curvature of circle

[logistic_guy]

here is the question

Given a circle of radius \(\displaystyle 2\) centered at the point \(\displaystyle (2,2)\). It is found immediately that its curvature \(\displaystyle \kappa = \frac{1}{2}\). Given the same circle centered at the point \(\displaystyle (3,5)\), what will be its curvature \(\displaystyle \kappa\) now?

a. \(\displaystyle \frac{1}{2}\)
b. \(\displaystyle \frac{1}{3}\)
c. \(\displaystyle \frac{1}{5}\)
d. non of the above.

initally i say d. non of the above but i'm not sure and i want to be sure by calculating the curvature from this formula \(\displaystyle \kappa = \frac{|\bold{T}'(t)|}{|\bold{r}'(t)|}\). if \(\displaystyle \bold{T}\) is a unit vector then its value is \(\displaystyle |\bold{T}| = 1\) and its derivative should be zero. so \(\displaystyle \kappa = \frac{|\bold{T}'(t)|}{|\bold{r}'(t)|} = \frac{|0|}{|\bold{r}'(t)|} = 0\). this tell me the curvature is zero, but this is impossible because zero curvature mean straight line and i've a circle in the question.

 

[khansaheb]

here is the question

Given a circle of radius \(\displaystyle 2\) centered at the point \(\displaystyle (2,2)\). It is found immediately that its curvature \(\displaystyle \kappa = \frac{1}{2}\). Given the same circle centered at the point \(\displaystyle (3,5)\), what will be its curvature \(\displaystyle \kappa\) now?

a. \(\displaystyle \frac{1}{2}\)
b. \(\displaystyle \frac{1}{3}\)
c. \(\displaystyle \frac{1}{5}\)
d. non of the above.

initally i say d. non of the above but i'm not sure and i want to be sure by calculating the curvature from this formula \(\displaystyle \kappa = \frac{|\bold{T}'(t)|}{|\bold{r}'(t)|}\). if \(\displaystyle \bold{T}\) is a unit vector then its value is \(\displaystyle |\bold{T}| = 1\) and its derivative should be zero. so \(\displaystyle \kappa = \frac{|\bold{T}'(t)|}{|\bold{r}'(t)|} = \frac{|0|}{|\bold{r}'(t)|} = 0\). this tell me the curvature is zero, but this is impossible because zero curvature mean straight line and i've a circle in the question.

A translation of the center inherently does not change any characteristic of the circle.

Hence the radius of the circle remain the same along with the curvature.

 

[khansaheb]

κ=∣T′(t)∣∣r′(t)∣\(\displaystyle \displaystyle \kappa = \frac{|\bold{T}'(t)|}{|\bold{r}'(t)|}κ=∣r′(t)∣∣T′(t)∣\)

Could you please provide a reference to the equation above.

 

[BigBeachBanana]

If \(\displaystyle \bold{T}\) is a unit vector then its value is \(\displaystyle |\bold{T}| = 1\) and its derivative should be zero.

This is incorrect. You find T then take the derivative then compute the magnitude so [imath]|T'(t)|=1[/imath].
PS: For circles, the curvature can be simplified to [imath]\kappa = \dfrac{1}{\text{radius}}[/imath].

 

[khansaheb]

Whathere is the question

Given a circle of radius \(\displaystyle 2\) centered at the point \(\displaystyle (2,2)\). It is found immediately that its curvature \(\displaystyle \kappa = \frac{1}{2}\). Given the same circle centered at the point \(\displaystyle (3,5)\), what will be its curvature \(\displaystyle \kappa\) now?

a. \(\displaystyle \frac{1}{2}\)
b. \(\displaystyle \frac{1}{3}\)
c. \(\displaystyle \frac{1}{5}\)
d. non of the above.

initally i say d. non of the above but i'm not sure and i want to be sure by calculating the curvature from this formula \(\displaystyle \kappa = \frac{|\bold{T}'(t)|}{|\bold{r}'(t)|}\). if \(\displaystyle \bold{T}\) is a unit vector then its value is \(\displaystyle |\bold{T}| = 1\) and its derivative should be zero. so \(\displaystyle \kappa = \frac{|\bold{T}'(t)|}{|\bold{r}'(t)|} = \frac{|0|}{|\bold{r}'(t)|} = 0\). this tell me the curvature is zero, but this is impossible because zero curvature mean straight line and i've a circle in the question

What if r'(t) = 0 ........ then you would get \(\displaystyle \kappa\) = undefined...................not 0

 

[blamocur]

if T\displaystyle \bold{T}T is a unit vector then its value is ∣T∣=1\displaystyle |\bold{T}| = 1∣T∣=1 and its derivative should be zero.

The derivative of [imath]|\mathbf T|[/imath] will be zero, but not of [imath]\mathbf T[/imath].

 

[mario99]

here is the question

Given a circle of radius \(\displaystyle 2\) centered at the point \(\displaystyle (2,2)\). It is found immediately that its curvature \(\displaystyle \kappa = \frac{1}{2}\). Given the same circle centered at the point \(\displaystyle (3,5)\), what will be its curvature \(\displaystyle \kappa\) now?

a. \(\displaystyle \frac{1}{2}\)
b. \(\displaystyle \frac{1}{3}\)
c. \(\displaystyle \frac{1}{5}\)
d. non of the above.

initally i say d. non of the above but i'm not sure and i want to be sure by calculating the curvature from this formula \(\displaystyle \kappa = \frac{|\bold{T}'(t)|}{|\bold{r}'(t)|}\). if \(\displaystyle \bold{T}\) is a unit vector then its value is \(\displaystyle |\bold{T}| = 1\) and its derivative should be zero. so \(\displaystyle \kappa = \frac{|\bold{T}'(t)|}{|\bold{r}'(t)|} = \frac{|0|}{|\bold{r}'(t)|} = 0\). this tell me the curvature is zero, but this is impossible because zero curvature mean straight line and i've a circle in the question.

It's true [imath]\displaystyle \frac{d}{dt}|\bold{T}| = 0[/imath]

But

[imath]\displaystyle \frac{d}{dt}|\bold{T}| \neq \bold{T}'(t)[/imath]

I think that you were confused and you mixed this [imath]\displaystyle \frac{d}{dt}\bold{T}(t) = \bold{T}'(t)[/imath] with the wrong expression above.

 

[logistic_guy]

A translation of the center inherently does not change any characteristic of the circle.

Hence the radius of the circle remain the same along with the curvature.

thank khansaheb

still not convince. proof it

Could you please provide a reference to the equation above.

i'll take a picture. it's better to proof it

Definition: The curvature \(\displaystyle \kappa\) of a curve is the scalar quantity \(\displaystyle \kappa = \left|\frac{d\bold{T}}{ds}\right|\)

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

\(\displaystyle \kappa = \left|\frac{d\bold{T}}{ds}\right| = \left|\frac{\bold{T}'(t)}{\frac{ds}{dt}}\right|\)

arc length \(\displaystyle = s(t) = \int_{a}^{t} |\bold{r}'(u)| \ du\)

\(\displaystyle \frac{d \ s(t)}{dt} = \frac{d}{dt}\int_{a}^{t} |\bold{r}'(u)| \ du = |\bold{r}'(t)|\)

\(\displaystyle \kappa = \left|\frac{d\bold{T}}{ds}\right| = \left|\frac{\bold{T}'(t)}{\frac{ds}{dt}}\right| = \left|\frac{\bold{T}'(t)}{\bold{r}'(t)}\right| = \frac{|\bold{T}'(t)|}{|\bold{r}'(t)|}\)

This is incorrect. You find T then take the derivative then compute the magnitude so [imath]|T'(t)|=1[/imath].
PS: For circles, the curvature can be simplified to [imath]\kappa = \dfrac{1}{\text{radius}}[/imath].

thank BigBeachBanana

not convince, proof it

What if r'(t) = 0 ........ then you would get \(\displaystyle \kappa\) = undefined...................not 0

read about smooth curves and their derivatives

The derivative of [imath]|\mathbf T|[/imath] will be zero, but not of [imath]\mathbf T[/imath].

thank blamocur

not convince, proof it

It's true [imath]\displaystyle \frac{d}{dt}|\bold{T}| = 0[/imath]

But

[imath]\displaystyle \frac{d}{dt}|\bold{T}| \neq \bold{T}'(t)[/imath]

I think that you were confused and you mixed this [imath]\displaystyle \frac{d}{dt}\bold{T}(t) = \bold{T}'(t)[/imath] with the wrong expression above.

thank Mario99

not convince, proof it

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

then

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

 

[BigBeachBanana]

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

 

[mario99]

thank Mario99

not convince, proof it

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

then

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

This means that every circle has a zero curvature. Well I agree with your analysis. You have convinced me.

 

[Agent Smith]

 

[blamocur]

thank khansaheb

still not convince. proof it


i'll take a picture. it's better to proof it

Definition: The curvature \(\displaystyle \kappa\) of a curve is the scalar quantity \(\displaystyle \kappa = \left|\frac{d\bold{T}}{ds}\right|\)

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

\(\displaystyle \kappa = \left|\frac{d\bold{T}}{ds}\right| = \left|\frac{\bold{T}'(t)}{\frac{ds}{dt}}\right|\)

arc length \(\displaystyle = s(t) = \int_{a}^{t} |\bold{r}'(u)| \ du\)

\(\displaystyle \frac{d \ s(t)}{dt} = \frac{d}{dt}\int_{a}^{t} |\bold{r}'(u)| \ du = |\bold{r}'(t)|\)

\(\displaystyle \kappa = \left|\frac{d\bold{T}}{ds}\right| = \left|\frac{\bold{T}'(t)}{\frac{ds}{dt}}\right| = \left|\frac{\bold{T}'(t)}{\bold{r}'(t)}\right| = \frac{|\bold{T}'(t)|}{|\bold{r}'(t)|}\)


thank BigBeachBanana

not convince, proof it


read about smooth curves and their derivatives


thank blamocur

not convince, proof it


thank Mario99

not convince, proof it

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

then

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

What is [imath]\mathbf T[/imath]? Is it different from [imath]\mathbf r[/imath] ?

 

[BigBeachBanana]

See Paul's Note.

 

[logistic_guy]

What is [imath]\mathbf T[/imath]? Is it different from [imath]\mathbf r[/imath] ?

\(\displaystyle \bold{T}\) is a unit tangent vector to the curve \(\displaystyle \bold{r}\)

 

[logistic_guy]

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

i calculate your resoning i get \(\displaystyle 1\) but i'm still confused why the formula give me zero?

 

[khansaheb]

why the formula give me zero?

The "formula" gave you 0/0 which is NOT 0......

 

[BigBeachBanana]

i calculate your resoning i get \(\displaystyle 1\) but i'm still confused why the formula give me zero?

Which formula gives you 0? I only see you're claiming it to be 0 but no work to support it.

 

[Agent Smith]

i calculate your resoning i get \(\displaystyle 1\) but i'm still confused why the formula give me zero?

The curvature of A is less than the curvature of B is less than the curvature of C
We can compute the [imath]|\text{Average Slope}|[/imath] (the red lines are from the ends of the curve to the midpoint of the curve). Since we're using the midpoint, our labor is halved. Oui?

I'm not quite comfortable doing that because curvature is basically a TURN and turns are best handled as angles.

 

[logistic_guy]

The "formula" gave you 0/0 which is NOT 0......

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

Which formula gives you 0? I only see you're claiming it to be 0 but no work to support it.

post #10 that Mario99 agree it's correct

View attachment 38283
The curvature of A is less than the curvature of B is less than the curvature of C
We can compute the [imath]|\text{Average Slope}|[/imath] (the red lines are from the ends of the curve to the midpoint of the curve). Since we're using the midpoint, our labor is halved. Oui?

I'm not quite comfortable doing that because curvature is basically a TURN and turns are best handled as angles.

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

 

[BigBeachBanana]

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

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?

post #10 that Mario99 agree it's correct

He's being sarcastic with the emoji.

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

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