Turn on thread page Beta

Will the real TeeEm please stand up! watch

  • View Poll Results: will the real TeeEm please stand up
    man on the desk
    19
    6.76%
    man in the library with lap top
    69
    24.56%
    man 3
    13
    4.63%
    bodybuilder
    43
    15.30%
    donkey
    52
    18.51%
    man eating sandwich
    25
    8.90%
    ginger
    60
    21.35%

    Offline

    1
    ReputationRep:
    Name:  1440386918875.jpg
Views: 52
Size:  18.5 KBName:  1440386937970.jpg
Views: 42
Size:  11.1 KBName:  1440386952795.jpg
Views: 48
Size:  6.5 KBName:  1440386988655.jpg
Views: 49
Size:  20.6 KBName:  1440387006470.jpg
Views: 43
Size:  21.3 KBName:  1440387034735.jpg
Views: 41
Size:  16.8 KBName:  1440387077451.jpg
Views: 49
Size:  16.9 KB
    Offline

    15
    ReputationRep:
    (Original post by Mutleybm1996)
    Is there a document showing how many candidates sat each exam?


    Posted from TSR Mobile
    Graham sent out a vast document with exam statistics, but it is mostly by percentages. The only figures involving numbers of candidates are for each qualification, not each unit.
    • Thread Starter
    Offline

    19
    ReputationRep:
    (Original post by fatart123)
    What's it on? There's no guarantee I'd even be able to do it if it's one of your horrible sequence product questions...
    (Original post by physicsmaths)
    I will try it if i can do it bro.


    Posted from TSR Mobile
    Here comes the beast.
    The finite region bounded by the curve with equation
    y = (6 - x)(x -2), the tangent to the curve at the point where x = 3 and the x axis, is rotated by a full turn about the tangent.
    Determine the volume generated.
    (Calculator is allowed)

    I will make another similar question later (easier) because when I made this little did I know how tough it was going to be
    Online

    11
    ReputationRep:
    (Original post by TeeEm)
    Here comes the beast.
    The finite region bounded by the curve with equation
    y = (6 - x)(x -2), the tangent to the curve at the point where x = 3 and the x axis, is rotated by a full turn about the tangent.
    Determine the volume generated.
    (Calculator is allowed)

    I will make another similar question later (easier) because when I made this little did I know how tough it was going to be
    Well, I've moved the region so that it's 'connected' to the origin. But rotating the axis/curve + tangent about the origin and keeping the equation of the curve correct is pretty horrible
    • Thread Starter
    Offline

    19
    ReputationRep:
    (Original post by fatart123)
    Well, I've moved the region so that it's 'connected' to the origin. But rotating the axis/curve + tangent about the origin and keeping the equation of the curve correct is pretty horrible
    I tried rotation matrices but the new equation was appalling ...
    I derived a volume element but I do not know if there is any silly or serious error in my workings... over 3 pages still
    Online

    11
    ReputationRep:
    (Original post by TeeEm)
    I tried rotation matrices but the new equation was appalling ...
    I derived a volume element but I do not know if there is any silly or serious error in my workings... over 3 pages still
    lol, I found this after thinking about matrices, but I'm not going to use/adapt it because it's kinda cheating now:
    http://www.emathhelp.net/notes/calcu...ut-slant-line/

    Trying to translate to polar coordinates and rotate once I've got it in that system... This quadratic is a pain, but it should be doable
    • Thread Starter
    Offline

    19
    ReputationRep:
    (Original post by fatart123)
    lol, I found this after thinking about matrices, but I'm not going to use/adapt it because it's kinda cheating now:
    http://www.emathhelp.net/notes/calcu...ut-slant-line/

    Trying to translate to polar coordinates and rotate once I've got it in that system... This quadratic is a pain, but it should be doable
    excellent research...
    I can now check my answer
    Offline

    4
    ReputationRep:
    (Original post by TeeEm)
    here it is
    Thanks so much, been looking for this for ages now.

    Is there any chance you have the OCR markschemes and examiner reports? I do psychology and chemistry and haven't been able to find a single thing on TSR.
    • Thread Starter
    Offline

    19
    ReputationRep:
    (Original post by Craig1998)
    Thanks so much, been looking for this for ages now.

    Is there any chance you have the OCR markschemes and examiner reports? I do psychology and chemistry and haven't been able to find a single thing on TSR.
    I am sorry but I am unable to.

    I have a friend in a school which teaches EDEXCEL maths and kindly lets me have access to Maths EDEXCEL material only
    Online

    11
    ReputationRep:
    (Original post by TeeEm)
    excellent research...
    I can now check my answer
    What was your answer? Would like to check mine tbh since I'm 50/50 as to whether I made a logic error in my work somewhere
    Offline

    11
    ReputationRep:
    (Original post by TeeEm)
    Here comes the beast.
    The finite region bounded by the curve with equation
    y = (6 - x)(x -2), the tangent to the curve at the point where x = 3 and the x axis, is rotated by a full turn about the tangent.
    Determine the volume generated.
    (Calculator is allowed)

    I will make another similar question later (easier) because when I made this little did I know how tough it was going to be
    It truly is a beastly question! I had an attempt although I haven't checked for mistakes yet (they are no doubt in plentiful supply with the integration, the transformations etc.). I ended up with
    7root 5 /6 pi
    Offline

    4
    ReputationRep:
    (Original post by TeeEm)
    I am sorry but I am unable to.

    I have a friend in a school which teaches EDEXCEL maths and kindly lets me have access to Maths EDEXCEL material only
    No problem, I waited the entire year to find out all of my GCSE exam markschemes and examiner reports.
    • Thread Starter
    Offline

    19
    ReputationRep:
    (Original post by fatart123)
    What was your answer? Would like to check mine tbh since I'm 50/50 as to whether I made a logic error in my work somewhere
    my answer is wrong

    I am going to go through the calculus method you supplied in a while (still doing other material)
    Offline

    11
    ReputationRep:
    (Original post by TeeEm)
    my answer is wrong

    I am going to go through the calculus method you supplied in a while (still doing other material)
    Did you not use calculus yourself?
    Online

    11
    ReputationRep:
    (Original post by A Slice of Pi)
    It truly is a beastly question! I had an attempt although I haven't checked for mistakes yet (they are no doubt in plentiful supply with the integration, the transformations etc.). I ended up with
    7root 5 /6 pi
    I'm pretty close to that and got lazy at the end by approximating my integral using Wolfram, so we could be correct. Did you translate and reflect the quadratic so that it makes applying the calculus from that link easier or not?
    • Thread Starter
    Offline

    19
    ReputationRep:
    (Original post by A Slice of Pi)
    Did you not use calculus yourself?
    I did but I got the thickness of the element wrong
    Offline

    11
    ReputationRep:
    (Original post by fatart123)
    I'm pretty close to that and got lazy at the end by approximating my integral using Wolfram, so we could be correct. Did you translate and reflect the quadratic so that it makes applying the calculus from that link easier or not?
    My method was a little more complicated I'd say (making it more likely incorrect). I drew a diagram showing the curve, the tangent and the required area. What I then did was introduce a second pair of axes (w and z, as shown) at (1.5, 0) and started trying to express w and z in terms of x and y. I came up with w = \frac{2x-3}{\sqrt{5}} and z = \frac{\sqrt{5}y}{2}. All I did then was make substitutions in the formula V = \pi\int_{0}^{\frac{3\sqrt{5}}{2}  }w^2 dz to get it in terms of x and dy. The limits etc. were explained in my full method. Excuse the poor diagram...
    Attached Images
     
    Offline

    19
    ReputationRep:
    I don't know if my decent results this year were luck or my maths skills deteriorate incredibly quickly without sufficient practice...I have absolutely no clue how to approach the last question given, or would not have had any without the ideas of others (still probably can't work towards a solution even given that). Starting to feel a maths degree is not for me..

    (Original post by fatart123)
    lol, I found this after thinking about matrices, but I'm not going to use/adapt it because it's kinda cheating now:
    http://www.emathhelp.net/notes/calcu...ut-slant-line/
    May have to resort to that...although it looks somewhat ugly
    Online

    11
    ReputationRep:
    (Original post by A Slice of Pi)
    My method was a little more complicated I'd say (making it more likely incorrect). I drew a diagram showing the curve, the tangent and the required area. What I then did was introduce a second pair of axes (w and z, as shown) at (1.5, 0) and started trying to express w and z in terms of x and y. I came up with w = \frac{2x-3}{\sqrt{5}} and z = \frac{\sqrt{5}y}{2}. All I did then was make substitutions in the formula V = \int_{0}^{\frac{3\sqrt{5}}{2}}w^  2 dz to get it in terms of x and dy. The limits etc. were explained in my full method. Excuse the poor diagram...
    Hah, I tried to do this. Instead of introducing new axis, I converted the translated equations of the line and curve to polar form, rotated them in that system (quite a bit easier) and then reconverted to cartesian. Sadly, it's seems actually impossible (Wolfram could only approximate) to explicitly express y in terms of x after doing so, so that didn't work for me.

    I like how you've tried it though, so I'll have a go at doing the question that way tbh.

    (Original post by 13 1 20 8 42)
    I don't know if my decent results this year were luck or my maths skills deteriorate incredibly quickly without sufficient practice...I have absolutely no clue how to approach the last question given, or would not have had any without the ideas of others (still probably can't work towards a solution even given that). Starting to feel a maths degree is not for me..

    May have to resort to that...although it looks somewhat ugly
    It's just lots and lots of boring algebra. I was too confused as to how to get it to work from underneath, so I just worked it out from the 'inverse' of the quadratic above the tangent
    • Thread Starter
    Offline

    19
    ReputationRep:
    Do you guys think this will be too much to go into one of my SPECIAL PAPERS?

    I made a far easier version which is to revolve the finite region bounded by y=x^2 and y=x about the line y=x
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: August 22, 2017

1,366

students online now

800,000+

Exam discussions

Find your exam discussion here

Poll
Should predicted grades be removed from the uni application process

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.