Edexcel FP2 June 2015 - Official Thread Watch

V0ldemort17
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#21
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#21
Urgh yeah that chapter is really doing my head in


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john122334455
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#22
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Polar coordinates really difficult, if someone can briefly explain it here, itl be really helpful thanks
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john122334455
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#23
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So im attempting this question.

Part a, when simplified i get: 4r-4 or 4(r-1)

However in B, my most simplified version comes as: 2n(n+3) after using methods of differences. Which does not match with the show that answer please help.
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Abel Demoz
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#24
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#24
(Original post by john122334455)
So im attempting this question.

Part a, when simplified i get: 4r-4 or 4(r-1)

However in B, my most simplified version comes as: 2n(n+3) after using methods of differences. Which does not match with the show that answer please help.
Part a simplified is 4(r+1) not 4(r-1)
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simonli2575
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#25
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I'm aiming for 100%. Who's with me?
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crashMATHS
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#26
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(Original post by simonli2575)
I'm aiming for 100%. Who's with me?
Me too!


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simonli2575
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#27
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#27
(Original post by kingaaran)
Me too!


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Quick! How would you expand cos^n(x) and cos(nx)?


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crashMATHS
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#28
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(Original post by simonli2575)
Quick! How would you expand cos^n(x) and cos(nx)?


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If I understand what you're asking...:

With the first one, you would use the fact that  z + \frac{1}{z} = 2 cosx. Whatever your value of n is, you would raise both sides of the equation to that power, and binomially expand the first part, collect the like terms and simplify. That way, your cos^n(x) will now be expressed in terms of cos(nx).

With the latter, you would use de Moivre's Theorem to rewrite cos(nx) as [cos(x) + isin(x)]^n, and binomially expand that. Then, you'd equate the real terms, if you wanted it in terms of cos(x), or you'd equate the imaginary terms if you wanted it in terms of sin(x).

(If you're asking in terms of Taylor series, then you'd use the standard result for cos(x) and replace all of the x's with (nx), when expanding the second one. And you would expand the first using Taylor's series by writing cos^n(x) as [cos(x)]^n and calculating the derivatives of that up to what you want to expand to and substituting it in.)

Hope it helps and I haven't confused you too much, lol.
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crashMATHS
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#29
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#29
I have a suggestion: why don't we create a Skype group revision session, revise FP2 content together and do practice questions, so we can help each other, hone our skills and (ideally) ace* the exam?

If anybody is interested, let me know :

*ace = full marks


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simonli2575
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#30
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#30
(Original post by kingaaran)
If I understand what you're asking...:

With the first one, you would use the fact that  z + \frac{1}{z} = 2 cosx. Whatever your value of n is, you would raise both sides of the equation to that power, and binomially expand the first part, collect the like terms and simplify. That way, your cos^n(x) will now be expressed in terms of cos(nx).

With the latter, you would use de Moivre's Theorem to rewrite cos(nx) as [cos(x) + isin(x)]^n, and binomially expand that. Then, you'd equate the real terms, if you wanted it in terms of cos(x), or you'd equate the imaginary terms if you wanted it in terms of sin(x).

(If you're asking in terms of Taylor series, then you'd use the standard result for cos(x) and replace all of the x's with (nx), when expanding the second one. And you would expand the first using Taylor's series by writing cos^n(x) as [cos(x)]^n and calculating the derivatives of that up to what you want to expand to and substituting it in.)

Hope it helps and I haven't confused you too much, lol.
Thanks, I almost forgot about the  z + \frac{1}{z} = 2 cosx thing.
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FN510
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#31
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How do you know when to add x to your particular integral?
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simonli2575
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#32
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(Original post by FN510)
How do you know when to add x to your particular integral?
If f(x) = ax+b, then the PI will be in the form of cx+d
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FN510
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#33
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(Original post by simonli2575)
If f(x) = ax+b, then the PI will be in the form of cx+d
Cheers I understand that, I was referring to when you have to multiply PI by x in some cases or x^2
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crashMATHS
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#34
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(Original post by FN510)
Cheers I understand that, I was referring to when you have to multiply PI by x in some cases or x^2
I had this confusion. You have to multiply your particular integral by x if your complementary function contains the term you were going to use as your particular integral.


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FN510
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#35
(Original post by kingaaran)
I had this confusion. You have to multiply your particular integral by x if your complementary function contains the term you were going to use as your particular integral.

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simonli2575
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#36
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(Original post by kingaaran)
I had this confusion. You have to multiply your particular integral by x if your complementary function contains the term you were going to use as your particular integral.


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Sometimes when b^2=4ac, you have to multiply it by x^2.
Does it always work though? And why does it work?


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imyimy
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#37
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Hey, we just did a mock in fp2 which was the june 14 paper and I got 69/75 which I was happy with, but that is mainly due to the fact that I could do the polar coordinate question fine, as it was just the area under the graph between 2 points on a curve. however in some you have to calculate the area under a tangent such as a triangle shape in polar coordinates and I don't know how to do this, I can find the area under the graph by using 1/2 r^2 but I don't know how you would do this for just a basic shape. Thanks.


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crashMATHS
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#38
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#38
(Original post by simonli2575)
Sometimes when b^2=4ac, you have to multiply it by x^2.
Does it always work though? And why does it work?


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You have to multiply what by x^2 - your particular integral?

That would depend on the differential equation, not the nature of the discriminant of your auxiliary equation.

You multiply by x when your complementary function contains what you would have otherwise used as your particular integral. If by multiplying by x, you would still get a particular integral that's part of the complementary function, you multiply by x^2 instead.

Hopefully, I make sense, haha!



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crashMATHS
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#39
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(Original post by imyimy)
Hey, we just did a mock in fp2 which was the june 14 paper and I got 69/75 which I was happy with, but that is mainly due to the fact that I could do the polar coordinate question fine, as it was just the area under the graph between 2 points on a curve. however in some you have to calculate the area under a tangent such as a triangle shape in polar coordinates and I don't know how to do this, I can find the area under the graph by using 1/2 r^2 but I don't know how you would do this for just a basic shape. Thanks.


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That's good, well done!

Do you have an example? Maybe I might be able to help


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imyimy
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(Original post by kingaaran)
That's good, well done!

Do you have an example? Maybe I might be able to help


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thanks! and something like this

http://www.examsolutions.net/a-level...estions/q6.gif

like if you had to find the area of the triangle and something else, in cartesian coord systems you'd just do 1/2 base x height or similar, but I'm not sure what you would do in polar. thanks!
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