Study Support Thread

[Karin]

Since the last one seems to have become deleted.

So, I just got my math midterm back. 18/30, but with a curve that's about a B-
Not too shabby, I guess.

This problem bothered me. e^ln limits make me tear out my hair.

Evaluate the limit: lim(x->0) of (cos x)^(5/x^2)

So the first thing I do is do a e^ln of (cos x)^(5/x^2), that way I can take down the exponent.
It turns into e^((5/x^2)(ln cos x))
But what am I suppose to do after that?

[Hikaru]

I think the thread still exits, just in the 'useful information section'

first we need to understand that the limit only exists if the power has a limit.

so:
lim(x->0) of (cos x)^(5/x^2) = e^lim(x->0-+) of ((5/x^2)(ln cos x))

since the function is symmetrical, it seems that we can only work on one side (and not check 0- too)
so we need to find the lim (x->0+) of

5* ln cos x
------------
    x^2

then we solve it with l'hopital I suppose, doesn't look like one of those cases where l'hopital makes things much worse.
                  - 5* sin x/cos x
= lim(x->0+) --------------- = -5/2    (sinx / x = 1 at 0, it's an important function in signal processing)
                           2x
of course, to each case it's own solution...








OK typed it down then saw it's unrelated, but I'm still going to post it because I went through the trouble of typing it.
well I kinda ponder about that a couple of days ago because of the Tailor Series and all.

we know that
(ln(y))' = y' * ln'(y)   <--->
(ln(y))' = y'  / (y)

because ln'x = 1/x

therefor, y' = (ln(y))' * y
and as it sometimes easier to calculate (ln(y))' in certain functions, then we can calculate y'.

[Karin]

D'oh. I knew it was a L'Hospital's Rule question, but I couldn't figure out how to get there. Thanks again Uriku.

[Zanibas]

Today I learned about cos, sin, and tan, plus what the -1 of those mean as well!

I feel so innocent looking at the work you guys do, knowing I'll have to do that eventually T_T

[Hikaru]

D'oh. I knew it was a L'Hospital's Rule question, but I couldn't figure out how to get there. Thanks again Uriku.

I can't remember what was the case where L'hopital fails horribly and makes things much worse when there's actually a much simpler way (probably involving e^x and the trigonometric functions), but keep a mental note not to always go to L'hopial directly

Today I learned about cos, sin, and tan, plus what the -1 of those mean as well!

I feel so innocent looking at the work you guys do, knowing I'll have to do that eventually T_T

but they don't tell you the real meaning!
cos x is the real part of e^ix, and sin is the imaginary part of e^ix
with that said, sinus is based on numbers that don't even exists!
(well, that is a very boxed image of the truth, but who cares)

[Karin]

The triangle was the easiest way for me to learn about the trig stuff.

Uriku, remind me. What's your major again?

[Hikaru]

computers, but I'm just messing with him to make him jump off a roof somewhere 2 feet above the ground
and they tend to shove in as much math as they can, so it's about 50% math

sorry, above was said under the influence of this person:
http://www.escapistmagazine.com/articles/view/editorials/zeropunctuation.1


and it seems I forgot a very significant 'not' in the last post :s

[Karin]

xD, so you're EECS? Or are you just CS?

Ya, I'm just an IB major so thankfully I don't have to deal with too much math there.

[Hikaru]

just CS, but I've taken the signal processing path(would have said course but could be interpreted differently), which means I'll be figuring out things out of cameras and such, hurray!

IB major? sorry I'm not familiar with the term (found EECS just because of google)

[Karin]

Integrative Biology.

AKA the lazy version of MCB (Molecular and Cellular Biology). Also known as the lazy path to med school.

[Zanibas]

What do the M buttons on a calculator do (or more specifically how you use each)...

I know they're for storing numbers, but I don't know how ;_;

[Karin]

"Memory. No calculator is complete without those confusing M buttons. For once, however, you'll know what they do, and you'll probably find them very useful once you get the hang of them. To store the displayed number, tap the MS button (an M appears in the box above the M buttons). Every time you tap this button, you replace the number you've stored there with the one currently displayed. If you want to add your displayed number to the one that is already in storage, press M+. When you're ready to take a look at the number you have stored in memory, tap the MR button. And when you're ready to clear out the memory, tap the MC button."

Taken from Smart Computing website.

[Orion]

http://www.mathway.com/

If you ever need help with your math.

[Karin]

Okay, clearly I don't know my physics. >.>




what is the distance that the spring needs to be displaced so that the marble will land in the box?

m, k, g, H, and d should be part of the answer.

Basically, how much do I need to compress the spring so that the marble of mass m lands in the box. I've been trying for 20 minutes now and I keep running into either gmd/2Hk or -gmd/2Hk as my answer for displacement of the spring..

So, we know that F = -kx for a spring. F = ma

v = v0 +at
 
The time it takes for the ball to hit the ground is squareroot(2H/g). I don't see anything wrong with that.

That means the ball needs to travel d distance in that amount of time.

Thus intial x velocity needs to be d/(squareroot(2H/g)).

and i'm not quite sure whether I'm correct past that point..

[kryss]

For what happens after the marble is fully accelerated by the spring, what you have there is perfectly right. There is no further acceleration in the x direction so the speed remains constant.

Now we know the final speed the marble attains from the acceleration provided by the spring.

Now, I haven't done this for a while so I might be wrong but now:

a=Kk/m where K is the spring constant
v0=0
v=d/sq(2H/g)
s=k
put all this into
v^2 = v0^2 +2as

(d/sq(2H/g))^2 = 0 + 2*(Kk/m)*k

d^2/(2H/g) = 2Kk^2\m
gd^2/2H = 2Kk^2/m
2Kk^2 = gmd^2/2H
k^2 = gmd^2/(4HK)

therefore

k = (d/2)sq(gm/HK)

...maybe.