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#1
I've noticed plenty of people posting threads asking questions about their math homework or whatever.... so i thought to myself,

"hey, why isn't there one big super thread where the highly intellectual users of UG can help those in need of some brain power?!"

And thus, I decided to create the Math/Science help thread!

Ask questions about any field of mathematics, physics, chemistry, biology, etc. and someone out there will try to help! This thread is where the brainiacs, nerds and geeks rule; brain power > muscle power; and no question shall go unanswered!

So post those ball breaking questions, problems, and equations that you simply cannot crack!

But please no dividing by zero posts, we all know that anything divided by zero equals 69, so there is no need for it


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#2
I like maths.

EDIT: To the user below me; Me too, we could be twins.
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#6
well here comes the must maniacly crazy thread ever
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#9
Here's an easy one.

Since I can't do the symbols, I'll have to do it in words. The integral from -1 to 1 of the square root of 1-x^2 with respect to x is... <---(dx if that helps any)
#11
That's not it. I hope I wrote it right...I'm so used to just writing the symbols etc., not saying it...
#12
Quote by Zackie EL
Here's an easy one.

Since I can't do the symbols, I'll have to do it in words. The integral from -1 to 1 of the square root of 1-x^2 with respect to x is... <---(dx if that helps any)


1.571
#13
Quote by Zackie EL
Here's an easy one.

Since I can't do the symbols, I'll have to do it in words. The integral from -1 to 1 of the square root of 1-x^2 with respect to x is... <---(dx if that helps any)


4/3
Quote by denizenz
I'll logic you right in the thyroid.

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#16
Quote by Zackie EL
Err...keep it in radians.


There's no radians. It's an integral of a 2nd degree polynomial from an integer to another.

EDIT: Ok, I take it back. I are a retard. I completely ignored the 'square root' part.
Quote by denizenz
I'll logic you right in the thyroid.

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Last edited by darkstar2466 at Aug 26, 2007,
#18
Ok can someone tell me how to plot those graphs that are like "y=x+2" except they dont tell you x?
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#19
Are you sure that's the right thing you're talking about? That's just a normal graph. are you talking about stuff like sqrt(y^2+x^2)= 1?

Edit: Or are you perhaps, joking?

Or are you perhaps not joking and I just insulted you? If so, I apologize .
Last edited by guywithguitar at Aug 26, 2007,
#21
I got (sin(4)/2 - 1/2) - (sin(-4)/2 + 1/2). I think it's right.

EDIT: F*ck, I forgot to change the bounds of integration. Whatever, it would end up right if I redid it.
Quote by denizenz
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Art & Lutherie
Last edited by darkstar2466 at Aug 26, 2007,
#22
Quote by Zackie EL
Bingo.


Haha, guys, it's just the area under half of a unit circle, no math involved.
#25
Quote by myself101
Ok can someone tell me how to plot those graphs that are like "y=x+2" except they dont tell you x?

im not quite sure what u mean, if u mean as in y=2, then you would just draw a straight line on the y=2 spaces, i.e. (0,2)(1,2)(-1,2) all the way across, so no matter what the x=?, the y=2 always
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#26
Quote by bigwillie
Which also equals 1.571. Which is what I said.


Yeah, but you actually did the integral .
#27
Quote by bigwillie
Which also equals 1.571. Which is what I said.


I know, but keep it in radians. Sorry I didn't mention it.
#28
Anyone done double/triple or flux/circulation integrals here?
Quote by denizenz
I'll logic you right in the thyroid.

Art & Lutherie
#29
Quote by guywithguitar
Yeah, but you actually did the integral .


But the long way was much more fun.

Quote by Zackie EL
I know, but keep it in radians. Sorry I didn't mention it.


Ah well, it's all good.
#30
Quote by myself101
Ok can someone tell me how to plot those graphs that are like "y=x+2" except they dont tell you x?


y = x+2

your slope is 1/1, and your y-intercept is (0,2). So start at that point and move up one and over one. Plot that point. then move up one and over one. and plot that point. you'll soon see what the line is supposed to look like.
#31
Quote by darkstar2466
Anyone done double/triple or flux/circulation integrals here?

I've had one double integral problem ever, and was kind of walked through it, but I think I understand how to do them. But do elaborate on the flux and circulation integrals
.
#32
Quote by darkstar2466
Anyone done double/triple or flux/circulation integrals here?


I've done a few triples, but I've never heard of the flux/circulation integrals before either.
#33
Ok here's another easy one. This is trig.

Give me the area of a triangle with sides a-9.12, b-4.67, c-10.83 and tell me how you did it.
#34
Quote by guywithguitar
I've had one double integral problem ever, and was kind of walked through it, but I think I understand how to do them. But do elaborate on the flux and circulation integrals
.


Flux is the amount of flow over a certain space, so you're usually given a scalar quantity and you integrate over that region to the flow from it. It's usually used for electricity and magnetism. Here's the basic form:



Circulation is similar, except it's over a closed loop and you figure out the integral of the velocity of a fluid or something circulating. Here's the basic form for that:



But before you do that, you must learn line integrals, which is the basis of these.
Quote by denizenz
I'll logic you right in the thyroid.

Art & Lutherie
#35
Ah, I'm taking E&M next semester, a.k.a. about a week from now, so I'll be familiar with those soon .
#36
Quote by guywithguitar
Ah, I'm taking E&M next semester, a.k.a. about a week from now, so I'll be familiar with those soon .


Yea, you should hit these pretty soon. The math series I'm taking does this first, to prepare for E&M, I guess.

What math are you taking?
Quote by denizenz
I'll logic you right in the thyroid.

Art & Lutherie
#37
im a dumbass when it comes to math

n science


n all school stuff



i like music class tho that count?
#39
Quote by Zackie EL
Ok here's another easy one. This is trig.

Give me the area of a triangle with sides a-9.12, b-4.67, c-10.83 and tell me how you did it.


Uh, I think I would use law of cosines to get the angle C, facing the hypotenuse c, then I would use the general formula of the triangle area A = 1/2 a * b * sin(C). I don't feel like plugging in the numbers.
Quote by denizenz
I'll logic you right in the thyroid.

Art & Lutherie
#40
Quote by Zackie EL
Ok here's another easy one. This is trig.

Give me the area of a triangle with sides a-9.12, b-4.67, c-10.83 and tell me how you did it.


21.07

I used the Law of Cosines to solve for angle A, (56.436), then solved for the height using sine equals opposite over hypotenuse. Then just basic A=b*h/2.

EDIT: I wrote 27.07, but meant 21.07. I can't even read my own writing.
Last edited by bigwillie at Aug 26, 2007,