Problem 38
Question
Convert the point given in cylindrical cocudinates to rectangular cocrdinates. $$ \left(4, \frac{7 \pi}{4}, 0\right) $$
Step-by-Step Solution
Verified Answer
The point in rectangular coordinates is \((2\sqrt{2}, -2\sqrt{2}, 0)\).
1Step 1: Understanding the Provided Point
The given point in cylindrical coordinates is \((r, \theta, z) = (4, \frac{7\pi}{4}, 0)\). Here, \(r\) is the radial distance from the origin, \(\theta\) is the angle in the xy-plane from the positive x-axis, and \(z\) is the height along the z-axis.
2Step 2: Applying Conversion Formulas
To convert cylindrical coordinates \((r, \theta, z)\) to rectangular coordinates \((x, y, z)\), we use the formulas: \[x = r \cos \theta,\]\[y = r \sin \theta,\]\[z = z.\]
3Step 3: Substituting Values for x
Substitute \(r = 4\) and \(\theta = \frac{7\pi}{4}\) into the formula for \(x\):\[x = 4 \cos \left(\frac{7\pi}{4}\right).\]
4Step 4: Calculating x
Calculate the cosine of \(\theta = \frac{7\pi}{4}\):\[\cos \left(\frac{7\pi}{4}\right) = \cos \left(\frac{8\pi}{4} - \frac{\pi}{4}\right) = \cos \left(2\pi - \frac{\pi}{4}\right) = \cos \left(-\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}.\]Therefore, \[x = 4 \times \frac{1}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}.\]
5Step 5: Substituting Values for y
Substitute \(r = 4\) and \(\theta = \frac{7\pi}{4}\) into the formula for \(y\):\[y = 4 \sin \left(\frac{7\pi}{4}\right).\]
6Step 6: Calculating y
Calculate the sine of \(\theta = \frac{7\pi}{4}\):\[\sin \left(\frac{7\pi}{4}\right) = \sin \left(\frac{8\pi}{4} - \frac{\pi}{4}\right) = \sin \left(2\pi - \frac{\pi}{4}\right) = \sin \left(-\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}}.\]Therefore, \[y = 4 \times -\frac{1}{\sqrt{2}} = -\frac{4}{\sqrt{2}} = -2\sqrt{2}.\]
7Step 7: Using the z-coordinate
The z-coordinate in cylindrical and rectangular coordinates is the same. Therefore, \[z = 0.\]
8Step 8: Final Rectangular Coordinates
Combining the results, the rectangular coordinates are: \((x, y, z) = (2\sqrt{2}, -2\sqrt{2}, 0)\).
Key Concepts
Cylindrical CoordinatesRectangular CoordinatesTrigonometric FunctionsCoordinate Systems
Cylindrical Coordinates
Cylindrical coordinates are a system used to define the position of a point in a three-dimensional space using three values. This system is somewhat like polar coordinates but with an added height measurement. It consists of:
- Radial distance \(r\): The distance from the origin (center of the coordinate system) to the point in the xy-plane.
- Angular coordinate \(\theta\): The angle measured from the positive x-axis to the line connecting the origin with the projection of the point in the xy-plane.
- Height \(z\): The vertical distance from the xy-plane to the point.
Rectangular Coordinates
Rectangular, or Cartesian coordinates, describe the position of points in space using orthogonal axes. These coordinates rely on three coordinates:
- x-coordinate: This is the horizontal distance from the yz-plane to the point.
- y-coordinate: The horizontal distance from the xz-plane to the point.
- z-coordinate: The vertical distance from the xy-plane to the point.
Trigonometric Functions
Trigonometric functions are essential tools for connecting angles with side ratios in right triangles and are used in coordinate conversions. The primary trigonometric functions that relate to coordinate conversions are:
- Cosine \(\cos\theta\): Represents the x-component of the unit circle corresponding to an angle \(\theta\).
- Sine \(\sin\theta\): Represents the y-component of the unit circle corresponding to an angle \(\theta\).
- \(x = r \cos\theta\)
- \(y = r \sin\theta\)
Coordinate Systems
Coordinate systems like cylindrical and rectangular ones provide frameworks for referencing points in space. Whether to describe simple lines or complex, three-dimensional objects:
- Cylindrical systems often suit situations where symmetry around a central axis is pivotal, such as circles or spiral patterns.
- Rectangular systems represent linear dimensions and are prevalent in everyday applications like maps and blueprints.
Other exercises in this chapter
Problem 38
The pressure \(P\) exerted by an enclosed ideal gas is given by \(P=k(T / V)\), where \(k\) is a constant, \(T\) is temperature, and \(V\) is volume. Find: (a)
View solution Problem 38
Suppose $$ \begin{gathered} D_{\mathrm{u}} f(a, b)=7, \quad D_{\mathrm{v}} f(a, b)=3 \\ \mathbf{u}=\frac{5}{13} \mathbf{i}-\frac{12}{13} \mathbf{j}, \quad \math
View solution Problem 38
Consider the three curves between \((0,0)\) and \((2,4)\) : $$ \begin{array}{lll} C_{1}: x=t, & y=2 t, & 0 \leq t \leq 2 \\ C_{2}: x=t, & y=t^{2}, & 0 \leq t \l
View solution Problem 38
Suppose $$ \begin{gathered} D_{\mathbf{u}} f(a, b)=7, \quad D_{\mathbf{v}} f(a, b)=3 \\ \mathbf{u}=\frac{5}{13} \mathbf{i}-\frac{12}{13} \mathbf{j}, \quad \math
View solution