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

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    (Original post by PhysicsIP2016)
    Work out the area of the triangle between OA and the line theta=pi/2. You would then use polar coordinate integration of the curve between the value of theta you got for the coordinates of A (from part a) and pi/2, then subtract this from the area of the triangle.
    Thank you!!
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    (Original post by SB0073)
    Try find the area of the triangle from OA(to the pi/2 line).

    Then integrate from A to the initial line to find the remaining area.

    Then add the two areas? Try that and let me know if it works :P

    Whoops misread the Q thought it was finding the total area!
    Thank you, got it
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    (Original post by AakashG)
    Hi, what book is that from?
    Further Pure 2 by Mark Rowland, my teacher gave it to me
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    https://8fd9eafbb84fdb32c73d8e44d980...lZweHc/CH3.pdf

    ExH 14 b I'm just checking they made a error where they equated real components it Should have been 4 and not 2?

    Don't get why they making so many errors I've seen atleast 4
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    (Original post by Rkai01)
    https://8fd9eafbb84fdb32c73d8e44d980...lZweHc/CH3.pdf

    ExH 14 b I'm just checking they made a error where they equated real components it Should have been 4 and not 2?

    Don't get why they making so many errors I've seen atleast 4
    Yes, it's an error.
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    (Original post by A Slice of Pi)
    The final answer is describing the equation of the line. The value of sin theta found in the first few lines is for where the line meets the curve. Points on the actual line have different values of r and theta, so you can't say that theta is constant.
    So is y=rsintheta any point on the graph because I thought it was for when it lies on the curve or on a line.
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    Q 4 b ex I we aren't given a range for theta so how would we know the solutions as they could be infinite ? The actual question isn't a problem though.

    https://8fd9eafbb84fdb32c73d8e44d980...lZweHc/CH3.pdf
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    (Original post by target21859)
    So is y=rsintheta any point on the graph because I thought it was for when it lies on the curve or on a line.
    It's just for a general point in polar coordinates. But that general point can be anywhere so theta can vary
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    is there a 2016 jan ial paper for fp2?
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    (Original post by 1asdfghjkl1)
    is there a 2016 jan ial paper for fp2?
    No.
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    (Original post by A Slice of Pi)
    It's just for a general point in polar coordinates. But that general point can be anywhere so theta can vary
    Thanks. It makes sense now. It's just using Pythagoras. So you're working out y on the line to plug it into y=rsintheta just like you would find c in y=mx+c, right?
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    Are there any other good resources for revision apart from the book questions and the edexcel past papers?
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    (Original post by Katiee224)
    Are there any other good resources for revision apart from the book questions and the edexcel past papers?
    http://www.madasmaths.com/archive_ma...s_various.html
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    (Original post by Katiee224)
    Are there any other good resources for revision apart from the book questions and the edexcel past papers?
    https://a4bcd373d3f1cf86182e63b18df2...P2%20Notes.pdf
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    Thanks i'l get stuck into those further complex numbers questions, why are the solutions so small though, i have to zoom the page right in :laugh:
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    (Original post by Katiee224)
    Thanks i'l get stuck into those further complex numbers questions, why are the solutions so small though, i have to zoom the page right in :laugh:
    Yeah that's good ol' TeeEm's style!
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    (Original post by target21859)
    Thanks. It makes sense now. It's just using Pythagoras. So you're working out y on the line to plug it into y=rsintheta just like you would find c in y=mx+c, right?
    Yeah I think so. From the graph it's clear that the equation must be y = 'constant' in Cartesian form. Then just use the definition of y in polars to get the equation into polar form. It's important to remember that you can 'superimpose' the Cartesian x and y axes on the polar diagram, with the pole being the origin, and then it becomes clear why x = r cos theta and y = r sin theta by simple trigonometry.
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    (Original post by Craig1998)
    They have used the formula for the area printed in the formula booklet.

    Area = \frac 12 \int_{\beta}^\alpha r^2 d\theta

    They have used this for each sector individually, and have split some of the sectors up so they do not cancel out. This is why the areas have been doubled, as, with polar coordinates, the graphs are very often symmetrical. The area for r= 2 + \cos\theta is calculated by squaring r and then dividing the total area by 2 (which cancels with the half from the formula). A similar thing is then done with r = 5\cos\theta.

    The angles for r = 2 + \cos\theta are \frac\pi3 and as the intersection at the top is at an angle of \frac\pi3 and this refers to the red curve in the diagram. The angles for r = 5\cos\theta are \frac\pi3 and \frac\pi2 as the curve goes from an angle of \frac\pi3 up to the angle of \frac\pi2. This is then doubled for the angle at the bottom

    If you want me to go into any more detail please ask, what follows on from here is relatively easy integration (the more you practice this style of integration, the more second nature it becomes), the double angle formula for \cos2\theta is often needed, where the 2 may be changed for another value.

    This is my first time using latex by the way .
    Thanks.
    I also got a couple more questions if you or anyone else wont mind helping me out.
    From the book I dont understand example 29b and example 30 and example 31 on pages 47 and 48.
    I also dont understand question 6 june 2011. I dont get the mark scheme.
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    (Original post by fpmaniac)
    Thanks.
    I also got a couple more questions if you or anyone else wont mind helping me out.
    From the book I dont understand example 29b and example 30 and example 31 on pages 47 and 48.
    I also dont understand question 6 june 2011. I dont get the mark scheme.
    I'm not sure about 29b, all they seem to have done is found an equation of the circle to which the locus is a major arc. I've never seen a question like that, maybe somebody else can help..

    For example 30, you have a point at (0, 0) and a point at (0, 4i), these will form the points of your arc. This question is unique as the angle given is \frac\pi2 which means that there will be a right angle between each point and your point P, which leads to your arc just being a semi circle. Therefore, it would lead you to draw a semi circle from (0, 0) to (0, 4i) or vice versa, where your point P forms a right angle. I hope this explains it.

    For example 31, the minimum and maximum points of |z| are the points which are nearest and furthest from the origin. As your circle is essentially given (radius 3 and centre (12, 5)), and the distance from the origin to the centre can be worked out using pythagoras (\sqrt{12^2 + 5^2} = 12, you can add 3 and subtract 3 (the radius) to find the minimum and maximum points.
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    guys has anyone come across a past paper question similar to example 13 from chapter 7?
 
 
 
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