Chapter 1: Problem 3
(a) The following points are given in cylindrical coordinates; express each in rectangular coordinates and spherical coordinates: \(\left(1,45^{\circ}, 1\right),(2, \pi / 2,-4)\) \(\left(0,45^{\circ}, 10\right),(3, \pi / 6,4),(1, \pi / 6,0),\) and \((2,3 \pi / 4,-2)\) (Only the first point is solved in the Study Guide.) (b) Change each of the following points from rectangular coordinates to spherical coordinates and to cylindrical coordinates: \((2,1,-2),(0,3,4),(\sqrt{2}, 1,1)\), \((-2 \sqrt{3},-2,3),\) (Only the first point is solved in the Study Guide.)
Short Answer
Step by step solution
Convert Cylindrical to Rectangular Coordinates
Convert to Spherical Coordinates
Repeat Step 1 for Remaining Points
Repeat Step 2 for Remaining Points
Convert Rectangular to Spherical Coordinates
Convert Rectangular to Cylindrical Coordinates
Repeat Steps 5 and 6 for Remaining Points
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cylindrical Coordinates
When converting from cylindrical to rectangular coordinates, remember these formulas:
- \(x = r \cos(\theta)\)
- \(y = r \sin(\theta)\)
- \(z = z\)
For example, if you want to convert the cylindrical point \( (1, 45^\circ, 1) \) into rectangular coordinates, calculate \(x = 1 \times \cos(45^\circ) = \frac{1}{\sqrt{2}}\), and \(y = 1 \times \sin(45^\circ) = \frac{1}{\sqrt{2}}\). The coordinate in rectangular form would be \( \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 1\right)\).
Rectangular Coordinates
When calculating the conversion to spherical coordinates, use these formulas:
- \( \rho = \sqrt{x^2 + y^2 + z^2} \)
- \( \phi = \arccos\left(\frac{z}{\rho}\right) \)
- \( \theta = \arctan\left(\frac{y}{x}\right) \)
- \( r = \sqrt{x^2 + y^2} \)
- \( \theta = \arctan\left(\frac{y}{x}\right) \)
- \( z = z \)
Spherical Coordinates
The conversion from cylindrical coordinates to spherical coordinates uses the following equations:
- \( \rho = \sqrt{r^2 + z^2} \)
- \( \phi = \arctan\left(\frac{r}{z}\right) \)
- \( \theta = \theta \)
This framework is immensely helpful in modeling fields like radio waves, gravitational analysis, and celestial mappings.