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    (Original post by TeeEm)
    around 70/75

    the parametric is not that hard (I am only joking)
    hmmm interesting.

    just opened it up and looked - didn't seem too bad
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    (Original post by Leking9)
    My teacher told us that when integrating 1/x , we should use the modulus lines but in the mark schemes, lt's simply ln(x). Which should I do?
    My teacher said the same thing. I normally just write ln(x) or lnx rather than the modulus lines though because the mark scheme does
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    (Original post by Gilo98)
    hmmm interesting.

    just opened it up and looked - didn't seem too bad
    it is not hard ... this is the A level version

    my version will be disgusting as I will insist in a particular method ("Shell Method")
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    (Original post by TeeEm)
    it is not hard ... this is the A level version

    my version will be disgusting as I will insist in a particular method ("Shell Method"
    apparently the question is not on current A Level spec - it seems we don't need volume of rev. about y axis let alone about any vertical line. Still interesting to do as an extension mind.

    hmm, what method would you call the approach to that question you gave me earlier on volume of rev about line x=4? Q56 to be precise.
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    (Original post by Gilo98)
    apparently the question is not on current A Level spec - it seems we don't need volume of rev. about y axis let alone about any vertical line. Still interesting to do as an extension mind.

    hmm, what method would you call the approach to that question you gave me earlier on volume of rev about line x=4? Q56 to be precise.
    Either create a formula (as in the solution I supplied) or translate the curve to the left by 4 units and revolve about the y axis ...

    I know it is not on the syllabus ... I though you were doing extension work ... people do all sort of things.
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    (Original post by TeeEm)
    Either create a formula (as in the solution I supplied) or translate the curve to the left by 4 units and revolve about the y axis ...

    I know it is not on the syllabus ... I though you were doing extension work ... people do all sort of things.
    second approach never occurred to me although seems to be the simplest way of going about it (would be unsure how to do this with parametrics though - just -4 from equation in x? seems quite crude).

    Ah OK, my confusion. I understand this is merely extension work; tbh the bore that is chugging through past papers has forced me to rekindle my enjoyment of maths though questions such as your own.
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    (Original post by Gilo98)
    maybe a sign slip? below is my workings with a positive answer.

    Attachment 427427

    quick question; why are you using the formula: 'x^2 dx/dtheta' ?

    i thought the formula for the volume for those kind of questions was always: 'y^2 dx/dtheta' ?

    i'm completely baffled. does this mean you can change the equation you would use in the 'squared' part of the formula? or have i succeed in completely muddling up my understanding of C4..
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    what is a position vector?
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    (Original post by theDanIdentity)
    quick question; why are you using the formula: 'x^2 dx/dtheta' ?

    i thought the formula for the volume for those kind of questions was always: 'y^2 dx/dtheta' ?

    i'm completely baffled. does this mean you can change the equation you would use in the 'squared' part of the formula? or have i succeed in completely muddling up my understanding of C4..
    don't worry this is extension work which is not required for C4. The question at hand was asking for the volume of revolution about the y axis not the x axis and the curve was given parametrically. It was parametric volume of rev (which we need to be able to do about the x axis but not the y axis) hence you use

    \pi\displaystyle\int^t_t x^2\frac{dy}{dt}\ dt\
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    (Original post by theDanIdentity)
    quick question; why are you using the formula: 'x^2 dx/dtheta' ?

    i thought the formula for the volume for those kind of questions was always: 'y^2 dx/dtheta' ?

    i'm completely baffled. does this mean you can change the equation you would use in the 'squared' part of the formula? or have i succeed in completely muddling up my understanding of C4..
    Don't worry, the y^2 version is all we have to know. The reason we use y (and x values in the limits) is because the revolutions are around the x-axis.

    The reason he's using x^2 is because he's dealing with a revolution around the y-axis instead, so the x and y values are in effect reversed.

    At least that's what I've inferred so far. It's interesting to follow and read but I dare not take part myself. Not anywhere near clever enough.
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    (Original post by ridirahman)
    what is a position vector?
    vector relative to the origin. If A has position vector (3,4,7) then

    \vec{OA} = \begin{pmatrix} 3 \\ 4 \\ 7 \end{pmatrix}
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    (Original post by Gilo98)
    don't worry this is extension work which is not required for C4. The question at hand was asking for the volume of revolution about the y axis not the x axis and the curve was given parametrically. It was parametric volume of rev (which we need to be able to do about the x axis but not the y axis) hence you use

    \pi\displaystyle\int^t_t x^2\frac{dy}{dt}\ dt\

    ahhhhh. i seee. cause i worked it out; but.. the answer i got, wasn't the same as what i was supposed to get. i even triple checked and quadrupled checked until i got a headache. got to a point i started seeing my life flash before my eyes like it did when i was doing C3.

    cheeers; should've noticed it was 'about the y-axis'.
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    (Original post by theDanIdentity)
    ahhhhh. i seee. cause i worked it out; but.. the answer i got, wasn't the same as what i was supposed to get. i even triple checked and quadrupled checked until i got a headache. got to a point i started seeing my life flash before my eyes like it did when i was doing C3.

    cheeers; should've noticed it was 'about the y-axis'.
    No worries: I presume you instead did...

     \pi\displaystyle\int^t_t y^2\frac{dx}{dt}\ dt\ \mathrm{?}
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    (Original post by Gilo98)
    vector relative to the origin. If A has position vector (3,4,7) then

    \vec{OA} = \begin{pmatrix} 3 \\ 4 \\ 7 \end{pmatrix}
    ohh okay so if you have: (ai+bj+ck)+x(di+ej+fk) [x is generally lamda]

    then (ai+bj+ck) is the position vector of a point on the line right?

    and (di+ej+fk) is the direction vector?
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    (Original post by ridirahman)
    ohh okay so if you have: (ai+bj+ck)+x(di+ej+fk) [x is generally lamda]

    then (ai+bj+ck) is the position vector of a point on the line right?

    and (di+ej+fk) is the direction vector?
    Yep. Thats why in vector Qs they always say 'position vector of A/B/Point D etc etc'.
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    (Original post by Gilo98)
    No worries: I presume you instead did...

     \pi\displaystyle\int^t_t y^2\frac{dx}{dt}\ dt\ \mathrm{?}

    yeah. and i got: -(Pi)^2 / 6.



    **i'm not sure if that's right or not btw; my C4 isn't at a leve/standard i'd like, especially when it comes to combining parametric equations / those surprise 'vector' questions that pop up in most past paper questions. :/ **
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    (Original post by theDanIdentity)
    yeah. and i got: -(Pi)^2 / 6.



    **i'm not sure if that's right or not btw; my C4 isn't at a leve/standard i'd like, especially when it comes to combining parametric equations / those surprise 'vector' questions that pop up in most past paper questions. :/ **
    I wouldn't even know what the answer that way would be - i think the hardest integration has gotten in past papers was volume of rev for curve given parametrically. Love vector Qs personally - did M1 last year which has definitely helped familiariser myself with them. Hate binomial expansion though - always seem to make a stupid sign slip.
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    (Original post by TeeEm)
    around 70/75

    the parametric is not that hard (I am only joking)
    How do I find limits if original limits are 2 and 3 and equation is x=sectheta please?
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    (Original post by games211)
    How do I find limits if original limits are 2 and 3 and equation is x=sectheta please?
    for x=2
    sec() = 2
    cos() = 1/2
    () = pi/3

    for x=3
    sec() = 3
    cos() = 1/3
    () = arccos(1/3) in radians
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    (Original post by TeeEm)
    for x=2
    sec() = 2
    cos() = 1/2
    () = pi/3

    for x=3
    sec() = 3
    cos() = 1/3
    () = arccos(1/3) in radians
    I got that but I still am not getting the correct answer :9
 
 
 
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