Length of Line between 2 Points on Circle

[quasar987]

Hi,

I'd like to know the name for the length in a circle that I will describe:

Consider 2 arbitrary points lying on the perimeter of a circle. I'm talking about the shortest distance distance between these those points. (i.e. the straight line joining them.)

Thx

 

[jamesrc]

It's called a chord.

 

[HallsofIvy]

Yep, a chord.

 

[quasar987]

Ok, I have a confusion involving a cord. In physics, we often make the transformation from an infinitesimal cord to an infinitesimal arc: d(cord) = d(arc). And I think it's used in some math proofs too.

For exemple, imagine a position vector $\vec{r}(t)$ of fixed norm, rotating around the z axis (i.e. $\vec{\omega} = \omega \hat{z}$). Let's say the radius of the circle described by the motion of its tip/head/arrow is R. After a time $\Delta t$, it has rotated an angle $\omega \Delta t$ and then they (the manuals) say that it can be seen that when $\Delta t$ is small, $||\Delta \vec{r}||$ (which is a cord), is very near the length of the arc $\Delta s$ subtended by the angle $\omega \Delta t$, and thus, poof, $||d\vec{r}|| = ds$.

And while this seems to be true intuitively, I have never seen a proof of this statement. And when I try to do it, here's what I get:

I start from a cercle of radius R and a cord $\delta$ subtended by an angle $\theta$. I find that the length of the cord is given by

$$\delta = 2Rsin\left(\frac{\theta}{2}\right)$$

Therefor,

$$d\delta=Rcos\left(\frac{\theta}{2}\right)d\theta$$

While

$$ds=rd\theta$$

So

$$d\delta=cos\left(\frac{\theta}{2}\right)ds$$

A result indicating that even the differential version is just an approximation because only true for a principal angle $\theta=0$.

 

[StatusX]

it follows from the identity

$$\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$$

Also, the mistake in your derivation is that theta is a function of the chord length.

 

[quasar987]

Thanks for your reply StatusX

it follows from the identity

$$\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$$

At which point does that identity fit in?

Also, the mistake in your derivation is that theta is a function of the chord length.

Could you elaborate?

 

[dextercioby]

Sure,if "y" is a function of "x",then "x" is a function of "y",right??So he basically didin't say anything new...You were right,though...Your calculations were corrrect.I guess u knew that,but u have my confirmation...

Daniel.

 

[StatusX]

At which point does that identity fit in?

A chord of length 2 R sin(theta) is subtended by an angle of 2 theta (in radians) and so has an arclength of 2 R theta. The identity means d(chordlength)/d(arclength) = 1.

Could you elaborate?

I'm sorry, I just glanced at your last step and assumed you had forgotten that if arclength is 0, then theta must be as well. You did forget that, but for a different reason. What you derived is how an chord of finite length changes with an infinitessimal change in arclength. What you wanted was the ratio of a differential chord to the differential arclength it subtends. To get the differential you're looking for, just take theta=0.