Cylindrical coordinates - Orthonormal system

In summary: C1CHBF_enUS747US747&source=lnms&tbm=isch&sa=X&ved=0ahUKEwibyZqMk4DoAhXXnOAKHdF8B3oQ_AUIEigB&biw=1366&bih=657#imgrc=_ In summary, we discussed using cylindrical coordinates and the orthonormal system of vectors $\overrightarrow{e}_r, \overrightarrow{e}_{\theta}, \overrightarrow{e}_z$ to describe each vector as a function of $\overrightarrow{i}, \overrightarrow{j}, \
  • #1
mathmari
Gold Member
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Hey! :eek:

Using cylindrical coordinates and the orthonormal system of vectors $\overrightarrow{e}_r, \overrightarrow{e}_{\theta}, \overrightarrow{e}_z$
  1. describe each of the $\overrightarrow{e}_r$, $\overrightarrow{e}_{\theta}$ and $\overrightarrow{e}_z$ as a function of $\overrightarrow{i}, \overrightarrow{j}, \overrightarrow{k}$ and $(x, y, z)$ and
  2. calculate $\overrightarrow{e}_{\theta} \times \overrightarrow{j}$ with two ways: analytically, using (1), and geometrically.

Could you give me some hints how I could do that?? (Wondering)
 
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  • #2
mathmari said:
Hey! :eek:

Using cylindrical coordinates and the orthonormal system of vectors $\overrightarrow{e}_r, \overrightarrow{e}_{\theta}, \overrightarrow{e}_z$
  1. describe each of the $\overrightarrow{e}_r$, $\overrightarrow{e}_{\theta}$ and $\overrightarrow{e}_z$ as a function of $\overrightarrow{i}, \overrightarrow{j}, \overrightarrow{k}$ and $(x, y, z)$ and
  2. calculate $\overrightarrow{e}_{\theta} \times \overrightarrow{j}$ with two ways: analytically, using (1), and geometrically.

Could you give me some hints how I could do that?? (Wondering)

Hi! ;)

We have the point $\overrightarrow r = x\overrightarrow{i} + y\overrightarrow{j} + z\overrightarrow{k}$.
Its spherical coordinates are $(r,\theta,\phi)$.

At the point we want to have a local orthonormal basis aligned with spherical coordinates.
The vector $\overrightarrow{e}_r$ is the unit vector with a direction that corresponds to how $\overrightarrow r$ changes if we increase $r$.
That is in the direction of $\overrightarrow r$.
Normalizing to unit length, we get:
$$\overrightarrow{e}_r = \frac x r\overrightarrow{i} + \frac y r\overrightarrow{j} + \frac z r\overrightarrow{k}$$
Alternatively we can say that:
$$\overrightarrow{e}_r = \frac{\d {\overrightarrow r}{r}}{\left|\d {\overrightarrow r}{r}\right|}$$
(Nerd)

Next is $\overrightarrow{e}_{\theta}$, which is the unit vector with a direction that corresponds to how $\overrightarrow r$ changes if we increase $\theta$.
What would its direction be? (Wondering)
 
  • #3
I like Serena said:
Next is $\overrightarrow{e}_{\theta}$, which is the unit vector with a direction that corresponds to how $\overrightarrow r$ changes if we increase $\theta$.
What would its direction be? (Wondering)

Is it $$\overrightarrow{e}_{\theta} = \frac{\d {\overrightarrow r}{\theta}}{\left|\d {\overrightarrow r}{\theta}\right|}$$ ?? (Wondering)

Do we have the following:

$$\overrightarrow{e}_{r}=(\cos \theta )\overrightarrow{i}+(\sin \theta )\overrightarrow{j} \\ \overrightarrow{e}_{\theta}=(-r \sin \theta )\overrightarrow{i}+(r \cos \theta )\overrightarrow{j} \\ \overrightarrow{e}_{z}=\overrightarrow{k}$$ ?? (Wondering)

So using $$x=r \cos \theta , y=r \sin \theta , z=z$$ we have the following:

$$\overrightarrow{e}_{r}=\frac{x\overrightarrow{i}+y \overrightarrow{j}}{\sqrt{x^2+y^2}} \\ \overrightarrow{e}_{\theta}=-y\overrightarrow{i}+x\overrightarrow{j} \\ \overrightarrow{e}_{z}=\overrightarrow{k}$$

Is this correct??
For the question 2. using 1. we have that $$\overrightarrow{e}_{\theta} \times \overrightarrow{j}=-y \overrightarrow{k}$$ right?? (Wondering)

But how could we show it geometrically?? (Wondering)
 
  • #4
mathmari said:
Is it $$\overrightarrow{e}_{\theta} = \frac{\d {\overrightarrow r}{\theta}}{\left|\d {\overrightarrow r}{\theta}\right|}$$ ?? (Wondering)

Do we have the following:

$$\overrightarrow{e}_{r}=(\cos \theta )\overrightarrow{i}+(\sin \theta )\overrightarrow{j} \\ \overrightarrow{e}_{\theta}=(-r \sin \theta )\overrightarrow{i}+(r \cos \theta )\overrightarrow{j} \\ \overrightarrow{e}_{z}=\overrightarrow{k}$$ ?? (Wondering)

So using $$x=r \cos \theta , y=r \sin \theta , z=z$$ we have the following:

$$\overrightarrow{e}_{r}=\frac{x\overrightarrow{i}+y \overrightarrow{j}}{\sqrt{x^2+y^2}} \\ \overrightarrow{e}_{\theta}=-y\overrightarrow{i}+x\overrightarrow{j} \\ \overrightarrow{e}_{z}=\overrightarrow{k}$$

For the question 2. using 1. we have that $$\overrightarrow{e}_{\theta} \times \overrightarrow{j}=-y \overrightarrow{k}$$ right?? (Wondering)

Is this correct??

Almost.
We still have to normalize $\overrightarrow{e}_{\theta}$ to unit length. (Wasntme)
But how could we show it geometrically?? (Wondering)

Did you make a drawing? (Wondering)
 
  • #5
I like Serena said:
Almost.
We still have to normalize $\overrightarrow{e}_{\theta}$ to unit length. (Wasntme)

Is it $$\overrightarrow{e}_{\theta}=-sin \theta \overrightarrow{i}+\cos \theta \overrightarrow{j}$$ ?? (Wondering)
I like Serena said:
Did you make a drawing? (Wondering)

No... How could I do that?? (Wondering)
 
  • #6
mathmari said:
Is it $$\overrightarrow{e}_{\theta}=-sin \theta \overrightarrow{i}+\cos \theta \overrightarrow{j}$$ ?? (Wondering)

Yep! (Nod)

No... How could I do that?? (Wondering)

Something like this:
CylindricalCoordinates_1001.gif
 

Related to Cylindrical coordinates - Orthonormal system

What are cylindrical coordinates?

Cylindrical coordinates are a type of coordinate system used in three-dimensional space to describe the location of a point. They consist of a radial distance, an angle measured from a reference direction, and a height or depth from a reference plane.

How are cylindrical coordinates related to Cartesian coordinates?

Cylindrical coordinates can be converted to Cartesian coordinates by using the following formulas:
x = r * cos(θ)
y = r * sin(θ)
z = h

What is the purpose of an orthonormal system?

An orthonormal system is a set of vectors that are orthogonal (perpendicular) to each other and have a length of 1. It is used to simplify calculations and make it easier to work with different coordinate systems.

What are the advantages of using cylindrical coordinates?

One advantage of using cylindrical coordinates is that they are well-suited for describing objects with cylindrical symmetry, such as cylinders, cones, and some types of curves. They also make it easier to calculate certain physical quantities, such as volume and surface area.

How do you plot points in cylindrical coordinates?

To plot a point in cylindrical coordinates, you first locate the point's distance from the origin (r), then its angle from a reference direction (θ), and finally its height or depth from a reference plane (h). Using these values, you can then plot the point on a three-dimensional graph.

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