Wait, what am I talking about.
I've always thought pi is a bit weird, but it really isn't that big of a hassle.
[QUOTE=JohnnyMo1;28611148]Really wouldn't simplify much. As long as pi is still around I think it would take more effort to change it than it would save in effort.[/QUOTE]
This.
She gave some wonderful ways of visualizing Pi. The whole Tau thing is arbitrary at best, though. While I won't argue that it is a nice idea, it would be better if people gave kids more visualization and hands-on over [i]memorize these charts and values.[/i]
yeah i think ill stick with pi
Her voice is annoying, she sounds like a nerd.
[QUOTE=-ZeeBo-;28614723]Her voice is annoying, she sounds like a nerd.[/QUOTE]
Gee, it's about math. I think you came to the wrong thread.
It's funny because even if using it did have advantages and Tau didn't already represent so many other things then it would still be impossible to change the way it's [b]tau[/b]ght.
Wow lets start riots.
I am fucking serious.
I am going to start a riot in front of local matphys university.
I don't get anything. Probably because I'm just utterly horrid at math.
I like pies.
[QUOTE=Jojje;28617195]I don't get anything. Probably because I'm just utterly horrid at math.[/QUOTE]
Basically Pi is stupid and assigning value of 2Pi to a new constant would make everything ever easier.
That makes a whole lot of sense. The problem with pi being defined through the diameter instead of the radius bugged the hell out of me when I was learning trigonometry. Tau is being used elsewhere ofcourse, but since when was science ever completely consistent with its labeling? Sure constants are often associated with only one letter, but variables in physics like velocity and length often get different labels so would it really be that much of a stretch to introduce this new definition of tau?
[QUOTE=ThePuska;28613879]1 tau = 1 rotation in radians. Seems like it would simplify some things.
Circumference of a circle:
[img_thumb]http://www.codecogs.com/gif.latex?C%20=%20\tau%20r[/img_thumb]
Area of a circle:
[img_thumb]http://www.codecogs.com/gif.latex?A%20=%20{1%20\over%202}%20\tau%20r^2[/img_thumb]
What odd resemblance to the antiderivative and second antiderivative of a constant![/QUOTE]
That's because
[img]http://gyazo.com/3ae401a373e5f8b171d043f0f132492f.png[/img]
Which is kinda cool because that's a shell integration-proof
that pie didnt really look yummy :v
[QUOTE=Number-41;28617335]That's because
[img_thumb]http://gyazo.com/3ae401a373e5f8b171d043f0f132492f.png[/img_thumb]
Which is kinda cool because that's a shell integration-proof[/QUOTE]
That was what I was thinking when I saw it last night. It is pretty much adding up all of the circumferences from R to 0, which would give you the area. In calc II we had to find the area of a circle via calc methods, and everybody ended up solving it in a much more difficult, but still valid way. I don't know why nobody in my class (including myself) thought of this method. The TA even showed us the more complex way of solving it.
Maybe you guys did it in Euclidian coordinates? That's a bit more complicated (really have to learn my basic antiderivatives)
It is kinda a brain-twist that you just multiply a circle with an infinitesimal value and suddenly it turns into a valid area. Maybe it has to do with small angles or something like that...
Isn't that a stupid idea?
I mean sine and cosine are equal to those long sums that are something like x - x^3/3! + x^5/5!... right? And when you use it like that you get sin(pi) to be 0, which would be half a rotation.
[QUOTE=Pavarotti;28618332]Isn't that a stupid idea?
I mean sine and cosine are equal to those long sums that are something like x - x^3/3! + x^5/5!... right? And when you use it like that you get sin(pi) to be 0, which would be half a rotation.[/QUOTE]
The series definitions of sine and cosine remain unchanged if you switch to using tau instead of pi. The definition works for radians, and you can easily express radians with either pi or tau.
[editline]15th March 2011[/editline]
[QUOTE=Pepin;28617885]That was what I was thinking when I saw it last night. It is pretty much adding up all of the circumferences from R to 0, which would give you the area. In calc II we had to find the area of a circle via calc methods, and everybody ended up solving it in a much more difficult, but still valid way. I don't know why nobody in my class (including myself) thought of this method. The TA even showed us the more complex way of solving it.[/QUOTE]
Here's a needlessly elaborate way of calculating the area of a circle (I was trying to explain the connection between the circumference and the area to myself).
Defining the circle's curve in parametric form:
[img]http://latex.codecogs.com/gif.latex?(x(t),%20y(t))[/img]
[img]http://latex.codecogs.com/gif.latex?x(t)%20=%20r%20%5Ccdot%20%5Ccos%20t[/img]
[img]http://latex.codecogs.com/gif.latex?y(t)%20=%20r%20%5Ccdot%20%5Csin%20t[/img]
The area enclosed by the curve is given by ([url=http://mathworld.wolfram.com/GreensTheorem.html#eqn3]1[/url]):
[img]http://latex.codecogs.com/gif.latex?A%20=%20{1%20%5Cover%202}%20%5Cint_{0}^{2%5Cpi}%20%5Cfrac{x(t)%20%5Ccdot%20%5Cpartial%20y(t)%20}{%5Cpartial%20t}%20-%20%5Cfrac{y(t)%20%5Ccdot%20%5Cpartial%20x(t)%20}{%5Cpartial%20t}%20dt[/img]
Substitution and derivation gives:
[img]http://latex.codecogs.com/gif.latex?A%20=%20{%5Cfrac{1}{2}}%20%5Cint_{0}^{2%5Cpi}%20{(r^2%20{%5Ccos^2t}%20+%20r^2%20{%5Csin^2t}%20)%20dt}[/img]
Integration of the trigonometric functions' squares leaves:
[img]http://latex.codecogs.com/gif.latex?A%20=%20{%5Cfrac{1}{2}%5Cint_{0}^{2%5Cpi}{r^2dt}}[/img]
Which simplifies to:
[img]http://latex.codecogs.com/gif.latex?A%20=%20%5Cpi%20r^2[/img]
This blew my fucking mind!
I feel smart being that I understood it all and I'm only 15 :o
The problem's that torque is also marked with Τ (tau).
[QUOTE=ThePuska;28613879]1 tau = 1 rotation in radians. Seems like it would simplify some things.
Circumference of a circle:
[img_thumb]http://www.codecogs.com/gif.latex?C%20=%20\tau%20r[/img_thumb]
Area of a circle:
[img_thumb]http://www.codecogs.com/gif.latex?A%20=%20{1%20\over%202}%20\tau%20r^2[/img_thumb]
What odd resemblance to the antiderivative and second antiderivative of a constant![/QUOTE]
That was the very first thing I noticed once I noticed when I learned how to calculate an antiderivative. It's not like using pi rather than tau has obscured our ability to see mathematical relations.
Wait, no, earlier. I noticed d/dr pi*r^2 was 2pi*r
[QUOTE=JohnnyMo1;28621691]That was the very first thing I noticed once I noticed when I learned how to calculate an antiderivative. It's not like using pi rather than tau has obscured our ability to see mathematical relations.
Wait, no, earlier. I noticed d/dr pi*r^2 was 2pi*r[/QUOTE]
I don't think it's about pi being too hard to understand, but more about how tau would be [i]even easier[/i]. It'd make things more uniform as far as I can see. Patterns are good.
Yeah I don't think it'd be easy or even worth it to switch, and if we did we would probably just change the value of Pi rather than make Tau the new constant (because of torque), but I still like this better.
I learned Euler's identity in school like last week and I'm confused as fuck
[QUOTE=Skunky;28617206]I like pies.[/QUOTE]
I like math.
[editline]15th March 2011[/editline]
We should hang out sometime.
[QUOTE=LazyBoy;28622975]I learned Euler's identity in school like last week and I'm confused as fuck[/QUOTE]
Some people think it's THE MOST BEAUTIFUL THING EVER OMG
I don't get that. It's just a special case of euler's formula, which I think is a lot more interesting. When you get your mind around the notion that the complex exponential e^(ix) makes a circle of radius 1 in the complex plane it seems obvious and boring that e^(i*pi) = -1. And that e^(i*tau) = 1. Euler's formula is a lot more interesting to me than the identity is.
I learned all about sinewaves just this morning in math class and now this video is telling me it could all be so much simpler D:
[QUOTE=Raygen;28623872]I learned all about sinewaves just this morning in math class and now this video is telling me it could all be so much simpler D:[/QUOTE]
It's still a sinewave.
I wish she was my tutor back in 9th grade. Our math teacher knew jack squat.
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