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Atomic Boo

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C4.

 

Ahhh, i'm only on Core 2 at the mo. I'm finding it abit on the hard side.

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Just you wait :)

 

Remember sin and cos? Well you learn to differentiate and intergrate them, and e and ln. Oh the joy you'll have.

 

Wheres supergrunch, and everyone else who usually laugh at my easy maths questions when you need them!

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Just saw that thread - if I have time I can help with maths. I finished all maths courses on university I need but one. I am also a bit into electronics ...

 

So analysis, linear algebra, discrete mathematics, probability, statistic, stochastic, differential equations... are okay with me. Just don't expect any miracles.

 

 

If someone likes try and help me with:

 

(2x + x^2*y − y^3)dx + (x*y^2 − x^3 + 2y)dy = 0 solve it with a integrating factor of the form mu(x*y).

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C4.

 

Its so annoying, I'm sure I've made a silly mistake. I was hoping to have it figured out for when I take it in tomorrow but I just can't see where I've gone wrong, nor understand the fish.

 

Also fish, when you said (t+1)(t+1) at the bottom = T^2+1, thats wrong. Its t^2+ 2t +1.

 

Scrap what I said, looking back at it now, it's all bullshit.

 

Oddly, I did a similar question about 10 minutes ago on a C4 practice paper, and got it perfectly right... I hate being inconsistant...

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Just saw that thread - if I have time I can help with maths. I finished all maths courses on university I need but one. I am also a bit into electronics ...

 

So analysis, linear algebra, discrete mathematics, probability, statistic, stochastic, differential equations... are okay with me. Just don't expect any miracles.

 

 

If someone likes try and help me with:

 

(2x + x^2*y − y^3)dx + (x*y^2 − x^3 + 2y)dy = 0 solve it with a integrating factor of the form mu(x*y).

 

Either this is a differential equation or I have no idea how to solve it. If it's the former, write it like this:

 

(2x + x^2*y − y^3)dx=-(x*y^2 − x^3 + 2y)dy

 

and Integrate both sides.

 

 

Haven't seen you for quite some time btw. How's everything going?

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'Why is a client server more economically viable than a peer to peer network??'

 

Its a 3 mark question, but I can only think of one point:

 

Because client server's share the use of expensive hardware, where as peer to peer needs several systems.

 

:s

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x= t-1/t y=t+1/t

 

show dy/dx= (t-1)(t+1)/t^2+1

 

 

This is really annoying me, because we've done the problem essentially in a maths exercise before (except y and x other way round) but I can't get the final answer, because I differentiate y to get dy/dt= -t^-2+1

 

which means I have a negative which I don't want. Can someone help me please :) thanks.

Ah, jolly parametrics.

 

well, dy/dx = (dy/dt)*(dt/dx) = (dy/dt)/(dx/dt)

 

dy/dt = 1 - 1/(t^2)

 

dx/dt = 1 + 1/(t^2)

 

thus:

 

dy/dx = {1 - 1/(t^2)}/{1 + 1/(t^2)}

dy/dx = {t^2 - 1}/{t^2 + 1}

dy/dx = (t+1)(t-1)/(t^2 + 1)

 

(using difference of two squares in that final line)

 

Just saw that thread - if I have time I can help with maths. I finished all maths courses on university I need but one. I am also a bit into electronics ...

 

So analysis, linear algebra, discrete mathematics, probability, statistic, stochastic, differential equations... are okay with me. Just don't expect any miracles.

 

 

If someone likes try and help me with:

 

(2x + x^2*y − y^3)dx + (x*y^2 − x^3 + 2y)dy = 0 solve it with a integrating factor of the form mu(x*y).

Either this is a differential equation or I have no idea how to solve it. If it's the former, write it like this:

 

(2x + x^2*y − y^3)dx=-(x*y^2 − x^3 + 2y)dy

 

and Integrate both sides.

 

 

Haven't seen you for quite some time btw. How's everything going?

It is a differential equation, but separation of variables isn't going to work and the question specifies use of the integrating factor method. Separation of variables is by no means the only way to solve differential equations; in fact it's merely the tip of the iceberg.

 

Anyway, I'm afraid your equation is too evil for me to solve...

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dy/dx = {1 - 1/(t^2)}/{1 + 1/(t^2)}

 

Its the working here which I needed =]

dy/dx = {t^2 - 1}/{t^2 + 1}

dy/dx = (t+1)(t-1)/(t^2 + 1)

 

 

Well I asked someone and they just did what you did mr supergrunch, but I said thats basically fudging it because the questions gives you the answer and you have to reach it. Your answer isn't wrong of course, but for the marks I needed to show each step, and I forgot to make it all one fraction on the top and all one on the bottom. Its so clear now!

 

Thanks guys :)

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Between those two steps I simply multiplied top and bottom by t^2.

 

You have to be careful in questions that give you the answer, because you need to make sure the examiner knows that you genuinely know how to reach the answer. This means it's generally best to try and work out the answer, then compare it with the one given in the question. However, provided your steps don't make too large jumps (i.e. you are able to reasonably do them in your head), then your working should be fine. This can be a problem if you like to do loads in your head, because then you have to remember to write steps that the examiner will believe. But means you're less likely to make errors, anyway.

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I have to make an E-portflio for ICT GCSE and I was thinking of some tips for how to use dreamweaver (8).

Ive learnt how to link, add backgrounds but I wondered if anyone here new some slightly more advanced techniques like Flash buttons and picture links etc (both I can do, but I was wondering if there were any others).

Maybe something like a drop-down selection bar?

 

(quoted from me in another thread)

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Either this is a differential equation or I have no idea how to solve it. If it's the former, write it like this:

 

(2x + x^2*y − y^3)dx=-(x*y^2 − x^3 + 2y)dy

 

and Integrate both sides.

 

 

Haven't seen you for quite some time btw. How's everything going?

 

I am mostly fine ... some minor drawbacks related to my studies ;-)

 

btw. I solved. In the end it was easier than I thought.

 

Now I am all over Riccati differential equations - so be prepared everyone!

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FP3 is a real bitch. Going through the complex numbers chapter at the moment, it's annoying when you're expected to know about hyperbolics when we haven't done FP2 (our school decided that it would be better for us to do FP3 instead of FP2). But meh, I should be more concerned with the vectors, they just pass straight over my head.

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FP3 is a real bitch. Going through the complex numbers chapter at the moment, it's annoying when you're expected to know about hyperbolics when we haven't done FP2 (our school decided that it would be better for us to do FP3 instead of FP2). But meh, I should be more concerned with the vectors, they just pass straight over my head.

FP3 is win, though I didn't take the exam and only learnt the juicy bits like group theory. Even so, the idea of doing it without first doing FP2 seems slightly insane to me... but then again, it sounds like I was on a different board to you. (OCR MEI)

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FP3 is win, though I didn't take the exam and only learnt the juicy bits like group theory. Even so, the idea of doing it without first doing FP2 seems slightly insane to me... but then again, it sounds like I was on a different board to you. (OCR MEI)

 

Damn, no group theory in the Edexcel syllabus. Guess I'll have to wait til uni. We're doing FP3 because I think it follows on better from FP1 and C4.

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Damn, no group theory in the Edexcel syllabus. Guess I'll have to wait til uni. We're doing FP3 because I think it follows on better from FP1 and C4.

 

Group theory is not win. Not as far as symmetry goes anyway. I'm sick of rotating octahedral basis sets in my head.

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Group theory is not win. Not as far as symmetry goes anyway. I'm sick of rotating octahedral basis sets in my head.

It's more fun than curly arrows... :wink:

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Just reading through this maths talk makes me think..."Why...WHY?!!"

 

To me it's always been one of those subjects that seems to have no practical use, just jargon (besides basic sums).

 

If a man came up to me and said "I have 2 apples, ? melons, totalling 5 fruits. How many melons??" - I'd probably kick him in the nuts and run for it!

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To me it's always been one of those subjects that seems to have no practical use, just jargon (besides basic sums).

 

I'm trying so hard right now not to take the bait...

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I'm trying so hard right now not to take the bait...

To be honest, applications of maths annoy me, and I'd rather it stayed protected from the nasty world in a little glass tower. (sorry, I know you're a physicist and all... :heh:) But then I think that about all subjects.

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Anyone do Higher Latin?

 

Comment on the following phrases and explain the effect you think that Cicero was trying to achieve by his use of these words.

 

(i) unus plures (line 34 - full sentence: unus plures praedonum duces vivos cepit P. Servilius quam omnes antea.)

(ii) hoc tu quam ob rem non fecisti? (line 44)

(iii) per truimphum, credo (line 50 - full sentence - per truimphum, credo, quem ante currum tuum duceres)

 

6 marks

 

 

I just hate the way the questions are phrased, because I know the jist of the story, and know what the words mean, but I can't explain why he used said words.

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If anyone can help me with some mathmatical problems; cheers.

 

1st Q: the surface area of a sphere is 1380cm^2. Calculate the voume of the sphere. I have the equations:

 

Voume of a sphere = 4/3(pi)r^3

Surface area of a sphere= 4(pi)r^2

 

I'm not sure if my ordering is correct when I work it out backwards, so I would be grateful if someone would work it out so I can be sure my method is right. I got 4826cm^3 btw.

 

Also; solve the equation:

 

3/y+7-1/y+6 = 1/2. I got about -0.4 or 7.6, is that correct? Thanks.

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If anyone can help me with some mathmatical problems; cheers.

 

1st Q: the surface area of a sphere is 1380cm^2. Calculate the voume of the sphere. I have the equations:

 

Voume of a sphere = 4/3(pi)r^3

Surface area of a sphere= 4(pi)r^2

 

I'm not sure if my ordering is correct when I work it out backwards, so I would be grateful if someone would work it out so I can be sure my method is right. I got 4826cm^3 btw.

 

Also; solve the equation:

 

3/y+7-1/y+6 = 1/2. I got about -0.4 or 7.6, is that correct? Thanks.

Just think about them logically, and you'll know whether you're right or wrong.

 

1. A = 4(pi)r^2 = 1380

r^2 = 345/(pi)

r = sqrt{345/(pi)}

 

V = 4/3(pi)r^3 = 460sqrt{345/(pi)}

= 4820 cm^3 (3sf)

 

2. Needs to be clearer in terms of notation. If you mean:

 

3/(y+7) - 1/(y+6) = 1/2

{3(y+6)-y-7}/((y+6)(y+7)) = 1/2

(y+6)(y+7) = 2{3(y+6)-y-7}

y^2 + 13y + 42 = 4y + 22

y^2 + 9y + 20 = 0

(y+4)(y+5) = 0

y = -4, -5

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Thanks alot. I've realised for the second question I multplied everything by 2 to make it easier, but didn't multiply the second fraction and left it as 1/ instead of 2/ which mucked me up. Silly mistake! :smile:

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Say I want to measure the power of a PV Cell, will this circuit work?

 

circuit.jpg

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Say I want to measure the power of a PV Cell, will this circuit work?

 

circuit.jpg

 

I think so but it's ages since I've done electicity.

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