curvature of circle

[Agent Smith]

I don't know what a "regular curve" is, do you mean smooth?
You're asserting that \(\displaystyle \bold{r}(t) = 0\) what is your reasoning for this assertion?


He's being sarcastic with the emoji.


I'd ignore @Agent Smith . He belongs to the internet troll category.

So rude!

 

[Agent Smith]

the circle is a regular curve and a regurlar curve \(\displaystyle \bold{r}(t)\) has \(\displaystyle \bold{r}'(t) \neq 0\), so \(\displaystyle \frac{0}{\bold{r}'(t)} = 0\)


post #10 that Mario99 agree it's correct


i don't understand what you're doing but i think \(\displaystyle B\) is the circle we're discussing now

I was just offering my views on what a curvature is and how we could measure it in a simple fashion, which in this case could mean no calculus. If you look at my diagrams, it's true that ... given an arc \(\displaystyle ABC\), the absolute value of the average slopes/gradients (call this \(\displaystyle M\)) of the straight lines \(\displaystyle AB\) and \(\displaystyle BC\) correlates with the curvature of \(\displaystyle ABC\): \(\displaystyle M \propto \text{Curvature}\). I see a formula in there somewhere. Can you take it from here?

NB: \(\displaystyle B\) should be midpoint of arc \(\displaystyle ABC\)

 

[BigBeachBanana]

So rude!

Not really after reading 1000's of your posts over the past few years.

 

[logistic_guy]

I don't know what a "regular curve" is, do you mean smooth?
You're asserting that \(\displaystyle \bold{r}(t) = 0\) what is your reasoning for this assertion?


He's being sarcastic with the emoji.


I'd ignore @Agent Smith . He belongs to the internet troll category.

regular mean smooth
i say \(\displaystyle \bold{T}’(t) = 0\)

I was just offering my views on what a curvature is and how we could measure it in a simple fashion, which in this case could mean no calculus. If you look at my diagrams, it's true that ... given an arc \(\displaystyle ABC\), the absolute value of the average slopes/gradients (call this \(\displaystyle M\)) of the straight lines \(\displaystyle AB\) and \(\displaystyle BC\) correlates with the curvature of \(\displaystyle ABC\): \(\displaystyle M \propto \text{Curvature}\). I see a formula in there somewhere. Can you take it from here?

NB: \(\displaystyle B\) should be midpoint of arc \(\displaystyle ABC\)

thank Agent Smith

according to your nice drawings, what is the answer?
a.
b.
c.
d.

 

[Agent Smith]

regular mean smooth
i say \(\displaystyle \bold{T}’(t) = 0\)

thank Agent Smith

according to your nice drawings, what is the answer?
a.
b.
c.
d.

Cogito ...
The transformation is isometric - translating the circle from (2, 3) to (3, 5) - and so all properties of the object/circle in this case, are preserved. There's no change in the curvature i.e. [imath]k = \frac{1}{2}[/imath].

However, from a physical POV, we have flat earthers, one reason being the curvature of the earth is imperceptible at the relative scales between homo sapiens and Gaia.

Question:
Is curvature
a) Objective
b) Subjective
c) Both a) and b)
d) Neither a) nor b)

?

 

[logistic_guy]

Cogito ...
The transformation is isometric - translating the circle from (2, 3) to (3, 5) - and so all properties of the object/circle in this case, are preserved. There's no change in the curvature i.e. [imath]k = \frac{1}{2}[/imath].

it seem you understand isomorphism. can you please proof the answer is \(\displaystyle \kappa = \frac{1}{2}\)?

However, from a physical POV, we have flat earthers, one reason being the curvature of the earth is imperceptible at the relative scales between homo sapiens and Gaia.

Question:
Is curvature
a) Objective
b) Subjective
c) Both a) and b)
d) Neither a) nor b)

?

what is this?

 

[BigBeachBanana]

i say \(\displaystyle \bold{T}’(t) = 0\)

1) T'(t) is a vector are you saying T'(t) is a vector of 0 or scalar of 0?
2) Again, you keep using the phrasing "I say", and "its value is..." without any of the "because...." to follow up with that. Nobody can help without understanding what you're thinking. I'll try again.
What is your rationale in claiming [imath]\bold{r}(t)=0[/imath], when you know [imath]\bold{r}'(t)=0[/imath] which violates the smooth curve assumption?

 

[Agent Smith]

it seem you understand isomorphism. can you please proof the answer is \(\displaystyle \kappa = \frac{1}{2}\)?


what is this?

1. All translations are isometric transformations that preserve all the object's properties
2. The circle in the OP has undergone only a translation
Ergo,
3. The circle in the OP's properties [shape & size (lengths, ratios, angles, curvature, etc.) are preserved] ... from 1 and 2
4. If the circle in the OP's properties are preserved then the circle in the OP's curvature is preserve i.e. remains the same
5. The circle in the OP's curvature is preserved i.e. IF \(\displaystyle k = \frac{1}{2} \text{ then after the translation } k = \frac{1}{2}\)... from 3 and 4

This is the best I can do.

Cogito, the way it's actually done is you start off with a formula for curvature (like the one you have in your OP).
First compute the curvature of the preimage. Then compute the curvature of the image
You should see that \(\displaystyle \text{Curvature(Preimage) = Curvature(Image)}\)

 

[logistic_guy]

1) T'(t) is a vector are you saying T'(t) is a vector of 0 or scalar of 0?
2) Again, you keep using the phrasing "I say", and "its value is..." without any of the "because...." to follow up with that. Nobody can help without understanding what you're thinking. I'll try again.
What is your rationale in claiming [imath]\bold{r}(t)=0[/imath], when you know [imath]\bold{r}'(t)=0[/imath] which violates the smooth curve assumption?

i don't say \(\displaystyle \bold{r}(t) = 0\) or \(\displaystyle \bold{r}'(t) = 0\), i say \(\displaystyle \bold{T}'(t) = 0\)

proof

\(\displaystyle \displaystyle \frac{d}{dt}|\bold{T}(t)| = |\bold{T}'(t)| = 0\)

if you think this formula is wrong, why? i don't see that

1. All translations are isometric transformations that preserve all the object's properties
2. The circle in the OP has undergone only a translation
Ergo,
3. The circle in the OP's properties [shape & size (lengths, ratios, angles, curvature, etc.) are preserved] ... from 1 and 2
4. If the circle in the OP's properties are preserved then the circle in the OP's curvature is preserve i.e. remains the same
5. The circle in the OP's curvature is preserved i.e. IF \(\displaystyle k = \frac{1}{2} \text{ then after the translation } k = \frac{1}{2}\)... from 3 and 4

This is the best I can do.

thank Agent Smith

i'm convinse now but i'm need to show this result by formula

Cogito, the way it's actually done is you start off with a formula for curvature (like the one you have in your OP).
First compute the curvature of the preimage. Then compute the curvature of the image
You should see that \(\displaystyle \text{Curvature(Preimage) = Curvature(Image)}\)

this is what i'm trying to do from the beginning of the post number \(\displaystyle 1\)

 

[BigBeachBanana]

i don't say \(\displaystyle \bold{r}(t) = 0\) or \(\displaystyle \bold{r}'(t) = 0\), i say \(\displaystyle \bold{T}'(t) = 0\)

proof

\(\displaystyle \displaystyle \frac{d}{dt}|\bold{T}(t)| = |\bold{T}'(t)| = 0\)

if you think this formula is wrong, why? i don't see that

The derivative of a scalar i.e. [imath]\frac{d}{dt}|\bold{T}(t)|[/imath] is 0, but not necessarily true for [imath]|\bold{T}'(t)|[/imath]. I"ll try this one last time. Why is [imath]|\bold{T}'(t)| = 0 [/imath] ?

 

[Agent Smith]

i don't say \(\displaystyle \bold{r}(t) = 0\) or \(\displaystyle \bold{r}'(t) = 0\), i say \(\displaystyle \bold{T}'(t) = 0\)

proof

\(\displaystyle \displaystyle \frac{d}{dt}|\bold{T}(t)| = |\bold{T}'(t)| = 0\)

if you think this formula is wrong, why? i don't see that


thank Agent Smith

i'm convinse now but i'm need to show this result by formula


this is what i'm trying to do from the beginning of the post number \(\displaystyle 1\)

This seems like a straightforward plug-n-play scenario (use the formula that is) then, si/non?

 

[logistic_guy]

The derivative of a scalar i.e. [imath]\frac{d}{dt}|\bold{T}(t)|[/imath] is 0, but not necessarily true for [imath]|\bold{T}'(t)|[/imath]. I"ll try this one last time. Why is [imath]|\bold{T}'(t)| = 0 [/imath] ?

if \(\displaystyle \displaystyle \frac{d}{dt}\bold{T}(t) = \bold{T}'(t)\)

mathematics rules allow us to take absolute of both sides, so

\(\displaystyle \displaystyle \frac{d}{dt}|\bold{T}(t)| = |\bold{T}'(t)|\)

i don't violate any rule. why suddenly i can't do that?

 

[BigBeachBanana]

if \(\displaystyle \displaystyle \frac{d}{dt}\bold{T}(t) = \bold{T}'(t)\)

mathematics rules allow us to take absolute of both sides, so

\(\displaystyle \displaystyle \frac{d}{dt}|\bold{T}(t)| = |\bold{T}'(t)|\)

i don't violate any rule. why suddenly i can't do that?

There is a misunderstanding about the notation. The vertical bars around vectors are not absolute values, but they indicate the magnitudes of the vectors. See this page.

 

[blamocur]

if you think this formula is wrong, why? i don't see that

The formula is wrong. But if you believe it is correct than you should try to prove it -- you might discover some misconceptions along the way.

BTW, by the same reasoning you would have [imath]\left|\frac{d}{dt}\mathbf r\right| = 0 \Rightarrow \frac{d}{dt} \mathbf r = 0[/imath]

 

[logistic_guy]

There is a misunderstanding about the notation. The vertical bars around vectors are not absolute values, but they indicate the magnitudes of the vectors. See this page.

i know the absolute value in this case mean magnitude of vector. if two things equal why their magnitude will be different?

The formula is wrong. But if you believe it is correct than you should try to prove it -- you might discover some misconceptions along the way.

BTW, by the same reasoning you would have [imath]\left|\frac{d}{dt}\mathbf r\right| = 0 \Rightarrow \frac{d}{dt} \mathbf r = 0[/imath]

so the formula is wrong. but nobody can't tell the reason why and i have to proof it is correct if i say it is correct. i don't think make sense

 

[BigBeachBanana]

i know the absolute value in this case mean magnitude of vector. if two things equal why their magnitude will be different?


so the formula is wrong. but nobody can't tell the reason why and i have to proof it is correct if i say it is correct. i don't think make sense

Order of operation. This is true
\(\displaystyle \displaystyle \left|\frac{d}{dt}\bold{T}(t)\right| = |\bold{T}'(t)|\)

but not this.
\(\displaystyle \displaystyle \frac{d}{dt}|\bold{T}(t)| = |\bold{T}'(t)|\)

The latter says to take the derivative of a scalar which is always 0, while the former takes the derivative of the vector first and then finds the magnitude. The operation is not commutative.

 

[logistic_guy]

Order of operation. This is true
\(\displaystyle \displaystyle \left|\frac{d}{dt}\bold{T}(t)\right| = |\bold{T}'(t)|\)

but not this.
\(\displaystyle \displaystyle \frac{d}{dt}|\bold{T}(t)| = |\bold{T}'(t)|\)

The latter says to take the derivative of a scalar which is always 0, while the former takes the derivative of the vector first and then finds the magnitude. The operation is not commutative.

thank BigBeachBanana

this answer my question. now i see my mistake

You can convince yourself.
Parameterize [imath]r(t)=\lbrace R\cos(t),R\sin(t) \rbrace[/imath]. Compute the necessary components.

i solve the problem in paper before when you give me the parametrization curve in post #9. i'll show my work again with latex.

\(\displaystyle \bold{r}(t) = (R\cos t, R\sin t)\) given in the problem \(\displaystyle R = 2\)

\(\displaystyle \bold{r}(t) = (2\cos t, 2\sin t)\)

\(\displaystyle \bold{r}'(t) = (-2\sin t, 2\cos t)\)

\(\displaystyle \bold{T}(t) = \frac{\bold{r}(t)}{|\bold{r}(t)|} = \frac{(2\cos t, 2\sin t)}{\sqrt{4}} = \frac{(2\cos t, 2\sin t)}{2} = (\cos t, \sin t)\)

\(\displaystyle \bold{T}'(t) = (-\sin t, \cos t)\)

i'll use the formula i proof in post#8

\(\displaystyle \kappa = \frac{|\bold{T}'(t)|}{|\bold{r}'(t)|} = \frac{\sqrt{1}}{\sqrt{4}} = \frac{1}{2}\)

this mean the answer is a. \(\displaystyle \frac{1}{2}\)

i get a lot of experience in thsi thread
thank again BigbeachBanana
thank blamocur
thank Agent Smith
thank khansaheb
thank Mario99

 

[BigBeachBanana]

thank BigBeachBanana

this answer my question. now i see my mistake


i solve the problem in paper before when you give me the parametrization curve in post #9. i'll show my work again with latex.

\(\displaystyle \bold{r}(t) = (R\cos t, R\sin t)\) given in the problem \(\displaystyle R = 2\)

\(\displaystyle \bold{r}(t) = (2\cos t, 2\sin t)\)

\(\displaystyle \bold{r}'(t) = (-2\sin t, 2\cos t)\)

\(\displaystyle \bold{T}(t) = \frac{\bold{r}(t)}{|\bold{r}(t)|} = \frac{(2\cos t, 2\sin t)}{\sqrt{4}} = \frac{(2\cos t, 2\sin t)}{2} = (\cos t, \sin t)\)

\(\displaystyle \bold{T}'(t) = (-\sin t, \cos t)\)

i'll use the formula i proof in post#8

\(\displaystyle \kappa = \frac{|\bold{T}'(t)|}{|\bold{r}'(t)|} = \frac{\sqrt{1}}{\sqrt{4}} = \frac{1}{2}\)

this mean the answer is a. \(\displaystyle \frac{1}{2}\)

i get a lot of experience in thsi thread
thank again BigbeachBanana
thank blamocur
thank Agent Smith
thank khansaheb
thank Mario99

You're welcome. If you leave the variable [imath]R[/imath] instead of setting it to 2. You'd effectively shown that the curvatures for circles are 1/R.