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Chapter 2: Problem 406

For the following exercises, the cylindrical coordinates of a point are given. Find its associated spherical coordinates, with the measure of the angle \(\varphi\) in radians rounded to four decimal places. $$\left(3,-\frac{\pi}{6}, 3\right)$$

Short Answer

Expert verified
The spherical coordinates are \((\sqrt{18}, 0.7854, -\frac{\pi}{6})\)."

Step by step solution

01

Understand the Cylindrical Coordinates

The cylindrical coordinates given are \((r, \theta, z) = \left( 3, -\frac{\pi}{6}, 3 \right)\). Here, \(r = 3\) is the radial distance from the z-axis, \(\theta = -\frac{\pi}{6}\) is the angular coordinate measure from the positive x-axis, and \(z = 3\) is the height along the z-axis.
02

Convert to Cartesian Coordinates

The conversion from cylindrical to Cartesian coordinates is given by the formulas: \(x = r\cos\theta\), \(y = r\sin\theta\), \(z = z\). Compute these as: - \(x = 3\cos\left(-\frac{\pi}{6}\right) = 3\left(\frac{\sqrt{3}}{2}\right) = \frac{3\sqrt{3}}{2}\)- \(y = 3\sin\left(-\frac{\pi}{6}\right) = 3\left(-\frac{1}{2}\right) = -\frac{3}{2}\)- \(z = 3\)Thus, the Cartesian coordinates are \(\left(\frac{3\sqrt{3}}{2}, -\frac{3}{2}, 3\right)\).
03

Convert to Spherical Coordinates

In spherical coordinates, \((\rho, \phi, \theta)\), we use the following conversions:- \(\rho = \sqrt{x^2 + y^2 + z^2}\)- \(\phi = \arccos\left(\frac{z}{\rho}\right)\)- Use the original \(\theta\) from cylindrical coordinatesCompute:- \(\rho = \sqrt{\left(\frac{3\sqrt{3}}{2}\right)^2 + \left(-\frac{3}{2}\right)^2 + 3^2} = \sqrt{\frac{27}{4} + \frac{9}{4} + 9} = \sqrt{18}\)- \(\phi = \arccos\left(\frac{3}{\sqrt{18}}\right) = \arccos\left(\frac{3}{3\sqrt{2}}\right) = \arccos\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}\)Thus, the spherical coordinates are \((\sqrt{18}, \frac{\pi}{4}, -\frac{\pi}{6})\).
04

Round \\( heta\\) to Four Decimal Places

We observe that \( \phi = \frac{\pi}{4} \approx 0.7854 \) radians when rounded to four decimal places.

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Key Concepts

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

Cylindrical Coordinates
Cylindrical coordinates provide an alternative way of locating a point in three-dimensional space. Picture differentiating points based on a blend of circular motion and elevation, which is exactly what this system does. You have three components to keep track of:
  • Radial distance (\(r\)): This is the straight-line distance from the z-axis to the point. It mirrors the radius of a circle.
  • Angular position (\(\theta\)): This angle, typically measured in radians, is between a reference direction (usually the positive x-axis) and the projection of the point onto the xy-plane.
  • Vertical height (\(z\)): Simply the distance of the point above or below the xy-plane, running parallel to the z-axis.
Given the cylindrical coordinates \((3, -\frac{\pi}{6}, 3)\), we can easily visualize this point having a distance of 3 units from the z-axis, an angle of negative \(\frac{\pi}{6}\) from the positive x-axis, and a height of 3 units above the xy-plane.
Spherical Coordinates
Spherical coordinates offer another method to describe a point in 3D space, where location is determined by encoding radius, polar angle, and azimuthal angle. Imagine earth's latitude and longitude combined with altitude鈥攖hey're quite similar.
  • Radial distance (\(\rho\)): The straight line from the origin to the point. It鈥檚 akin to our original radial distance in cylindrical coordinates but spatial instead of planar.
  • Polar angle (\(\phi\)): This is the angle measured down from the positive z-axis toward the point. It emphasizes elevation.
  • Azimuthal angle (\(\theta\)): Identical to its counterpart in cylindrical coordinates. It specifies direction in the xy-plane.
Translating a point from cylindrical to spherical coordinates involves recalculating its position based on these parameters. The spherical coordinates for our point could be determined as \((\sqrt{18}, \frac{\pi}{4}, -\frac{\pi}{6})\).
Angle in Radians
Radians provide a more natural measure for angles in mathematics, especially trigonometry and calculus. They're derived directly from the properties of circles鈥攁 full circle encompasses an angle of \(2\pi\) radians.
  • When you measure an angle in radians, you're essentially calculating the length of the arc mapped out by the angle across a unit circle.
  • An angle measured in degrees is converted to radians by multiplying it by \(\frac{\pi}{180}\). Conversely, to go from radians to degrees, multiply by \(\frac{180}{\pi}\).
Thus, the angular component \(-\frac{\pi}{6}\) in our exercise represents an angle that spans \(-30^\circ\). This helps us understand point placement in a real-world context and is key for conversion, as angles should remain consistent between coordinate systems.
Coordinate Conversion
Coordinate conversion is a crucial process in mathematics involving the change from one coordinate system to another, allowing for different perspectives and applications.
  • The transition from cylindrical to Cartesian coordinates involves straightforward trigonometric relations: \(x = r \cos\theta\), \(y = r \sin\theta\), and \(z = z\).
  • To switch from cylindrical to spherical, a recalibration using all three spatial dimensions is needed: employing \(\rho = \sqrt{x^2 + y^2 + z^2}\) and \(\phi = \arccos\left(\frac{z}{\rho}\right)\).
These transformations are not just mathematical exercises鈥攖hey're essential in fields like physics, engineering, and computer graphics, allowing seamless transitions between coordinate systems to solve practical problems.
Cartesian Coordinates
Cartesian coordinates are probably the most familiar coordinate system, often used in basic mathematics and physics. They describe a point in space using orthogonal axes.
  • x-coordinate: How far to the right (or left) a point lies relative to the origin along the horizontal axis.
  • y-coordinate: The position up or down along the vertical (in 2D) or depth (in 3D) axis.
  • z-coordinate: The placement forth and back along the axis perpendicular to both the x and y axes in 3D.
When moving from cylindrical to Cartesian, one uses the relationships involving trigonometric calculation with the radial distance and angle. By converting our cylindrical point \((3, -\frac{\pi}{6}, 3)\) to Cartesian, we derive \((\frac{3\sqrt{3}}{2}, -\frac{3}{2}, 3)\). This new form is instrumental for situations demanding analysis using linear algebra or calculus. The Cartesian representation keeps things straightforward and universally applicable across numerous scientific applications.

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Most popular questions from this chapter

The following problems consider your unsuccessful attempt to take the tire off your car using a wrench to loosen the bolts. Assume the wrench is 0.3 \(\mathrm{m}\) long and you are able to apply a 200 -N force. Someone lends you a tire jack and you are now able to apply a 200 -N force at an \(80^{\circ}\) angle. Is your resulting torque going to be more or less? What is the new resulting torque at the center of the bolt? Assume this force is not enough to loosen the bolt.

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Consider points \(P(2,1), Q(4,2),\) and \(R(1,2).\) a. Find the area of triangle \(P, Q,\) and \(R.\) b. Determine the distance from point \(R\) to the line passing through \(P\) and \(Q .\)
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