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Magazine Chapter 13: Problem 5
Points are given in either the rectangular, cylindrical or spherical coordinate systems. Find the coordinates of the points in the other systems. (a) Points in rectangular coordinates: (2,2,1) and \((-\sqrt{3}, 1,0)\) (b) Points in cylindrical coordinates: \((2, \pi / 4,2)\) and \((3,3 \pi / 2,-4)\) (c) Points in spherical coordinates: \((2, \pi / 4, \pi / 4)\) and (1,0,0)
Short Answer
Expert verified
All coordinates converted for each point: first set to cylindrical/spherical, second set to rectangular/spherical, and third set to rectangular/cylindrical.
Step by step solution
01
Rectangular to Cylindrical/Spherical (Point 1)
For the point (2,2,1), convert to cylindrical coordinates using the formulas:- Radial distance: \(r = \sqrt{x^2 + y^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}\)- Angle: \(\theta = \tan^{-1}(\frac{y}{x}) = \tan^{-1}(\frac{2}{2}) = \frac{\pi}{4}\)- Height (z remains unchanged): \(z = 1\)Cylindrical coordinates: \((2\sqrt{2}, \frac{\pi}{4}, 1)\)Convert to spherical coordinates:- Radius: \(\rho = \sqrt{x^2 + y^2 + z^2} = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{9} = 3\)- Polar angle: \(\phi = \cos^{-1}(\frac{z}{\rho}) = \cos^{-1}(\frac{1}{3})\)- Azimuthal angle: \(\theta = \tan^{-1}(\frac{y}{x}) = \frac{\pi}{4}\)Spherical coordinates: \((3, \frac{\pi}{4}, \cos^{-1}(\frac{1}{3}))\)
02
Rectangular to Cylindrical/Spherical (Point 2)
For the point \((-\sqrt{3}, 1, 0)\), convert to cylindrical coordinates:- Radial distance: \(r = \sqrt{x^2 + y^2} = \sqrt{(-\sqrt{3})^2 + 1^2} = 2\)- Angle: \(\theta = \tan^{-1}(\frac{y}{x}) = \tan^{-1}(\frac{1}{-sqrt{3}}) = -\frac{pi}{6}\) (in the correct quadrant: \(\theta = \frac{5\pi}{6}\))- Height: \(z = 0\)Cylindrical coordinates: \((2, \frac{5\pi}{6}, 0)\)Convert to spherical coordinates:- Radius: \(\rho = \sqrt{x^2 + y^2 + z^2} = \sqrt{3 + 1} = 2\)- Polar angle: \(\phi = \cos^{-1}(\frac{z}{\rho}) = \cos^{-1}(0) = \frac{\pi}{2}\)- Azimuthal angle: \(\theta = \frac{5\pi}{6}\)Spherical coordinates: \((2, \frac{5\pi}{6}, \frac{\pi}{2})\)
03
Cylindrical to Rectangular/Spherical (Point 1)
For the point \((2, \pi/4, 2)\) in cylindrical coordinates, convert to rectangular coordinates:- \(x = r\cos\theta = 2\cos(\pi/4) = 2 \cdot \frac{\sqrt{2}}{2} = \sqrt{2}\)- \(y = r\sin\theta = 2\sin(\pi/4) = 2 \cdot \frac{\sqrt{2}}{2} = \sqrt{2}\)- \(z = 2\)Rectangular coordinates: \((\sqrt{2}, \sqrt{2}, 2)\)Convert to spherical coordinates:- \(\rho = \sqrt{x^2 + y^2 + z^2} = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2 + 2^2} = \sqrt{8} = 2\sqrt{2}\)- \(\theta = \pi/4\)- \(\phi = \cos^{-1}(\frac{z}{\rho}) = \cos^{-1}(\frac{2}{2\sqrt{2}}) = \frac{\pi}{4}\)Spherical coordinates: \((2\sqrt{2}, \pi/4, \pi/4)\)
04
Cylindrical to Rectangular/Spherical (Point 2)
For the point \((3, 3\pi/2, -4)\) in cylindrical coordinates, convert to rectangular coordinates:- \(x = r\cos\theta = 3\cos(3\pi/2) = 3 \cdot 0 = 0\)- \(y = r\sin\theta = 3\sin(3\pi/2) = 3 \cdot (-1) = -3\)- \(z = -4\)Rectangular coordinates: \((0, -3, -4)\)Convert to spherical coordinates:- \(\rho = \sqrt{x^2 + y^2 + z^2} = \sqrt{0^2 + (-3)^2 + (-4)^2} = 5\)- \(\theta = 3\pi/2\)- \(\phi = \cos^{-1}(\frac{z}{\rho}) = \cos^{-1}(\frac{-4}{5})\)Spherical coordinates: \((5, 3\pi/2, \cos^{-1}(-4/5))\)
05
Spherical to Rectangular/Cylindrical (Point 1)
For the point \((2, \pi/4, \pi/4)\) in spherical coordinates, convert to rectangular coordinates:- \(x = \rho\sin\phi\cos\theta = 2\sin(\pi/4)\cos(\pi/4) = 2 \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = 1\)- \(y = \rho\sin\phi\sin\theta = 2\sin(\pi/4)\sin(\pi/4) = 2 \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = 1\)- \(z = \rho\cos\phi = 2\cos(\pi/4) = 2 \cdot \frac{\sqrt{2}}{2} = \sqrt{2}\)Rectangular coordinates: \((1, 1, \sqrt{2})\)Convert to cylindrical coordinates:- \(r = \sqrt{x^2 + y^2} = \sqrt{1^2 + 1^2} = \sqrt{2}\)- \(\theta = \pi/4\)- \(z = \sqrt{2}\)Cylindrical coordinates: \((\sqrt{2}, \pi/4, \sqrt{2})\)
06
Spherical to Rectangular/Cylindrical (Point 2)
For the point \((1, 0, 0)\) in spherical coordinates, convert to rectangular coordinates:- \(x = \rho\sin\phi\cos\theta = 1\cdot0\cdot1 = 0\)- \(y = \rho\sin\phi\sin\theta = 1\cdot0\cdot0 = 0\)- \(z = \rho\cos\phi = 1\cdot1 = 1\)Rectangular coordinates: \((0, 0, 1)\)Convert to cylindrical coordinates:- \(r = \sqrt{x^2 + y^2} = 0\)- \(\theta\) can be any value because \( r = 0 \), conventionally 0- \(z = 1\)Cylindrical coordinates: \((0, 0, 1)\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rectangular Coordinates
Rectangular coordinates, also known as Cartesian coordinates, are the most common type of coordinate system. They allow us to specify the position of a point in three-dimensional space using three values: \(x\), \(y\), and \(z\). This system is particularly intuitive because it uses perpendicular axes to represent dimensions. For example:
This system is widely used because it aligns so well with our natural understanding of up, down, left, right, and depth.
- \(x\) represents the horizontal distance along the x-axis.
- \(y\) represents the horizontal distance along the y-axis.
- \(z\) represents the vertical distance from the xy-plane.
This system is widely used because it aligns so well with our natural understanding of up, down, left, right, and depth.
Cylindrical Coordinates
Cylindrical coordinates provide a way to specify points in three-dimensional space by combining the polar coordinate system and a height value. They are particularly useful for problems with rotational symmetry around a central axis. Cylindrical coordinates are denoted by \( (r, \theta, z) \), where:
\[ r = \sqrt{x^2 + y^2} \]
\[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \]
This system efficiently tackles 3D problems like determining the position of a point on a cylinder, where the height and radial distance are easily defined.
- \(r\) is the radial distance from the z-axis.
- \(\theta\) is the angle between the projection of the point in the xy-plane and the positive x-axis.
- \(z\) is the height above the xy-plane, which is the same as in rectangular coordinates.
\[ r = \sqrt{x^2 + y^2} \]
\[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \]
This system efficiently tackles 3D problems like determining the position of a point on a cylinder, where the height and radial distance are easily defined.
Spherical Coordinates
The spherical coordinate system allows us to describe a point in 3D space with three values based on total distance and angles from a reference point. Usual notation for spherical coordinates is \( (\rho, \theta, \phi) \), where:
\[ \rho = \sqrt{x^2 + y^2 + z^2} \]
\[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \]
\[ \phi = \cos^{-1}\left(\frac{z}{\rho}\right) \]
This system is beneficial for applications involving radii and directions from a central point, like celestial coordinates in astronomy.
- \(\rho\) is the radial or spherical distance from the origin to the point.
- \(\theta\) is the azimuthal angle, taken from the positive x-axis within the xy-plane.
- \(\phi\) is the polar angle or inclination measured from the positive z-axis.
\[ \rho = \sqrt{x^2 + y^2 + z^2} \]
\[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \]
\[ \phi = \cos^{-1}\left(\frac{z}{\rho}\right) \]
This system is beneficial for applications involving radii and directions from a central point, like celestial coordinates in astronomy.
