My boring morning, in which I nearly pass out at the gym...
So, yesterday I managed my eating schedule poorly and didn't eat nearly enough all day.
I ate a sausage muffin at 9, then didn't eat anything 'till 9pm, and then when I did eat some rice and chicken, I didn't eat enough, so at 11 when I was over studying at a friend's, I was hungry again.
5 cups of coffee pretty much messed things all up, what with caffeine being an appetite suppressant. I was busy from 2 'till 9, so really didn't notice.
So ... this morning I get up, go to the gym with bashamer.
We warm up, it's great, have a great warmup, kicking ass for 10m on the elliptical cranked up ...
then we bench, and I'm doing great, pretty solid 10, 9, 8 reps at 135 ...
then we do leg curls and leg extensions at 180, 10 reps each, twice, no probs ...
I get a little notice from my brain saying "Hi. This is a similar message to when you drink too much. This is your only warning. Fix things now, or things will shut down"
So, I go sit down for a few minuites, then quickly wolfed down a bag of cherry gummies and a gatorade, then I took a shower ending with a significant cold blast, then went to the hill and ate a roast beef sub.
So, I recovered.
The rest of my afternoon and the second order derivative thereof...
So, I park it upstairs at Prufrock's coffeeshop on the hill near campus, hammer thru 8 (8.5?) of my 10 homework problems, then stomp all the way across campus to class ...
I understand this notation:f'(x) = derivative of f(x)I was having trouble with this notation:dx -- dy
It honestly seemed that a significant percentage of the class was also having a rough time with the chain rule. Fortunately, I now get it. Even though I still prefer looking at:y = sin(((2+x^2)/4x)^4) y = sin ( u ) u = ( v ^ 4 ) v = t / s t = 2 + x^2 s = 4x s' = 4 t' = 2x v' = ( s t' - t s' ) / s^2 = ( (( 4x ) ( 2x )) - (( 2 + x^2 ) (4)) ) / ( 4x )^2 = ( ( 4 x^2 - 8 ) / ( 16 x^2 ) ) ( t' ) ( s' ) = ( ( 4 x^2 - 8 ) ( 2x ) ( 4 ) ) / (16 x^2) = ( 32 x^3 - 64x )/ (16x) (x) = ( 2 x^2 - 4 ) / x u' = ( ( 4v )^3 ) ( v' ) = ( ( 4 ( ( 2 + x^2 ) / 4x ) ^ 3 ) ) ( ( 2 x^2 - 4 ) / x ) = ( x^8 + 4 x^6 - 16 x^2 - 16 ) / 32 x^4 y' = cos(u) ( v' ) y' = cos((2+x^2)/4x)^4)((x^8+4x^6-16x^2-16)/32x^4)
Hmm. $5 to anyone who finds a bug in my math. :-)
Is there some software that'll do this? (and I mean really do it, not brute force it with numbers)?
Anyway, back [5 hours later!] to what I was saying, I really prefer f'(g(x))g'(x) type notation to df/dg dg/dx = df/x notation. But I now understand it. Now, what I have to figure out, are .. umm... implicit differentiation? Something like that. Something that lets you determine the instantaneous tangent slope for any pair of polar coordinates on a given circle.
petit_baiser explained it to me last night and it all made sense at the time. Dammit.
Oh, and the HTML I used in this post is so abysmal I fully expect mrothermel to point and laugh hysterically.
...and LJ's mad at me for starting to blog before midnight and finishing after