Edexcel FP2 June 2015 - Official Thread Watch

imyimy
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Please could someone help me with this? Thanks Name:  ImageUploadedByStudent Room1432799767.148823.jpg
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bobabob
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(Original post by mmms95)
omg this is me! i liked fp2 throughout the whole year (self-studying) and i understood everything, I only did the questions in the book though, and then i started past papers and i at first i was like ???!! why am i finding this difficult?? but then i kept doing a lot, and now i got the hang of it the more you'll do the better you'll get! also for the complex numbers, the transformations can be a bit tricky but once you get the hang of it it's fine!

if you need any help, don't be afraid to ask! good luck, it'll get better trust me!
Ok thanj you very much mate!

The only problem is that only about 5 days left )))
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crashMATHS
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(Original post by imyimy)
Please could someone help me with this? Thanks Name:  ImageUploadedByStudent Room1432799767.148823.jpg
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*Make z the subject
*Sub w= u + iv into your equation
*Multiply top and bottom by the complex conjugate of your denominator
*Let x = real part, y = imaginary part.

If x=y, make the real part equal to the imaginary part.

If x + y + 1 = 0, sub your values of x and y into here and then manipulate the algebra.




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jf1994
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Is FP2 generally harder than M2, D2 and S2? I can get A*'s in C3 and C4 but lucky to get a C in FP2...
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crashMATHS
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(Original post by jf1994)
Is FP2 generally harder than M2, D2 and S2? I can get A*'s in C3 and C4 but lucky to get a C in FP2...
It's hard until you start understanding it. I mean I picked up differential equations very quickly, finding them very simple, but complex transformations took a very long time.

I guess, putting the work in and practising A LOT is the key to FP2.


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TeeEm
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useful material for the upcoming exams

http://www.thestudentroom.co.uk/show....php?t=3353445

http://www.thestudentroom.co.uk/show....php?t=3361867

http://www.thestudentroom.co.uk/show....php?t=3361905
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jf1994
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For differentiating (dy/dx)^2dx:

Letting t = dy/dx, t^2dx

So 2dy/dx * d^2y/dx^2

Is this correct?
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crashMATHS
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(Original post by jf1994)
For differentiating (dy/dx)^2dx:

Letting t = dy/dx, t^2dx

So 2dy/dx * d^2y/dx^2

Is this correct?
Yes


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jf1994
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Thanks

What about if I want to differentiate dy/dx with respect to z?

Do we essentially just treat dy/dx type terms as any other term when using the chain and product rule?
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crashMATHS
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(Original post by jf1994)
Thanks

What about if I want to differentiate dy/dx with respect to z?

Do we essentially just treat dy/dx type terms as any other term when using the chain and product rule?
Yeah you would use implicit differentiation.

 \frac{dy}{dx} dz = \frac{d^2y}{dx^2} \frac{dx}{dz}

I think that's right - pretty sure


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toddle1
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Is there a method to follow that works for all complex numbers transformation questions? I really struggle with them
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crashMATHS
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(Original post by toddle1)
Is there a method to follow that works for all complex numbers transformation questions? I really struggle with them
Yes, I posted it earlier.

If you want to go from the z plane to the w plane, you make z the subject and use either the bisector or component method.

If you want to go from the w plane to the z plane, you make w the subject and use either the bisector or component method.


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jf1994
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(Original post by kingaaran)
Depends on what you're trying to achieve. If you are given two moduli, then you should sub x + iy in and apply Pythagoras, because it's well faster, but if not, then you can either use the component method or the perpendicular bisector method, depending on which you prefer.


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What's the perpendicular bisector method? :coma:
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crashMATHS
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(Original post by jf1994)
What's the perpendicular bisector method? :coma:
If you have something lying on the real axis, then you can either use the fact that y=0 or that the real axis is the perpendicular bisector of | z - i | = | z + i |.

So, when you have  z = \frac{2}{w} , you can either sub w = u + iv into there, make y=0 and so on, or you can sub z into | z - i | = | z + i | and rearrange.


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jf1994
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(Original post by kingaaran)
If you have something lying on the real axis, then you can either use the fact that y=0 or that the real axis is the perpendicular bisector of | z - i | = | z + i |.

So, when you have  z = \frac{2}{w} , you can either sub w = u + iv into there, make y=0 and so on, or you can sub z into | z - i | = | z + i | and rearrange.


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Didn't know this, thanks!
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jf1994
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Name:  de moivres problem.png
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Can anyone explain this to me? Part B, when it asks you to find the solutions, I equate the RHS of the equation obtained in part (a) to 5 times the RHS of the equation given at the start of part (b). Subtracted RHS, setting equal to zero, pulled sin theta out as a common factor, to get sin theta = 0 or sin theta = +/- the 4th root of 5/8. I discounted sin theta = - the 4th root of 5/8 because theta turns out to be negative and theta must be between 0 and 2pi, but the mark scheme says I shouldn't have done that.

In the end I got theta = 0, 1.10, 2.05 and pi, which were all correct, but I was missing 4.237 and 5.188
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jf1994
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Also

Name:  approximation.png
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Part c, how do you go about finding approximation values for something like that?
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math42
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(Original post by jf1994)
Name:  de moivres problem.png
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Can anyone explain this to me? Part B, when it asks you to find the solutions, I equate the RHS of the equation obtained in part (a) to 5 times the RHS of the equation given at the start of part (b). Subtracted RHS, setting equal to zero, pulled sin theta out as a common factor, to get sin theta = 0 or sin theta = +/- the 4th root of 5/8. I discounted sin theta = - the 4th root of 5/8 because theta turns out to be negative and theta must be between 0 and 2pi, but the mark scheme says I shouldn't have done that.

In the end I got theta = 0, 1.10, 2.05 and pi, which were all correct, but I was missing 4.237 and 5.188
Although arcsin(-(5/8)^(1/4)) will give you a negative answer, there are still values within the range 0 to 2pi which will give you -(5/8)^(1/4). Try drawing the sin graph or using a cast diagram
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Maths degree
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everything you done correct up until u discounted the - 4th root of 5/8. as yes thats not in the region however that also means on the cast diagram theta can also be in the cos quadrant and tan quadrant, due to you also getting the negative thetha and thats where your 4.237 and 5.188 come from. hope this helps!
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Maths degree
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(Original post by jf1994)
Name:  de moivres problem.png
Views: 151
Size:  24.9 KB

Can anyone explain this to me? Part B, when it asks you to find the solutions, I equate the RHS of the equation obtained in part (a) to 5 times the RHS of the equation given at the start of part (b). Subtracted RHS, setting equal to zero, pulled sin theta out as a common factor, to get sin theta = 0 or sin theta = +/- the 4th root of 5/8. I discounted sin theta = - the 4th root of 5/8 because theta turns out to be negative and theta must be between 0 and 2pi, but the mark scheme says I shouldn't have done that.

In the end I got theta = 0, 1.10, 2.05 and pi, which were all correct, but I was missing 4.237 and 5.188
everything you done correct up until u discounted the - 4th root of 5/8. as yes thats not in the region however that also means on the cast diagram theta can also be in the cos quadrant and tan quadrant, due to you also getting the negative thetha and thats where your 4.237 and 5.188 come from. hope this helps
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