Normal vector to a polar curve
I'm struggling with working through a proof. Suppose I have a polar curve of the form $r = f(phi)$.
How do I find the $textbf{normal vector} $ to this curve? The end result I need should be in terms of $hat{r}$ and $hat{phi}$ but I'm unsure of what I should do.
All I know so far is how to find the tangent slope and normal slope at any given $phi$. Any help would be very appreciated. Thanks!
differential-geometry polar-coordinates
edited Nov 29 at 18:27
Robert Lewis
43.2k22863
asked Nov 29 at 17:51
R Thompson
526
add a comment |
I'm struggling with working through a proof. Suppose I have a polar curve of the form $r = f(phi)$.
How do I find the $textbf{normal vector} $ to this curve? The end result I need should be in terms of $hat{r}$ and $hat{phi}$ but I'm unsure of what I should do.
All I know so far is how to find the tangent slope and normal slope at any given $phi$. Any help would be very appreciated. Thanks!
differential-geometry polar-coordinates
edited Nov 29 at 18:27
Robert Lewis
43.2k22863
asked Nov 29 at 17:51
R Thompson
526
If you know the slope of the normal, then you can make a normal vector out of it.
– Berci
Nov 29 at 18:06
I added the "differential-geometry" tag to your post. Cheers!
– Robert Lewis
Nov 29 at 18:27
add a comment |
I'm struggling with working through a proof. Suppose I have a polar curve of the form $r = f(phi)$.
How do I find the $textbf{normal vector} $ to this curve? The end result I need should be in terms of $hat{r}$ and $hat{phi}$ but I'm unsure of what I should do.
All I know so far is how to find the tangent slope and normal slope at any given $phi$. Any help would be very appreciated. Thanks!
differential-geometry polar-coordinates
edited Nov 29 at 18:27
Robert Lewis
43.2k22863
asked Nov 29 at 17:51
R Thompson
526
I'm struggling with working through a proof. Suppose I have a polar curve of the form $r = f(phi)$.
How do I find the $textbf{normal vector} $ to this curve? The end result I need should be in terms of $hat{r}$ and $hat{phi}$ but I'm unsure of what I should do.
All I know so far is how to find the tangent slope and normal slope at any given $phi$. Any help would be very appreciated. Thanks!
differential-geometry polar-coordinates
differential-geometry polar-coordinates
edited Nov 29 at 18:27
Robert Lewis
43.2k22863
asked Nov 29 at 17:51
R Thompson
526
edited Nov 29 at 18:27
Robert Lewis
43.2k22863
asked Nov 29 at 17:51
R Thompson
526
edited Nov 29 at 18:27
Robert Lewis
43.2k22863
edited Nov 29 at 18:27
Robert Lewis
43.2k22863
edited Nov 29 at 18:27
Robert Lewis
43.2k22863
43.2k22863
asked Nov 29 at 17:51
R Thompson
526
asked Nov 29 at 17:51
R Thompson
526
asked Nov 29 at 17:51
R Thompson
526
526
If you know the slope of the normal, then you can make a normal vector out of it.
– Berci
Nov 29 at 18:06
I added the "differential-geometry" tag to your post. Cheers!
– Robert Lewis
Nov 29 at 18:27
add a comment |
If you know the slope of the normal, then you can make a normal vector out of it.
– Berci
Nov 29 at 18:06
I added the "differential-geometry" tag to your post. Cheers!
– Robert Lewis
Nov 29 at 18:27
If you know the slope of the normal, then you can make a normal vector out of it.
– Berci
Nov 29 at 18:06
If you know the slope of the normal, then you can make a normal vector out of it.
– Berci
Nov 29 at 18:06
I added the "differential-geometry" tag to your post. Cheers!
– Robert Lewis
Nov 29 at 18:27
I added the "differential-geometry" tag to your post. Cheers!
– Robert Lewis
Nov 29 at 18:27
add a comment |
1 Answer
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You have the vector of position:
$$vec{r}=hat{e}_r r=hat{e}_r f(phi)$$
differentiate over $phi$ to get the tangent:
$$vec{t}=frac{d}{dphi}vec{r}=frac{dhat{e}_r}{dphi}f(phi)+hat{e}_rf'(phi)$$
But in polar coordinates, we know how unit vector in the radial direction changes with angle:
$$frac{dhat{e}_r}{dphi}=hat{e}_phi$$
(if unsure, write it as $hat{e}_r=(cosphi,sinphi)$ and take the derivative.
So you have
$$vec{t}=f(phi)hat{e}_phi+f'(phi)hat{e}_r$$
You can normalize it you want.
A normal is just a $90^circ$ rotation of this vector, and because $hat{e}_r$ and $hat{e}_phi$ are orthogonal and unit sized, you can do the same as in the cartesian coordinates: $90^circ$ positive rotation is just exchanging components and negating the one that gets the "y" (in this case, $phi$) component.
$$vec{n}=-f(phi)hat{e}_r+f'(phi)hat{e}_phi$$
Orthogonality is easily verified with dot product.
answered Nov 29 at 18:04
orion
12.9k11836
add a comment |
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1 Answer
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You have the vector of position:
$$vec{r}=hat{e}_r r=hat{e}_r f(phi)$$
differentiate over $phi$ to get the tangent:
$$vec{t}=frac{d}{dphi}vec{r}=frac{dhat{e}_r}{dphi}f(phi)+hat{e}_rf'(phi)$$
But in polar coordinates, we know how unit vector in the radial direction changes with angle:
$$frac{dhat{e}_r}{dphi}=hat{e}_phi$$
(if unsure, write it as $hat{e}_r=(cosphi,sinphi)$ and take the derivative.
So you have
$$vec{t}=f(phi)hat{e}_phi+f'(phi)hat{e}_r$$
You can normalize it you want.
A normal is just a $90^circ$ rotation of this vector, and because $hat{e}_r$ and $hat{e}_phi$ are orthogonal and unit sized, you can do the same as in the cartesian coordinates: $90^circ$ positive rotation is just exchanging components and negating the one that gets the "y" (in this case, $phi$) component.
$$vec{n}=-f(phi)hat{e}_r+f'(phi)hat{e}_phi$$
Orthogonality is easily verified with dot product.
answered Nov 29 at 18:04
orion
12.9k11836
add a comment |
You have the vector of position:
$$vec{r}=hat{e}_r r=hat{e}_r f(phi)$$
differentiate over $phi$ to get the tangent:
$$vec{t}=frac{d}{dphi}vec{r}=frac{dhat{e}_r}{dphi}f(phi)+hat{e}_rf'(phi)$$
But in polar coordinates, we know how unit vector in the radial direction changes with angle:
$$frac{dhat{e}_r}{dphi}=hat{e}_phi$$
(if unsure, write it as $hat{e}_r=(cosphi,sinphi)$ and take the derivative.
So you have
$$vec{t}=f(phi)hat{e}_phi+f'(phi)hat{e}_r$$
You can normalize it you want.
A normal is just a $90^circ$ rotation of this vector, and because $hat{e}_r$ and $hat{e}_phi$ are orthogonal and unit sized, you can do the same as in the cartesian coordinates: $90^circ$ positive rotation is just exchanging components and negating the one that gets the "y" (in this case, $phi$) component.
$$vec{n}=-f(phi)hat{e}_r+f'(phi)hat{e}_phi$$
Orthogonality is easily verified with dot product.
answered Nov 29 at 18:04
orion
12.9k11836
add a comment |
You have the vector of position:
$$vec{r}=hat{e}_r r=hat{e}_r f(phi)$$
differentiate over $phi$ to get the tangent:
$$vec{t}=frac{d}{dphi}vec{r}=frac{dhat{e}_r}{dphi}f(phi)+hat{e}_rf'(phi)$$
But in polar coordinates, we know how unit vector in the radial direction changes with angle:
$$frac{dhat{e}_r}{dphi}=hat{e}_phi$$
(if unsure, write it as $hat{e}_r=(cosphi,sinphi)$ and take the derivative.
So you have
$$vec{t}=f(phi)hat{e}_phi+f'(phi)hat{e}_r$$
You can normalize it you want.
A normal is just a $90^circ$ rotation of this vector, and because $hat{e}_r$ and $hat{e}_phi$ are orthogonal and unit sized, you can do the same as in the cartesian coordinates: $90^circ$ positive rotation is just exchanging components and negating the one that gets the "y" (in this case, $phi$) component.
$$vec{n}=-f(phi)hat{e}_r+f'(phi)hat{e}_phi$$
Orthogonality is easily verified with dot product.
answered Nov 29 at 18:04
orion
12.9k11836
You have the vector of position:
$$vec{r}=hat{e}_r r=hat{e}_r f(phi)$$
differentiate over $phi$ to get the tangent:
$$vec{t}=frac{d}{dphi}vec{r}=frac{dhat{e}_r}{dphi}f(phi)+hat{e}_rf'(phi)$$
But in polar coordinates, we know how unit vector in the radial direction changes with angle:
$$frac{dhat{e}_r}{dphi}=hat{e}_phi$$
(if unsure, write it as $hat{e}_r=(cosphi,sinphi)$ and take the derivative.
So you have
$$vec{t}=f(phi)hat{e}_phi+f'(phi)hat{e}_r$$
You can normalize it you want.
A normal is just a $90^circ$ rotation of this vector, and because $hat{e}_r$ and $hat{e}_phi$ are orthogonal and unit sized, you can do the same as in the cartesian coordinates: $90^circ$ positive rotation is just exchanging components and negating the one that gets the "y" (in this case, $phi$) component.
$$vec{n}=-f(phi)hat{e}_r+f'(phi)hat{e}_phi$$
Orthogonality is easily verified with dot product.
answered Nov 29 at 18:04
orion
12.9k11836
answered Nov 29 at 18:04
orion
12.9k11836
answered Nov 29 at 18:04
orion
12.9k11836
answered Nov 29 at 18:04
orion
12.9k11836
12.9k11836
add a comment |
add a comment |
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If you know the slope of the normal, then you can make a normal vector out of it.
– Berci
Nov 29 at 18:06
I added the "differential-geometry" tag to your post. Cheers!
– Robert Lewis
Nov 29 at 18:27