Understanding Direction of Unit Vectors r roof & phi roof

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SUMMARY

The discussion focuses on the unit vectors r roof (##\hat{r}##) and phi roof (##\hat{\phi}##) in polar coordinates, emphasizing their roles in directional derivatives. The differentiation of the position vector ##\vec{r} = r(\cos \phi\,\hat{i} + \sin \phi\,\hat{j})## yields ##\frac{\partial \vec{r}}{\partial \phi} = r (\cos \phi \hat{j} - \sin \phi \hat{i})##, demonstrating that phi roof is orthogonal to r roof through the dot product. Participants express confusion about visualizing the orthogonality and the implications of the dot product in this context.

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Istiak
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Homework Statement
Why angle direction is perpendicular?
Relevant Equations
vector
Screenshot from 2021-08-21 00-50-39.png

The unit vector r roof points in the direction of
increasing r with phi fixed; phi roof points in the direction of increasing phi
with r fixed. Unlike x roof, the vectors r roof and phi roof change as the position
vector r moves.
What I was thinking of the image is
Screenshot from 2021-08-21 01-15-23.png


Although, I was thinking why phi roof is perpendicular.
Screenshot from 2021-08-21 01-17-29.png
I was trying to understand that direction by the vector direction. I can't figure out that direction. Usually, I didn't do any vector of angle. I was wondering I didn't find any tutorial of angle direction in YT.
 
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What's the question exactly?
 
Start with ##\vec r = r(\cos \phi\,\hat i + \sin\phi\,\hat j)## and find the direction of ##\partial \vec r / \partial \phi##.
 
vela said:
Start with ##\vec r = r(\cos \phi\,\hat i + \sin\phi\,\hat j)## and find the direction of ##\partial \vec r / \partial \phi##.
If I differentiate that then, I get

$$\frac{\partial \vec r }{\partial \phi}=r (\cos \phi \hat j - \sin \phi \hat i)$$

But, I was thinking what I should do with the equation. 🤔
 
Istiakshovon said:
If I differentiate that then, I get

$$\frac{\partial \vec r }{\partial \phi}=r (\cos \phi \hat j - \sin \phi \hat i)$$

But, I was thinking what I should do with the equation. 🤔
What happens if you take the dot product of that with ##\hat r##?
 
Istiakshovon said:
If I differentiate that then, I get

$$\frac{\partial \vec r }{\partial \phi}=r (\cos \phi \hat j - \sin \phi \hat i)$$

But, I was thinking what I should do with the equation. 🤔
Show that vector is orthogonal to ##\vec r##
 
haruspex said:
What happens if you take the dot product of that with ##\hat r##?
$$\hat r \cdot \frac{\partial \vec r }{\partial \phi} = \hat r \frac{\partial \vec r }{\partial \phi} \cos \theta$$
$$=\hat r r (\cos \phi \hat j - \sin \phi \hat i) \cos \theta$$
 
  • Wow
Likes   Reactions: PeroK
PeroK said:
Show that vector is orthogonal to ##\vec r##
How? :thinking Did you mean to graph?
 
Istiakshovon said:
How? :thinking Did you mean to graph?
You could use the dot product - correctly, of course.
 
  • #10
Istiakshovon said:
$$\hat r \cdot \frac{\partial \vec r }{\partial \phi} = \hat r \frac{\partial \vec r }{\partial \phi} \cos \theta$$
$$=\hat r r (\cos \phi \hat j - \sin \phi \hat i) \cos \theta$$
That's an original approach to say the least!
 
  • #11
PeroK said:
You could use the dot product - correctly, of course.
To me $$\hat r r$$ represents the direction of phi is toward r. But, I think $$(\cos \phi \hat j - \sin \phi \hat i) \cos \theta$$ this are representing direction of phi is perpendicular. But, I am saying that by looking at original picture. Without that, I can't say that. So, how can I see that that's really perpendicular. 🤔
 
  • #12
Istiakshovon said:
To me $$\hat r r$$ represents the direction of phi is toward r. But, I think $$(\cos \phi \hat j - \sin \phi \hat i) \cos \theta$$ this are representing direction of phi is perpendicular. But, I am saying that by looking at original picture. Without that, I can't say that. So, how can I see that that's really perpendicular. 🤔
Note that it's either ##\phi## or ##\theta##, not both. Also, look up the dot product in Cartesian coordinates and note that:

vela said:
Start with ##\vec r = r(\cos \phi\,\hat i + \sin\phi\,\hat j)##
And
Istiakshovon said:
$$\frac{\partial \vec r }{\partial \phi}=r (\cos \phi \hat j - \sin \phi \hat i)$$
 

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