from rectangular to cylindrical coordinates.
For converting a point from rectangular to cylindrical coordinates, we use the following formulas:

• **Radius (r)**: Calculated using \(r = \sqrt{x^2 + y^2}\). This represents the point's distance from the z-axis.
• **Angle (\(\theta\))**: Determined with \(\theta = \tan^{-1}(\frac{y}{x})\). It indicates the angle between the positive x-axis and the line from the origin to the point.
• **Height (z)**: Same in both cylindrical and rectangular coordinates; in essence, translating vertically does not change.

In our exercise, employing these formulas helps to smoothly convert \((-\sqrt{2}, \sqrt{6}, 2)\) into the cylindrical form \((2\sqrt{2}, \frac{5\pi}{3}, 2)\). Understanding these conversions enhances flexibility in mathematical problem solving.

Trigonometric Functions

Trigonometric functions play a pivotal role in coordinate conversion by providing the relationship between angles and distances in a plane. There are some key functions used in our problems:

• **Sine (sin \(\theta\))**: Represents the ratio of the opposite side to the hypotenuse in a right triangle.
• **Cosine (cos \(\theta\))**: Shows the ratio of the adjacent side to the hypotenuse.
• **Tangent (tan \(\theta\))**: Defined as \(\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}\) or equivalently \(\frac{y}{x}\) in coordinate terms.

For angle determination, the exercise used the tangent function, derived by \(\tan^{-1}(\frac{\sqrt{6}}{-\sqrt{2}})\), providing an angle in the fourth quadrant. Trigonometric functions are essential as they allow the conversion from distance measurements to angles, a crucial step in shifting between coordinate systems. Thus, mastering these functions offers a deeper understanding of geometric relationships and spatial reasoning.