Problem 2 - Calculus Early Transcendentals: Pearson New International Edition

Question

Change the following from cylindrical to spherical coordinates. (a) \((1, \pi / 2,1)\) (b) \((-2, \pi / 4,2)\)

Step-by-Step Solution

Answer

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

Step 1: Understanding the Problem

We need to convert from cylindrical coordinates \((r, \theta, z)\) to spherical coordinates \((\rho, \phi, \theta)\). The formulas for conversion are: \(\rho = \sqrt{r^2 + z^2}\), \(\phi = \arctan\left(\frac{r}{z}\right)\), and \(\theta\) remains the same.

Step 2: Applying Formulas for Point (a)

Given cylindrical coordinates \((r, \theta, z) = (1, \frac{\pi}{2}, 1)\), we apply the formulas:- \(\rho = \sqrt{1^2 + 1^2} = \sqrt{2}\).- \(\phi = \arctan\left(\frac{1}{1}\right) = \frac{\pi}{4}\).- \(\theta = \frac{\pi}{2}\).Thus the spherical coordinates are \((\sqrt{2}, \frac{\pi}{4}, \frac{\pi}{2})\).

Step 3: Applying Formulas for Point (b)

Given cylindrical coordinates \((r, \theta, z) = (-2, \frac{\pi}{4}, 2)\), we apply the formulas:- \(\rho = \sqrt{(-2)^2 + (2)^2} = \sqrt{8} = 2\sqrt{2}\).- \(\phi = \arctan\left(\frac{-2}{2}\right) = \arctan(-1) = -\frac{\pi}{4}\).- \(\theta = \frac{\pi}{4}\).Thus the spherical coordinates are \((2\sqrt{2}, -\frac{\pi}{4}, \frac{\pi}{4})\).

Key Concepts

Cylindrical Coordinates

Cylindrical coordinates provide a way of representing points in 3D space using a combination of both circular and linear components. These coordinates are defined by three values:

• \( r \) - the radial distance from the origin in the xy-plane. This measures how far away the point is from the z-axis.
• \( \theta \) - the angular coordinate, or azimuthal angle, which represents the counterclockwise angle from the positive x-axis in the xy-plane.
• \( z \) - the height above or below the xy-plane, essentially representing the point's elevation.

Imagine wrapping the xy-plane around a pole. The angle \( \theta \) tells you which direction you're facing around that pole, while \( r \) tells you how far you are from the pole. Finally, \( z \) adds the third dimension by indicating how high or low you are from the xy-plane. This system is particularly useful in contexts where symmetry around an axis is present, such as in engineering and physics.

Coordinate Conversion

Coordinate conversion allows you to change from one system of coordinates to another. In the context of this exercise, we are converting from cylindrical coordinates \((r, \theta, z)\) to spherical coordinates \((\rho, \phi, \theta)\). Spherical coordinates are another way to describe the location of a point in 3D space using:

• \( \rho \) - the radial distance from the origin to the point in space.
• \( \phi \) - the polar angle, which is the angle from the positive z-axis down to the point.
• \( \theta \) - similar to that in cylindrical coordinates, it is the same azimuthal angle.

The formulas for converting between cylindrical and spherical coordinates are:- \( \rho = \sqrt{r^2 + z^2} \)- \( \phi = \arctan\left(\frac{r}{z}\right) \)- \( \theta = \theta \)Using these conversion formulas, we can translate points between these two systems. Each conversion provides a different perspective on the point's positioning and is helpful in various fields, such as computer graphics, navigation, and even astronomy.

Trigonometric Functions

Trigonometric functions are mathematical functions that relate the angles of a right triangle to the ratios of its sides. In coordinate conversion, these functions help describe angles and relationships between directions in space. The key trigonometric functions used in this exercise are sine, cosine, and arctan (or arctangent).

• The arctangent function, represented as \( \arctan(x) \), gives the angle whose tangent is \( x \). This is essential in converting between cylindrical \( r \) and \( z \) values to spherical \( \phi \).

Consider a right triangle with sides \( a \) opposite the angle \( \theta \), and\( b \) adjacent to \( \theta \). The tangent of \( \theta \) is the ratio of the opposite side to the adjacent side, \( \tan(\theta) = \frac{a}{b} \). The arctangent function does the inverse, calculating the angle from the ratio of these two sides.These trigonometric understandings simplify complex spatial relationships, making them practical tools for a variety of scientific calculations. Once you grasp the basic functions of sine, cosine, and arctangent, you can unlock a new layer of understanding of angles and distances in multiple coordinate systems.