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Old 11-06-2008, 10:45 PM #85
Strawberryblonde.
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I have no clue what the hell you are saying. Please speak in coherent sentences. I was already helping that other guy with calculus, and now I'm going to my gf's house.

If you're looking for help posting, pretend you're trying to communicate to somebody by speaking, and just type those words you would say.
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Old 11-06-2008, 10:48 PM #86
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that dont help me if anyone could i would be so thankful
I suggest you grow a ****ing cerebrum you pathetic dolt. Actually a better suggestion would be for you to leave this thread, sub-forum and forum entirely. Good-day.
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Old 11-06-2008, 10:49 PM #87
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I tried it again, using u=3x and v=sec(3x), then du=3 and dv=3*sec(3x)tan(3x), then applying the product rule:
f'(x)=udv+vdu
=(3x)(3*sec(3x)tan(3x))+(3)(sec(3x))

=9x*sec(3x)tan(3x)+3sec(3x)

That's the exact same answer I got the first time, so I think I got it right.
Ah, my bad. I didn't mean that one. I mean f(x) = sqrt(sin2x)

Its first on my WS.
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Old 11-06-2008, 10:51 PM #88
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I'm pretty sure I did that right, but it would be really helpful if you would post what you got, so I could use it as a reference. Alas, I must be going now, I'll be back around noon or so, my calc II class is cancelled for tomorrow. (yay.)
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Old 11-06-2008, 10:55 PM #89
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alright if anyone could help me in a bcs problem solution paper i would be thankful. i need three solutions and a counter argument please help. thanks
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Old 11-06-2008, 10:58 PM #90
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I solved the derivative and then took the antiderivative and got the original function, so I'm pretty sure I did that one right, too.



There we go.

I converted the square root to the exponent 1/2, then just applied the power rule, and used the chain rule twice.

After that, it was simple algebra.

Last edited by Strawberryblonde. : 11-06-2008 at 11:00 PM.
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Old 11-06-2008, 10:58 PM #91
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I'm pretty sure I did that right, but it would be really helpful if you would post what you got, so I could use it as a reference. Alas, I must be going now, I'll be back around noon or so, my calc II class is cancelled for tomorrow. (yay.)
f(x) = (sin2x)^(1/2)
= (1/2)(sin2x)*cos2x
= (1/2)(cos2x)(sin2x)

Thats what i got...lol.
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Old 11-06-2008, 11:02 PM #92
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f(x) = (sin2x)^(1/2)
= (1/2)(sin2x)*cos2x
= (1/2)(cos2x)(sin2x)

Thats what i got...lol.
Your problem is that you only subtracted 1/2 from the exponent, and you need to subtract 1.
So d/dx[sin(2x)^(1/2)]=(1/2)[sin(2x)^(1/2 - 1)]*cos(2x)*2=(1/2)[sin(2x)^(-1/2)]*cos(2x)*2
=cos(2x)/[sin(2x)^(1/2)]

You have to use the chain rule twice, I believe.

Last edited by Strawberryblonde. : 11-06-2008 at 11:05 PM.
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Old 11-06-2008, 11:10 PM #93
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Your problem is that you only subtracted 1/2 from the exponent, and you need to subtract 1.
So d/dx[sin(2x)^(1/2)]=(1/2)[sin(2x)^(1/2 - 1)]*cos(2x)*2=(1/2)[sin(2x)^(-1/2)]*cos(2x)*2
=cos(2x)/[sin(2x)^(1/2)]

You have to use the chain rule twice, I believe.
Ahh, I completely forgot that. I'll rework it and see what I get.
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Old 11-09-2008, 04:04 PM #94
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does anyone know the organelles that are in a melanocyte?
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Old 11-09-2008, 04:21 PM #95
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hey does anyone have a good problem in sports right now? I dont want to do steroids because some one in my class is doing that already. The paper is a problem-solution paper and i need three solutions to my problem. I want to do something with sports because I usually do the best on my papers when I am actually interested in what im writing about. Any help would be greatly apprectiated.
Should college athletes be paid or not, easy plenty of things to consider. Just take a stance. Do some research on how much some schools make on college games, you'd be shocked, think millions. I do not think they should be payed though, I don't have the breath to explain.
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Old 11-09-2008, 05:02 PM #96
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I'm trying to solve this circuit for Isupply, Vbc, and I1. The circuit I was given is drawn in black and anything I solved is in red... Idk if that is even right haha.

Any and all help appreciated.
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Old 11-09-2008, 08:12 PM #97
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A steel belted radial automobile tire is inflated to a gauge pressure of 30.0 lb/in^2 when the temperature is 61 F. Later in the day, the temperature rises to 100 F. Assuming that the volume of the tire remains constant, what is the tire's pressure at the elevated temperature? [hint: remember that the ideal gas law uses absolute pressure.]

The answer is 33.4 lb/in^2, but I haven't been able to get that.

Edit: I got it now. I wasn't subtracting 1 atm from the final pressure so it would read as the gauge pressure.
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Old 11-10-2008, 09:51 PM #98
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approximate the intergral(2x^2-7x+1 dx) using n=6 subdivisions and midpoints for sample points. a=0(bottom),b=2(top)

The answer is -20/3, I need an explanation of how to solve it in the above method,I've used the anti-derivative substitution and limit to find and confirm the answer so I just need an explanation of how to get there.
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Old 11-10-2008, 09:55 PM #99
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Looks like you're using Riemann sums. Lemme look up how to do those again.
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Old 11-10-2008, 09:58 PM #100
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Looks like you're using Riemann sums. Lemme look up how to do those again.
Yeah,thanks in advance
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Old 11-10-2008, 09:59 PM #101
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My damn laptop power cord is about to break.

Are you looking for the approximate integral from 0 to 2?
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Old 11-10-2008, 10:16 PM #102
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If it's from 0 to 2, you want to find ∆x, the length of each rectangle under the curve. The equation for this is ∆x=(b-a)/n.
b-a = 2-0
n=6
So, (2-0)/6 = 1/3.

Now, you want to imagine the function divided into 6 rectangles of equal length from 0 to 2, or each 1/3 units long. At the midpoint of each of these rectangles, you want to find the value of ƒ(x), so you know the height of the rectangle and can find the area of each rectangle and eventually the approximate area underneath the curve.

So you take R1, it stretches from 0 to 1/3, the midpoint is at (1/3)/2 = 1/6.
The midpoint of R2 is at 1/6+(∆x)=1/6+1/3=1/2
I guess the general equation for this would be R(N)=(N)*∆x-∆x/2=∆x(N-1/2). But I just made that up, so I could be wrong.

Now you find the value at each midpoint, and all these added up and multiplied by the length of each rectangle should give you the sum of the areas of the rectangles.
ƒ(x)=2(x^2)-7x+1
∆x[ƒ(1/6)+ƒ(1/2)+ƒ(5/6)+...+ƒ(11/6)]
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Old 11-11-2008, 09:11 AM #103
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Thanks,you're a lifesaver. It was a take-home quiz. I did two other methods of solving it and got the same answer. Thanks.
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Old 11-11-2008, 01:38 PM #104
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Hey Strawberry, although I don't need your help I think it's awesome you are helping people out. Great to know there are still good people out there.

My hat is off to you, sir.
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Old 11-11-2008, 02:17 PM #105
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Thanks a lot, I love to help when I can, and I love to help with basic calc. I would help that other guy up there, but I'm terrible at circuit analysis due to complete laziness during a full semester of electricity and magnetism.

Facepalm to me.
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