Problem 5 - Calculus Early Transcendentals: Pearson New International Edition

Question

Change the following from Cartesian to spherical coordinates. (a) \((2,-2 \sqrt{3}, 4)\) (b) \((-\sqrt{2}, \sqrt{2}, 2 \sqrt{3})\)

Step-by-Step Solution

Answer

(a) \((4\sqrt{2}, -\frac{\pi}{3}, \frac{\pi}{4})\); (b) \((4, -\frac{\pi}{4}, \frac{\pi}{6})\)."

Step 1: Understand the Conversion Formulas

To convert Cartesian coordinates (x, y, z) to spherical coordinates (\rho, \theta, \phi), we use the formulas: - \(\rho = \sqrt{x^2 + y^2 + z^2}\) - \(\theta = \tan^{-1}(\frac{y}{x})\) - \(\phi = \cos^{-1}(\frac{z}{\rho})\).

Step 2: Calculate \(\rho\) for Point (a)

Given the point \((2, -2 \sqrt{3}, 4)\), calculate \(\rho\):\[\rho = \sqrt{2^2 + (-2\sqrt{3})^2 + 4^2} = \sqrt{4 + 12 + 16} = \sqrt{32} = 4\sqrt{2}.\]

Step 3: Calculate \(\theta\) for Point (a)

For the same point, calculate \(\theta\):\[\theta = \tan^{-1}\left(\frac{-2\sqrt{3}}{2}\right) = \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}.\]

Step 4: Calculate \(\phi\) for Point (a)

Now calculate \(\phi\):\[\phi = \cos^{-1}\left(\frac{4}{4\sqrt{2}}\right) = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}.\]

Step 5: Calculate \(\rho\) for Point (b)

Given the point \((-\sqrt{2}, \sqrt{2}, 2\sqrt{3})\), calculate \(\rho\):\[\rho = \sqrt{(-\sqrt{2})^2 + (\sqrt{2})^2 + (2\sqrt{3})^2} = \sqrt{2 + 2 + 12} = \sqrt{16} = 4.\]

Step 6: Calculate \(\theta\) for Point (b)

For the point \((-\sqrt{2}, \sqrt{2}, 2\sqrt{3})\), calculate \(\theta\):\[\theta = \tan^{-1}\left(\frac{\sqrt{2}}{-\sqrt{2}}\right) = \tan^{-1}(-1) = -\frac{\pi}{4}.\]

Step 7: Calculate \(\phi\) for Point (b)

Now calculate \(\phi\):\[\phi = \cos^{-1}\left(\frac{2\sqrt{3}}{4}\right) = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}.\]

Step 8: Finalize Spherical Coordinates

The spherical coordinates for the points are:- (a) \((4\sqrt{2}, -\frac{\pi}{3}, \frac{\pi}{4})\).- (b) \((4, -\frac{\pi}{4}, \frac{\pi}{6})\).

Key Concepts

Cartesian coordinates

Cartesian coordinates are a common way to identify the position of a point in three-dimensional space. We use three values: \(x\), \(y\), and \(z\) to describe a point. Each of these values represents a distance along one of the three perpendicular axes: the x-axis, the y-axis, and the z-axis.
• **x-axis**: The horizontal axis.
• **y-axis**: The vertical axis.
• **z-axis**: Represents depth, the point's height or depth from the x-y plane.

In our exercise, our points are given as \( (x, y, z) \), which means we are starting with Cartesian coordinates. For example, the point \( (2, -2\sqrt{3}, 4) \) tells us directly about its position relative to these axes. This straightforward geometric representation is fundamental in engineering, physics, computer graphics, and many other fields.

coordinate conversion

The process of coordinate conversion is crucial in mathematics and physics for transforming coordinates between different systems, each useful for different applications. The conversion formulas are necessary to go from one system to another, in this case from Cartesian to spherical coordinates.

We convert Cartesian coordinates \( (x, y, z) \) to spherical coordinates \( (\rho, \theta, \phi) \) using these relationships:

• \( \rho = \sqrt{x^2 + y^2 + z^2} \)
• \( \theta = \tan^{-1}(\frac{y}{x}) \)
• \( \phi = \cos^{-1}(\frac{z}{\rho}) \)

These formulas allow us to interpret the 3D space in terms of radius from the origin, azimuthal angle (usually from x), and polar angle from the vertical axis.

Coordinate conversion assists in representing points in a way that simplifies calculations or helps visualize problems better, making operations such as integration and differentiation more feasible in dealing with spherical symmetry.

trigonometric functions

Trigonometric functions are essential in converting coordinates, especially when calculating angles like \( \theta \) and \( \phi \) during the process of coordinate conversion. Let's briefly discuss these functions:

• **Tangent function**: \( \tan^{-1}(\frac{y}{x}) \) computes angle \( \theta \). This function helps determine the angle in the x-y plane relative to the x-axis.
• **Cosine function**: The function \( \cos^{-1}(\frac{z}{\rho}) \) calculates \( \phi \). This angle represents the elevation above the x-y plane.

It's important to note how angles are derived:

• \( \theta \) is essentially the direction within the horizontal plane, \( y/x \) reflecting the angle.
• Meanwhile, \( \phi \) involves elevation, i.e., how high from the base plane the point ascends, determined by the ratio of vertical distance to the overall radial distance.
These trigonometric relationships are powerful tools for simplifying problems involving spherical coordinates.

polar coordinates

Polar coordinates are similar but distinct from spherical coordinates, primarily used for two-dimensional space. In the polar coordinate system, we identify the location of points using a radius and an angle, generally denoted as \( (r, \theta) \).

• **Radius \( r \)**: The distance from the origin to the point.
• **Angle \( \theta \)**: The angle from the positive x-axis, measured counter-clockwise.

In spherical coordinates, we extend this concept into the third dimension by adding a radial component \( \rho \) and a polar angle \( \phi \). Thus, the spherical coordinate system includes aspects of both polar and Cartesian coordinates.

The connection between polar and spherical systems helps us translate our understanding from two dimensions to three, illustrating the natural progression from simple to complex geometric interpretations. This shows how mathematics builds upon itself, extending concepts into new domains.