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For the following exercises, the spherical coordinates \((\rho, \theta, \varphi)\) of a point are given. Find the rectangular coordinates \((x, y, z)\) of the point.\(\left(3, \frac{\pi}{4}, \frac{\pi}{6}\right)\)

Short Answer

Expert verified
The rectangular coordinates are \(\left(\frac{3\sqrt{2}}{4}, \frac{3\sqrt{2}}{4}, \frac{3\sqrt{3}}{2}\right)\).

Step by step solution

01

Understand Spherical Coordinates

Spherical coordinates are given as \((\rho, \theta, \varphi)\), where \(\rho\) is the radial distance from the origin \(\theta\) is the angle in the xy-plane from the positive x-axis, and \(\varphi\) is the angle from the positive z-axis. Our coordinates are \((3, \frac{\pi}{4}, \frac{\pi}{6})\).
02

Convert to Cartesian Coordinates Formulas

To convert from spherical coordinates to rectangular coordinates, we use these formulas:\[x = \rho \sin\varphi \cos\theta\]\[y = \rho \sin\varphi \sin\theta\]\[z = \rho \cos\varphi\]
03

Calculate x-coordinate

Insert the given values into the formula for x:\[x = 3 \sin\left(\frac{\pi}{6}\right) \cos\left(\frac{\pi}{4}\right)\]Since \(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\) and \(\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\), we have:\[x = 3 \times \frac{1}{2} \times \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{4}\]
04

Calculate y-coordinate

Insert the given values into the formula for y:\[y = 3 \sin\left(\frac{\pi}{6}\right) \sin\left(\frac{\pi}{4}\right)\]Since \(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\) and \(\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\), we have:\[y = 3 \times \frac{1}{2} \times \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{4}\]
05

Calculate z-coordinate

Insert the given values into the formula for z:\[z = 3 \cos\left(\frac{\pi}{6}\right)\]Since \(\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\), we have:\[z = 3 \times \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}\]
06

Conclude with Rectangular Coordinates

The rectangular coordinates \((x, y, z)\) are \(\left(\frac{3\sqrt{2}}{4}, \frac{3\sqrt{2}}{4}, \frac{3\sqrt{3}}{2}\right)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Spherical Coordinates
Spherical coordinates are an alternative way to specify the position of a point in three-dimensional space. Unlike rectangular coordinates, which use three perpendicular axes to locate a point, spherical coordinates use a combination of radial distance and angular measurements. Here’s how spherical coordinates work:

  • \( \rho \) (rho): This is the radial distance from the origin to the point. It is always a positive number or zero.
  • \( \theta \) (theta): This is the angle measured in the \(xy\)-plane starting from the positive x-axis. It is akin to the polar angle in two-dimensional polar coordinates.
  • \( \varphi \) (phi): This angle is measured from the positive z-axis toward the point. It ranges from 0 to \( \pi \).
The spherical coordinates are particularly useful for problems involving symmetry around a point, such as situations in physics or engineering.Understanding spherical coordinates allows you to easily transition to rectangular or cylindrical coordinates when needed.
Rectangular Coordinates
Rectangular coordinates, also known as Cartesian coordinates, are one of the most common coordinate systems used to describe the position of points in a plane or space. The coordinates specify the position of a point by using three values that denote distances along the x, y, and z axes.

  • \( x \): Represents the horizontal position and is the distance along the x-axis.
  • \( y \): Represents the vertical position and is the distance along the y-axis.
  • \( z \): Represents the height and is the distance along the z-axis.
In the conversion from spherical to rectangular coordinates, we utilize the equations:
  • \( x = \rho \sin \varphi \cos \theta \)
  • \( y = \rho \sin \varphi \sin \theta \)
  • \( z = \rho \cos \varphi \)
These transformations make it possible to describe the same point in a different coordinate system, offering flexibility in mathematical calculations and real-world applications.
Trigonometric Functions
Trigonometric functions are cornerstone mathematical functions that relate the angles of a triangle to the lengths of its sides. These functions are crucial in coordinate conversions, particularly when moving between spherical and rectangular coordinates.

Each of the primary trigonometric functions—sine, cosine, and tangent—corresponds to a ratio of sides in a right triangle.
  • \( \sin(\theta) \): This function equals the opposite side over the hypotenuse in a right triangle. It’s used in our formulas to convert spherical to rectangular coordinates.
  • \( \cos(\theta) \): This is the ratio of the adjacent side over the hypotenuse. It's crucial for determining the x-component of a point in 3D space.
  • \( \tan(\theta) \): Though not used directly in spherical conversion, tangent represents opposite over adjacent and is handy for angle calculations.
For our specific conversion exercise, we utilize these functions to find precise rectangular coordinates by inserting the given angle values. This strategic use simplifies the calculations, providing a bridge connecting spherical coordinates to their rectangular equivalents.

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Most popular questions from this chapter

A telephone pole guy-wire has an angle of elevation of \(35^{\circ}\) with respect to the ground. The force of tension in the guy-wire is \(120 \mathrm{lb}\). Find the horizontal and vertical components of the force of tension. (Round to the nearest integer.)

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Sketch and describe the cylindrical surface of the given equation. $$ x^{2}+z^{2}=1 $$

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For the following exercises, the vectors \(\mathbf{u}\) and \(\mathbf{v}\) are given. Find the vector projection \(\mathrm{w}=\operatorname{proj}_{\mathbf{u}} \mathbf{v}\) of vector \(\mathbf{v}\) onto vector \(\mathbf{u}\). Express your answer in component form. Find the scalar projection \(\operatorname{comp}_{\mathrm{u}} \mathrm{v}\) of vector \(\mathrm{v}\) onto vector \(\mathrm{u}\). $$ \mathbf{u}=\langle 4,4,0\rangle, \mathbf{v}=\langle 0,4,1\rangle $$

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