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22:00
I googled too
just did =P
vzn
vzn
Duke Forrest: That's a martin-eye, Frank.
Hawkeye Pierce: Finest kind. We're training Ho Jon to be a bartender. Would you care to embribe, sir?
@DanielSank Don't worry, I don't see the relation between the quote it gives me and the situation here anyway
martin-eye?
vzn
vzn
lol saw the MASH thing with my google fu also but then thought it spurious until ACM helped out :P
22:01
ah, nevermind, circumference
@heather Strange pronounciation of martini, I think
@ACuriousMind No relation. It was just a reference.
@ACuriousMind, ah, I see
vzn
vzn
@DanielSank lol as if ppl havent spoken fondly of their alcoholic beverages in here before incl some present :P
>.>
<.<
22:07
lol
So heather, what's $dx^2 = (-R \sin(\theta) d\theta)^2$?
$= R^2 \sin(\theta)^2 d\theta^2$
So do the same for $dy^2$.
vzn
vzn
@Balarka what kind of math are you studying these days anyway
you know; a bit of this, a bit of that
mostly topology, i guess
oh, so I just forgot to square the $d\theta$ in my original answer?
@heather Not sure you've given an answer yet :-)
Previously I think there was a typo where you had $dx$ instead of $d\theta^2$.
22:17
ah, okay
So what do you get now?
so it'd be $\sqrt{R^2\sin(\theta)^2+R^2\cos(\theta)^2d\theta^2}$
?
YES!
Now factor out the stuff you can factor out...
Oh, and you forgot the $d\theta^2$ in the first term.
I think there's a pair of brackets missing :x
um, let's see, $R^2$, and $d\theta^2$
vzn
vzn
22:19
@BalarkaSen do you still have any feelings for number theory? :)
so then we're left with $Rd\theta\sqrt{\sin(\theta)^2+\cos(\theta)^2}$?
no. Forgot to square-root the stuff you pulled from the square root :-)
@vzn of course; but I know nothing of it
What's $\sin^2 + \cos^2$?
vzn
vzn
@BalarkaSen yeah know the feeling me either :|
22:21
oh duh, 1, Pythagorean identity
so $Rd\theta$
because sqrt 1 = 1
and 1*anything = anything
right?
Yes.
So you have $ds = R d\theta$.
This is a very nice result: the length of an arc of a circle is equal to the radius times the angle of that arc.
This is, actually, how angle is usually defined, but here we derived it from the starting point that a circle is defined by $x = R \cos(\theta)$ and $y = R \sin(\theta)$.
Now @heather, how much arc length do you get if the angle goes from 0 to $2\pi$?
not sure...$2\pi$?
@heather Well, let's just do it!
$$s_\text{total} = \int_{\theta=0}^{2\pi} R d\theta$$
$$=R \int_{\theta=0}^{2\pi} d \theta = ?$$
$2\pi R$
?
the circumference of the circle!
Yay!
\(^.^)/
22:28
which makes sense because $2\pi$ is a full revolution around!
\o/
=D
You actually just did some multivariable calculus.
Remember we started with the Pythagorean thingy $ds = \sqrt{dx^2 + dy^2}$.
That tells you the length of a bit of curve in the 2D plane.
Then you used a parametrization of the circle, i.e. $x = R \cos(\theta)$ and $y = R \sin(\theta)$ to figure out $dx$ and $dy$ in terms of a single variable $\theta$, and then you did the integral to get the total path length.
oh...wow, that's cool!
Yes it is.
Good job!
So now you've used calculus to figure out the circumference of a circle and the area of a disk.
Nice.
=D
thank you!
Sure.
ok gotta go
ciao
22:37
good night =)

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