Jon (j_b) wrote,
Jon
j_b

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Jon's Monday

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 ...
  then whammy.

  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.

...and LJ's mad at me for starting to blog before midnight and finishing after

Error updating journal:

  • Incorrect time value: Your most recent journal entry is dated 2003-11-03 22:35, but you're trying to post one at 2003-11-03 16:21 without the backdate option turned on. Please check your computer's clock. Or, if you really mean to post in the past, use the backdate option.
Oh, and the HTML I used in this post is so abysmal I fully expect mrothermel to point and laugh hysterically.
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