Test Woes

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I know that you all are wondering: what on earth is Erica up to that's distracting her from the truly important things in her life (like her blog)?

Glad you asked! Apparently, taking 20 credit hours in one semester is a really, really bad idea. Like, land-war-in-Asia bad. I'd get started on the general evil that is [insert one of my classes here], but I'd actually rather not get into that, as I do have to get to bed and get up for my 8:00 am class.

So instead, I'll relate something I've noticed in a class I like. Namely, calculus. Now, I know what you're thinking. "Oh my god!" you're thinking, "she just said she likes calculus?!"

It's true. Calculus is something that evil math nerds who desperately need a self-esteem boost try to pass off as really difficult to make themselves feel better. It's not easy as in, 2 + 2 = 4 easy, but hey, what is?

But I've noticed a tendency during my tests, one that fellow students in high school complained of and I never really understood until recently. Basically, I'll be sitting there, and I'll look at a problem. Five minutes later, after having... sat there, and looked at it for five minutes, I insert my index finger between my lips, and move up and down repeatedly. This, in spite of the fact that I know how to do the damn problems. For example, one problem on my math test from this morning asked me to prove the following:

Which says, in English, that the derivative of the sum of two functions equals the sum of the derivatives of two functions. I stared at this problem. And then I stared at this problem. And then, I said "fuck it," because I needed to do the rest of the test before I wasted a third of my time on one problem.

The stupid thing is, I did know how to do this problem. I realized what I should have done thirty minutes later, in my car. You just have to use the formal definition of a limit:

Plain English version: The derivative of function f with respect to x is the limit of the difference between the function with an eeeeeeensy bit added to x and the function of x, all over that eensy number, as that eensy number approaches zero (hey, I said it was eensy, didn't I?)

Then, you can actually fill get from one side to the other, like so:

Basically, I just tossed the left side into that definition, rearranged using some standard algebra tricks, and came out with (last line) what I'd have gotten if I stuck the right side into that definition. (The f'(x) thing is the same as that d/dx f(x) bit - just a different way of writing it, called prime notation.)

Now that you've all had your yawn inspiring Erica goodness for the day, I'm going to bed - secure in the knowledge that I have correctly done the problem - even if I won't be getting any points for it.

Hmm, the math I am familar with usually has numbers in it. Formulas are more fun with numbers. Besides, when was the last time my Cap'n Crunch cost \$f(x+h)+g(x+h)-f(x)-g(x)/h?

Well, theoretically, if the price of your Cap'n Crunch fluctuated according to a certain function (which it does, kinda) and you wanted to be able to predict what it was going to cost at any given time (which you might, if you were planning your shopping a few years in advance), you might have to use these nasty equations. In fact, the fluctuation of the price of your two flavors (one with Crunchberries and one without) could be what f(x) and g(x) represented.

Calculus actually *is* useful. Who'd'a thunk?

There was plain English somewhere in there?

It's a DAMNED good thing I don't need significantly higher math to get into the medical profession....I'd be SCREWED! :/

I'm taking Calc right now, and I think you'll get a kick out of what I had to do on my last test. Prove the derivative of sinx is cosx.

Basically you just use the formal definition of the limit,
f`(x)= lim (h->0) f(x+h) - f(x)
--------------------
h

put sin in there...

f`(x)= lim (h->0) sin(x+h) - sinx
----------------------
h

use sin(x+h)= sinxcosh + cosxsinh

f`(x)= lim(h->0) sinxcosh + cosxsinh -sinx
----------------------
h

factor out sinx

f`(x)= lim(h->0) sinx(cosh -1) +cosx(sinh)
------- -----
h h

use cosh -1 = 0 and sinh = 1
------- ---
h h

f`(x) = sinx(0) + cosx(1)
f`(x) = cosx

I can also do that for the derivatives of cosx and tanx, as well as the inverse of all the trig functions.

I probably should have formatted that better. Sorry

can you prove the the derivative of cosx is -sinx