User:GFauxPas/Sandbox

Welcome to my sandbox, you are free to play here as long as you don't track sand onto the main wiki. --GFauxPas 09:28, 7 November 2011 (CST)

Thoughts on the picture, anyone? --GFauxPas 14:37, 9 November 2011 (CST)


 * Looks okay to me. I was going to get round to doing something similar myself at one point.


 * Mind, if you're planning on using it to illustrate sine and cosine, you might want to add the actual distances as lines. Oh, and purists may wince when they see x and y used to define the axes and the point on it, but I wouldn't be too fussed. --prime mover 14:47, 9 November 2011 (CST)

Try 2. I see I lost too many colors by saving it as a .gif, try 3 will be a .png or something.

File:Unitcirclev2.gif

\Let $P = (x,y)$ be a point on the unit circle centered at the origin.

Let $\theta$ be the angle formed by the arc $(1,0)$, $(x,y)$ subtending the origin, measured counterclockwise.

The unit circle definition of the trigonometric functions are $\cos \theta := x$

$\sin \theta := y$

That is, the directed distance between $P$ and the $x$-axis is the cosine, and the directed distance between $P$ and the $y$-axis is the sine.

Sources: khan academy "tau versus pi", wolfram mathworld "trigonometry"

After this is set up I can do a proof of the consistency between the right triangle definition and the circle definition. --GFauxPas 07:22, 17 November 2011 (CST)

What to Call this Theorem?
Hello friends, I'd like to add a proof for:


 * $\displaystyle \int \frac {1}{x^2 + a^2} \ \mathrm dx$

but I don't know what to call it. "Integral of a Rational Function"? "Integral Involving Arctangent"? "Integral of 1 Over (x^2 + a^2)"? What should I name the page? --GFauxPas 06:55, 15 December 2011 (CST)

Moved to Integral Involving Arctangent, please change the name if you think of something better. --GFauxPas 14:02, 15 December 2011 (CST)

where $a$ is a strictly positive constant and $a^2 > x^2$.

Moved to Integral Involving Arcsine --GFauxPas 08:26, 16 December 2011 (CST)


 * I appreciate the work you have been doing on these integrals. The only thing bothering me slightly is that you write equations like $\mathrm d x = a \cos\theta \mathrm d\theta$ while I suspect that you do not really understand what this means (the theory of differentials is really quite delicate and technical to deal with formally). Therefore, I suggest you stick with the substitution theorem instead. --Lord_Farin 08:38, 16 December 2011 (CST)

Sure. I'm not sure exactly how you want me to write it though, what do you mean by the substitution theorem? --GFauxPas 08:45, 16 December 2011 (CST)
 * Well, it is quite easy to show that one may also use the substitution theorem (that is, Integration by Substitution) for indefinite integrals by plugging in values temporarily (this might need a separate page). Then you can just use this route instead of writing the equation with differentials (not that the equality is false; it is just a bit of an intuitive, physicist's shorthand which is hard to state formally). --Lord_Farin 08:58, 16 December 2011 (CST)
 * You seem to have bad experiences with physics :) . The other day in Calculus class we were challenged to solve an integral, I don't remember what it was exactly, that would have been simple were it not for a $ + 6$ floating around. So I said to my Calc II professor, "can't we assume it's negligible?" --GFauxPas 09:01, 16 December 2011 (CST)

Something like this? This looks... bizarre:

--GFauxPas 10:57, 16 December 2011 (CST)

Or maybe:

There's probably a more elegant way to do this ... --GFauxPas 11:25, 16 December 2011 (CST)


 * I'd mention Derivative of an Inverse Function to write down $a\cos\theta \frac{\mathrm d \theta}{\mathrm dx} = 1$ separately, and then just plug it in. Maybe by putting that equation in the middle, using $\implies$ to signify your conclusion. This would yield:


 * Hope that makes a bit more sense. --Lord_Farin 11:44, 16 December 2011 (CST)