Problem 42
Question
In Problems \(39-42\), convert the point given in rectangular coardinates to cylindrical coordinates. $$ (1,2,7) $$
Step-by-Step Solution
Verified Answer
The point in cylindrical coordinates is \((\sqrt{5}, \arctan(2), 7)\).
1Step 1: Identify Components
The point assigned in rectangular coordinates is given as \((x, y, z) = (1, 2, 7)\). In cylindrical coordinates, the point is expressed as \((r, \theta, z)\). The third coordinate, \(z\), remains the same as in rectangular coordinates.
2Step 2: Calculate r (Radial Distance)
The formula to calculate \(r\) (the radial distance from the z-axis) is \(r = \sqrt{x^2 + y^2}\). Use this formula with \(x = 1\) and \(y = 2\): \[r = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}\]. So, \(r = \sqrt{5}\).
3Step 3: Calculate θ (Angular Component)
Determine \(\theta\), the angle in radians from the positive x-axis to the projection of the point in the xy-plane, using the formula \(\theta = \arctan\left(\frac{y}{x}\right)\). Plug in \(x = 1\) and \(y = 2\): \[\theta = \arctan\left(\frac{2}{1}\right) = \arctan(2)\]. This is the value for \(\theta\).
4Step 4: Present the Cylindrical Coordinates
The cylindrical coordinates are presented as \((r, \theta, z)\). From the previous steps, \(r = \sqrt{5}\), \(\theta = \arctan(2)\), and \(z = 7\). Therefore, the point in cylindrical coordinates is \((\sqrt{5}, \arctan(2), 7)\).
Key Concepts
Rectangular CoordinatesCoordinate ConversionRadial DistanceAngular Component
Rectangular Coordinates
Rectangular coordinates, also known as Cartesian coordinates, are a way to specify the position of a point in space using three values:
- the x-coordinate, representing the horizontal distance along the x-axis,
- the y-coordinate, representing the vertical distance along the y-axis,
- the z-coordinate, which denotes the depth or height above the xy-plane.
Coordinate Conversion
Coordinate conversion is the process of switching from one coordinate system to another, in this case from rectangular to cylindrical coordinates. This conversion is crucial in fields that involve rotational or circular motion, such as fluid dynamics or electromagnetism.
Our goal was to convert the point (1, 2, 7) from rectangular to cylindrical coordinates.
- Firstly, we keep the z-coordinate the same.
- We then calculate the radial distance (r) and angular component (θ) to express the point in cylindrical coordinates.
Radial Distance
Radial distance, noted as \(r\), measures how far a point is from the z-axis. It's one of the key parameters in cylindrical coordinates. We use the formula \(r = \sqrt{x^2 + y^2}\) to find this distance using the x and y components of a given point.
For the point (1, 2, 7), we applied this formula as follows:
\[r = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}\]Hence, the radial distance \(r\) is \(\sqrt{5}\). Understanding \(r\) is fundamental in determining how a point appears when projected onto the z-plane. It plays a significant role in simplifying the analysis of systems with symmetrical structures, as it represents the exact circular radius from the central axis.
For the point (1, 2, 7), we applied this formula as follows:
\[r = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}\]Hence, the radial distance \(r\) is \(\sqrt{5}\). Understanding \(r\) is fundamental in determining how a point appears when projected onto the z-plane. It plays a significant role in simplifying the analysis of systems with symmetrical structures, as it represents the exact circular radius from the central axis.
Angular Component
The angular component \(\theta\) in cylindrical coordinates represents the angle measured in radians from the positive x-axis towards the projection of the vector onto the xy-plane. It is a critical component to fully describe the position of a point in a rotational context.To compute \(\theta\), we use the formula \(\theta = \arctan\left(\frac{y}{x}\right)\). For the point (1, 2, 7):
\[\theta = \arctan\left(\frac{2}{1}\right) = \arctan(2)\]This means \(\theta\) is the angle that describes how the point lies in relation to the x-axis in the xy-plane. Understanding this component is beneficial in various technical fields as it assists in analyzing and solving problems involving circular movements or patterns within a plane.
\[\theta = \arctan\left(\frac{2}{1}\right) = \arctan(2)\]This means \(\theta\) is the angle that describes how the point lies in relation to the x-axis in the xy-plane. Understanding this component is beneficial in various technical fields as it assists in analyzing and solving problems involving circular movements or patterns within a plane.
Other exercises in this chapter
Problem 41
Assume that \(f\) and \(g\) have continuous second partial derivatives. Show that the given vector field is solenoidal. $$ \mathbf{F}=\nabla f \times \nabla g $
View solution Problem 41
Find the length of the curve traced by the given vector function on the indicated interval. $$ \mathbf{r}(t)=a \cos t \mathbf{i}+a \sin t \mathbf{j}+c t \mathbf
View solution Problem 42
Find the center of mass of the lamina that has the given shape and density. $$ x=0, y=0,2 x+y=4 ; \rho(x, y)=x^{2} $$
View solution Problem 42
In Problems, find the length of the curve traced by the given vector function on the indicated interval. $$ \mathbf{r}(t)=t \mathbf{i}+t \cos t \mathbf{j}+t \si
View solution