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    I'm doing m3. I've self taught it and there is no one to help me. If anyone finds anything hard or has any advice, please share as I would really appreciate any help or advice.
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    (Original post by JenniS)
    M3 june 2012 paper, I can't work out how to find any of the angles in question 4b ?
    That paper was the hardest I've done, by faaaar. Question 6b was very confusing. How likely are we to get something like that????
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    (Original post by LShirley95)
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    Angle OVA and OVQ add up to be the 45 degree angle you're given. Q being the C.o.M.
    I still don't get it.... :/
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    (Original post by mashmammad)
    That paper was the hardest I've done, by faaaar. Question 6b was very confusing. How likely are we to get something like that????
    6a is bad enough, I didn't know how to f****** integrate a triangle
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    (Original post by JenniS)
    I still don't get it.... :/
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    So the weirdly shaped thing hangs down so the line VQ is vertical, yeah? [I'm assuming you've found the location of the centre of mass as co-ordinates in k and a where O or V is the origin and the x and y directions are OB and OV? This is what you would usually do with a question like this so yeah.] You can find the angle OVQ as tan(OVQ) = x/y if you happened to use V as your origin or tan(OVQ) = x/(2a - y) if you used O. You can find the angle OVA as obviously tan(OVA) = a/2a. You are given that AV hangs 45deg from the vertical, and the vertical is VQ. Therefore AVQ = 45deg and thus OVQ + OVA = 45deg, solve for k.
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    (Original post by LShirley95)
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    So the weirdly shaped thing hangs down so the line VQ is vertical, yeah? [I'm assuming you've found the location of the centre of mass as co-ordinates in k and a where O or V is the origin and the x and y directions are OB and OV? This is what you would usually do with a question like this so yeah.] You can find the angle OVQ as tan(OVQ) = x/y if you happened to use V as your origin or tan(OVQ) = x/((sqrt3)a - y) if you used O. You can find the angle OVA as obviously sin(OVA) = a/2a. You are given that AV hangs 45deg from the vertical, and the vertical is VQ. Therefore AVQ = 45deg and thus OVQ + OVA = 45deg, solve for k.
    how do you find what I think you've called x for the tan equation?
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    (Original post by JenniS)
    how do you find what I think you've called x for the tan equation?
    (M+kM)x = kMa I believe. If you can find the mark scheme they did it a different way with moments that may make more sense to you, I tend to do things in funny ways...
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    Hello! Just wondering for M3 - which of the standard results for centres of masses of 3D objects do we need to be able to prove?

    I know one of the past paper questions asked to prove the result for a solid cone, but some of the proofs in the textbook have labels saying "You will not be required to prove this result in your exam" or simply have nonsensical tidbits such as "This result is in the formula book, but you should learn it." which as you can see, is both self contradictory and does not specify what "it" actually is. Is it the result or the proof? And if the result is in the formula book, why should one learn it? (No really, that isn't a rhetorical question.)

    Thanks in advance for your help! It's much appreciated.
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    (Original post by MathsForFun)
    Hello! Just wondering for M3 - which of the standard results for centres of masses of 3D objects do we need to be able to prove?

    I know one of the past paper questions asked to prove the result for a solid cone, but some of the proofs in the textbook have labels saying "You will not be required to prove this result in your exam" or simply have nonsensical tidbits such as "This result is in the formula book, but you should learn it." which as you can see, is both self contradictory and does not specify what "it" actually is. Is it the result or the proof? And if the result is in the formula book, why should one learn it? (No really, that isn't a rhetorical question.)

    Thanks in advance for your help! It's much appreciated.
    my teacher advised that if it comes up in the book, it's possible it could be in the exam. I think we may need to know all proofs apart from the hemisphere shell and the arc of a cirle :O
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    please help, how do you answer part b?Name:  WP_000752.jpg
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    (Original post by Elinor.m)
    please help, how do you answer part b?Name:  WP_000752.jpg
Views: 126
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    Name:  Snap_2013.06.08_20h38m51s_008.png
Views: 143
Size:  19.5 KB

    I was just doing the same question! If you can help me by explaining why angle CBD is the same as the one at the vertex, then I can explain the rest. But I can't seem to figure out why those angles are the same...?
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    (Original post by Tanmayee)
    Name:  Snap_2013.06.08_20h38m51s_008.png
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    I was just doing the same question! If you can help me by explaining why angle CBD is the same as the one at the vertex, then I can explain the rest. But I can't seem to figure out why those angles are the same...?
    I had the same problem just now, but if you think of it as at the point of moving then they have to be the same angle because BD has to be vertical. I think hehe
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    (Original post by Tanmayee)
    Name:  Snap_2013.06.08_20h38m51s_008.png
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    I was just doing the same question! If you can help me by explaining why angle CBD is the same as the one at the vertex, then I can explain the rest. But I can't seem to figure out why those angles are the same...?
    Though feel free to explain! lol Im a bit confused about the rest of it and I don't really get the solution bank answer
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    (Original post by MathsForFun)
    I had the same problem just now, but if you think of it as at the point of moving then they have to be the same angle because BD has to be vertical. I think hehe
    Oh! Silly me, I forgot all about BD being vertical.

    (Original post by Elinor.m)
    please help, how do you answer part b?Name:  WP_000752.jpg
Views: 126
Size:  70.8 KB
    If you look at the diagram I posted above, the line BD must go through OB for the toy to stay in equilibrium (same idea as in Ex5C in the textbook). The diagram shows the point before it topples, and for the conditions in the diagram, CD is (r/3). You can work that out using the fact that the angles are the same. For the toy to move, CD needs to be greater than (r/3) so that BD no longer goes through OB. CD is the value that you worked out in part a, so if you substitute that into CD > (r/3), you get [3(M-2m)] / [8(M+m)] > (r/3) and once you simplify this, you should get M > 26m.
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    (Original post by Tanmayee)
    Oh! Silly me, I forgot all about BD being vertical.



    If you look at the diagram I posted above, the line BD must go through OB for the toy to stay in equilibrium (same idea as in Ex5C in the textbook). The diagram shows the point before it topples, and for the conditions in the diagram, CD is (r/3). You can work that out using the fact that the angles are the same. For the toy to move, CD needs to be greater than (r/3) so that BD no longer goes through OB. CD is the value that you worked out in part a, so if you substitute that into CD > (r/3), you get [3(M-2m)] / [8(M+m)] > (r/3) and once you simplify this, you should get M > 26m.
    Thank you so much!!
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    (Original post by Tanmayee)
    Name:  Snap_2013.06.08_20h38m51s_008.png
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    I was just doing the same question! If you can help me by explaining why angle CBD is the same as the one at the vertex, then I can explain the rest. But I can't seem to figure out why those angles are the same...?
    The triangles BCD and BCO are similar. Shared angle BCO is of course \pi - \alpha and they both have right angles, meaning the final angle in each has to be \alpha? I'm just looking at this particular picture so the logic may be circular.
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    (Original post by mashmammad)
    That paper was the hardest I've done, by faaaar. Question 6b was very confusing. How likely are we to get something like that????
    Definitely the hardest M3 paper. I read the examiner's report and it basically said that it was very hard so I'd imagine we'll be unlikely to get one like that, I got an E on it first time round (but did it again a few days back and did fine)

    The moments one is all about finding perpendicular distances, its really annoying because it is so inaccurate due to the angles that you find but it is possible. But definitely worth 7 or 8 rather than 5 marks.

    Q6b is using the standard results for circle sectors, and again finding the perpendicular distance to the line of O, ie the height of the centre of mass of each circle above the bottom line. Hope that helped
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    Hello Fellow math prodigies--I have a problem(its long but please help) I just got released from the hospital for health issues that I had, and it hit me that I am completely unprepared for FP3,!!!not even the first chapter, and I have other exams like FP2, C4 C3 and unit 4 and 5 of aqa physics that I need to pump up for...is it possible for me to withdraw from the exam? I could cram everything in but I will not do good, and my other modules will be dragged down too from being obsessed over fp3... am taking a gap year next year because of health reasons but I realized that you cant do fp3 in jan(or can you?) What other A2 maths modules can you do in jan(can you do m3)...jan next year is the only time I can take exams as ill be in hospital most of the summer...

    I am also freaking out that there has to be a specific combo of maths modules you choose is this right?
    So far I have C1 C2 S1 S2 D1 FP1 M1 M2 and will be sitting fp2 and c3 4 within the next two weeks

    I know that if I sat fp3 from less than 2 weeks revision (for the hardest maths module!!) I will end up retaking which isnt favoured by unis

    Thank you for reading my ramble
    any help/advice appreciated.

    Love from London
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    Another M3 question sorry, it's about finding acceleration. How do you decide when to differentiate the velocity and when to use a=v(dv/dx)? They give different answers and I'm freaking out because I always did it by differentiating v before. Help! ~
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    (Original post by MathsForFun)
    Another M3 question sorry, it's about finding acceleration. How do you decide when to differentiate the velocity and when to use a=v(dv/dx)? They give different answers and I'm freaking out because I always did it by differentiating v before. Help! ~
    It depends whether the velocity is given in terms of t or x. If in terms of t you differentiate with respect to t, if in terms of x, use a = v(dv/dx), and go from there
 
 
 
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