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[Skyrunner]

If I set up a birthdate survey on bay12 and have maybe 50 people reply, would I really see at least one pair of matching birthdates? O.o I want to try it out.... I suppose I could use Google Forms.

[Tibbz2]

I'm not having a good time with A2 maths (C3, C4, M2) at the moment

[MagmaMcFry]

I'm not having a good time with A2 maths (C3, C4, M2) at the moment

Would you mind being slightly more specific? I assume you came here to get help, right?

[Tibbz2]

I'm not having a good time with A2 maths (C3, C4, M2) at the moment

Would you mind being slightly more specific? I assume you came here to get help, right?

I'm just generally stuck on understanding some of the stuff and was wondering if there was something I could do to help me understand some of the concepts, rather than just practicing stuff I don't understand then forgetting it?

[da_nang]

If I set up a birthdate survey on bay12 and have maybe 50 people reply, would I really see at least one pair of matching birthdates? O.o I want to try it out.... I suppose I could use Google Forms.

~97 % chance, yes.

[MagmaMcFry]

I'm not having a good time with A2 maths (C3, C4, M2) at the moment

Would you mind being slightly more specific? I assume you came here to get help, right?

I'm just generally stuck on understanding some of the stuff and was wondering if there was something I could do to help me understand some of the concepts, rather than just practicing stuff I don't understand then forgetting it?

There is stuff you can do to help you understand some of the concepts. Either you simply Google for those concepts, or you tell us the names of the concepts (as well as some example problems using it) and we'll try our best to help you understand it.

[Tibbz2]

I'm not having a good time with A2 maths (C3, C4, M2) at the moment

Would you mind being slightly more specific? I assume you came here to get help, right?

I'm just generally stuck on understanding some of the stuff and was wondering if there was something I could do to help me understand some of the concepts, rather than just practicing stuff I don't understand then forgetting it?

There is stuff you can do to help you understand some of the concepts. Either you simply Google for those concepts, or you tell us the names of the concepts (as well as some example problems using it) and we'll try our best to help you understand it.

I've been trying to go through my textbooks/class notes and using a maths site so far but I'm still not 100% there. I'll keep going through all the content we've covered and try some more, and I'll come back and ask if there is still stuff I don't fully understand.

Cheerd

[Pnx]

Ok, so I'm really exhausted right now and about to pass out, but if someone could help me find a solution to this problem while I'm snoozing it would really help me.

I have to find the arc length of x=(3/5)y^5/3-(3/4)y^1/3 from y=1 to y=8. I know about using the sqrt(1+(dy/dx)²) formula, but I'm having trouble integrating what I get from that. Does anyone think they can help?

[Karlito]

Well, the first thing to recognize is that x=f(y), rather than the other way around like it usually is. We'll integrate sqrt(1+(dx/dy)²) with respect to y to find the answer.

dx/dy = y^2/3 - (1/4)y^-2/3 is pretty easy to calculate

The next thing is just simplifying the integral. We'll start with the bit inside the radical.
1 + (dx/dy)²
1 + (y^2/3 - (1/4)y^-2/3)²
1 + y^4/3 - 1/2 + y^-4/3
y^4/3 + 1/2 + y^-4/3          This looks almost exactly like (dx/dy)², and in fact, it's also a perfect square
 (y^2/3 + (1/4)y^-2/3)²

Remember this whole thing is under a square root sign, which cancels out with the square, leaving you with a fairly simple polynomial to integrate (I'll leave the actual legwork to you). It's positive over the relevant range too, so you don't even have to worry about complications from the absolute value you'd usually get with such an operation.

[Pnx]

Thanks a bunch for that.

I really should have seen the perfect square going on.

[Karlito]

Hey, I'm trying to help my brother with his math homework. What's the best way to solve something like x³=8 for all complex roots without using de Moivre's theorem? He hasn't even learned the unit circle yet.

[Skyrunner]

x³-8 = 0
(x-2)(x²+2x+4) = 0

It's a formula:

(a-b)(a²+ab+b²) = a³-b³

de Moivre's theorem is a tad overkill, you know Heck, -I- don't know what that is either.

[Karlito]

Oooh right, I remember that thing.

[vagel7]

So guys. Do you have any tips or so on how to learn to solve word problems? I have checked the internet for tips and so, but everything is like pre-school level and really doesn't help me. I can understand the texts fine but I have a problem modelling the right equation.

This is like the only thing that I have a problem with in maths, everything is comes easily and mechanically.

(I think the problems that I have to do are possibly equal to A levels or so)

[Helgoland]

Depending on how severe your troubles with word problems are, you may want to try just mapping out the whole thing - all the variables, how they are connected, what you know, what you want to calculate. Mind-map style. With colors, possibly.