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Edexcel FP2 Official 2016 Exam Thread - 8th June 2016 watch

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    (Original post by economicss)
    You could use implicit differentiation but normally I would just square root to get plus minus r and go from there like usual, hope that helps
    How would you solve this question:
    Find the polar coordinates of the points on r²=a²sin2θ where the tangent is perp. to the initial line
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    (Original post by Patrick2810)
    I think you do this

    find x^2 or y^2, where x^2= r^2sin^2(theta) and y^2=r^2sin^2(theta)

    then you can differentiate implicitly, so for example you'd have 2y dy/d(theta) = f'(theta) => dy/d(theta) = f'(theta)/2y, and since it equals 0 the numerator must equal zero so you just put f'(theta)=0 and forget about the 2y
    Thanks ill try that
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    I can give a decent stab at them but would you say drawing polar graphs is fairly unlikely?
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    (Original post by Patrick2810)
    I think you do this

    find x^2 or y^2, where x^2= r^2sin^2(theta) and y^2=r^2sin^2(theta)

    then you can differentiate implicitly, so for example you'd have 2y dy/d(theta) = f'(theta) => dy/d(theta) = f'(theta)/2y, and since it equals 0 the numerator must equal zero so you just put f'(theta)=0 and forget about the 2y
    (Original post by Patrick2810)
    I think you do this

    find x^2 or y^2, where x^2= r^2sin^2(theta) and y^2=r^2sin^2(theta)

    then you can differentiate implicitly, so for example you'd have 2y dy/d(theta) = f'(theta) => dy/d(theta) = f'(theta)/2y, and since it equals 0 the numerator must equal zero so you just put f'(theta)=0 and forget about the 2y
    Actually i dont know if that works having tried that.
    The question is:Find the polar coordinates of the points on r²=a²sin2θ where the tangent is perp. to the initial line
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    (Original post by fpmaniac)
    How would you solve this question:
    Find the polar coordinates of the points on r²=a²sin2θ where the tangent is perp. to the initial line
    Hope this helps, sorry it's so messy! Didn't know what the range of theta values was so haven't written a final answer, let me know if you need any more help Name:  image.jpg
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    (Original post by economicss)
    Hope this helps, sorry it's so messy! Didn't know what the range of theta values was so haven't written a final answer, let me know if you need any more help Name:  image.jpg
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    Theyve got the answers for θ to be pi/6 and pi/2. Thats what you got but how do you know which values to choose. Also what did you do with r =,-a root (sin2θ). Thanks
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    (Original post by fpmaniac)
    Theyve got the answers for θ to be pi/6 and pi/2. Thats what you got but how do you know which values to choose. Also what did you do with r =,-a root (sin2θ). Thanks
    You would choose the values in the principal argument range normally so between -pi and pi! I think the negative value of r is ignored because r is always assumed to be positive in FP2 and is ignored otherwise, hope that helps
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    (Original post by Kvothe the arcane)
    I can give a decent stab at them but would you say drawing polar graphs is fairly unlikely?
    Haven't seen much of these, I would like to say that it's unlikely..
    I think its partially because if they get you to sketch it, there would be no way for them to ask to find the area of a certain shaded bit. Missing out on an opportunity to put a ~8 mark question in the paper
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    3 DAYS TO GO!

    Good luck all
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    (Original post by Rkai01)
    Attachment 543305
    Would you happen to know how to go about 11b. I feel dumb because it's a hence question but I can't integrate with the sqrt and by substitution becomes awkward.
    This is what I think you're meant to do since it asks to estimate. Name:  image.jpg
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    anyone know where to find complex transformation questions, ive done all the ones in the book and past papers but theyre usually the hardest part for me so i want to practice more
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    (Original post by SeanFM)
    Not easy to explain other than looking at what they've done.

    the 'middle term' for r = k is cancelled out by combining the 'last term' in r = k-1 and the 'first term' in r = k +1, so you work through up to r = 4 and you see that the only term at the start that isn't cancelling out is 1/2.

    Then you work towards the end (starting from n-2).

    So for r=n-2 everything cancels out (as you have the 'first term' below it and the 'last term' above it so the middle term is cancelled out, the 'first term' in r=n-2 is used up in cancelling out the previous 'middle term' and the last one is used in cancelling out the 'middle term' in r = n-1.



    Similar logic in r=n-1 for the first and second term, but the third term is only half of the middle term in r = n so it only 'half cancels' it.

    which is why you get 1/(n+1) - 2/(n+1) = -1/(n+1) as one of the terms in the summation.



    and then for r=n the 'middle term' half cancels (as before) and the first one is used up, and the last one does not do anything as there needs to be something below it.

    wow, thanks a lot!
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    Which IAL papers are available??

    Is it just
    june 14
    june 15
    jan 16
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    (Original post by AmarPatel98)
    Which IAL papers are available??

    Is it just
    june 14
    june 15
    jan 16
    I think its just june 14 and june 15
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    Hi guys, I don't understand question 6a of June 2014 R. I've looked at the mark scheme but I still don't get it. Could someone show me the working please? I would be extremely grateful

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    https://8dedc505ac3fba908c50836f5905...%20Edexcel.pdf

    Question 6b, so |z| = 3 maps to the circle and now we have to sketch |z| < 3 - if anything id thought itd be inside the circle but its every but inside. Could someone explain this please?
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    (Original post by imnoteinstein)
    https://8dedc505ac3fba908c50836f5905...%20Edexcel.pdf

    Question 6b, so |z| = 3 maps to the circle and now we have to sketch |z| < 3 - if anything id thought itd be inside the circle but its every but inside. Could someone explain this please?
    but what is the equation of the circle C? (after the transformation)
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    (Original post by ddbrain)
    Hi guys, I don't understand question 6a of June 2014 R. I've looked at the mark scheme but I still don't get it. Could someone show me the working please? I would be extremely grateful

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    is it y=1 ?
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    (Original post by ddbrain)
    Hi guys, I don't understand question 6a of June 2014 R. I've looked at the mark scheme but I still don't get it. Could someone show me the working please? I would be extremely grateful

    Name:  fp2 question.png
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    Hey! Which part don't you understand?

    Personally, I went with the first alternative in the mark scheme:
    1. Isolate z in the transformation
    2. Let z = x + iy, let w = u + iv, then let v = -1 as required
    3. Simplify down to have Re + Im = Re + Im on both sides
    4. Solve to find y (hint: on both sides of the equation, the real parts must be equal and the imaginary parts must be equal).
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    (Original post by target21859)
    This is what I think you're meant to do since it asks to estimate. Name:  image.jpg
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    No that's correct. Why did it shuffle down to that expression?
 
 
 
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