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Edexcel FP3 - 27th June, 2016 watch

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    (Original post by physicsmaths)
    Very nice indeed.
    I cna see the method but can't be bothered to put pen to paper.


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    What method have you chosen for (b)?
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    (Original post by A Slice of Pi)
    Given that \displaystyle I_n = \int_{0}^{2}x^n \sqrt{2x-x^2}\hspace{4pt}\mathrm{d}x, \hspace{15} n\geqslant0

    (a) find I_0

    (b) Show that \displaystyle I_n = \frac{2n+1}{n+2}I_{n-1}

    (c) Hence show that I_n = \displaystyle\frac{(2n+1)! \pi}{n!(n+2)!2^n}
    a) Write \int_0^2 \sqrt{1-(x-1)^2} \, \mathrm{d}x, then shift it over to get I_0 = \int_{-1}^{1} \sqrt{1-x^2} \, \mathrm{d}x where a straightforward u=\sin x sub sorts it out, gives \frac{\pi}{2}.

    b) Write x^n\sqrt{2x-x^2} = x^{n + 1/2} \sqrt{2-x} and use IBP with u = x^{n + 1/2} and \mathrm{d}v = \sqrt{2-x}. The uv portion clearly disappears and we are left with:

    \displaystyle 

\begin{align*}I_n &= \frac{(2n+1)}{3} \int_0^2 x^{n-1} \sqrt{2x-x^2} (2-x) \, \mathrm{d}x \\ \\ & \iff = 3I_n = 2(2n+1)I_{n-1} - (2n+1)I_n \\ \\ & \iff 2(n+2)I_n = 2(2n+1)I_{n-1} \\ \\ & \iff I_n = \frac{2n+1}{n+2}I_{n-1}\end{align*}

    as required.

    c) It follows straightforwardly that we end up with I_n = \frac{2(2n+1)!!}{(n+2)!} I_0 = \frac{(2n+1)!}{2^{n-1}n!(n+2)!} \times \frac{\pi}{2} = \frac{(2n+1)! \pi}{n! (n+2)! 2^n} using the well known double factorial identity.
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    FP3 June 2009

    this paper, question 7 c). shortest distance between two skew vectors. mark scheme says you have to use a formula (link), but i've never seen this formula and i can't see it in the formula booklet?

    could someone please explain how you would find the solution?
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    (Original post by A Slice of Pi)
    How did this turn out for you? Did you get the result?
    Couldn't figure out a substitution for a, so I tried to complete the square but it hasn't worked as that . Guessing there's a sneaky little thing which I cannot be arsed waiting for myself to figure out. I moved onto b at this point.

    As for b, I tried to do the method as I suggested last night, but this fails to work as I get \int{\frac{x}{2-2x}} . \sqrt{u} du for v. I tried to take a x^\frac12 outside of the square root which surprisingly lead me quite far, but at the end I was stuck with a (2x-x^2) which I wouldn't be able to multiply out, as it would lead to I_{n+1} stuff.
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    (Original post by Craig1998)
    Couldn't figure out a substitution for a, so I tried to complete the square but it hasn't worked as that . Guessing there's a sneaky little thing which I cannot be arsed waiting for myself to figure out. I moved onto b at this point.

    As for b, I tried to do the method as I suggested last night, but this fails to work as I get \int{\frac{x}{2-2x}} . \sqrt{u} du for v. I tried to take a x^\frac12 outside of the square root which surprisingly lead me quite far, but at the end I was stuck with a (2x-x^2) which I wouldn't be able to multiply out, as it would lead to I_{n+1} stuff.
    See two posts above for my solution.
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    (Original post by Zacken)
    See two posts above for my solution.
    Dammit. Forgot the power was 3/2 when I turned my piece of paper over. Thank you for figuring out the stupidness of part a which I'm going to be screwed over by if they ask something like it in the exam.
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    A Slice of Pi where did you get the reduction integration from btw? Would those sorts of ones appear in FP3 (even as a really hard question).
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    (Original post by Craig1998)
    A Slice of Pi where did you get the reduction integration from btw? Would those sorts of ones appear in FP3 (even as a really hard question).
    It seemed like a standard FP3 question to me. (and I think he made it up himself).
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    (Original post by Craig1998)
    A Slice of Pi where did you get the reduction integration from btw? Would those sorts of ones appear in FP3 (even as a really hard question).
    I based the part b question on a reduction formula I came across a few months ago, then put it into the 'Edexcel style'. It's one I could remember because the splitting of the integral wasn't obvious. I can't remember the source of the original question, which had a slightly different integral but the principle was the same. I'm not 100% sure whether something like this will come up. I'm with AQA for A-level so I don't know what level of difficulty the Edexcel FP3 papers go to.
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    (Original post by Zacken)
    It seemed like a standard FP3 question to me. (and I think he made it up himself).
    Most of the ones I've seen tend to be either:
    Generic easy ones like I_n = \int_\beta^\alpha x^n e^x dx
    Trig ones like I_n = \int_\beta^\alpha \cos^n x dx
    Weird ones where the thought comes after the ibp \int_\beta^\alpha x^n \sqrt{2-x^2} dx - this was similar to the other one, which is why my first thought was to split it into x . x^{n-1} then do a substitution.

    They seem to stop after that and don't really have ones that require a lot of thought before the ibp. Not in the past papers I've seen anyway.

    The only other one's I have seen are one's that are like I_{m,n}, but weve never been taught them so I assume they are not on the spec.
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    (Original post by Zacken)
    a) Write \int_0^2 \sqrt{1-(x-1)^2} \, \mathrm{d}x, then shift it over to get I_0 = \int_{-1}^{1} \sqrt{1-x^2} \, \mathrm{d}x where a straightforward u=\sin x sub sorts it out, gives \frac{\pi}{2}.

    b) Write x^n\sqrt{2x-x^2} = x^{n + 1/2} \sqrt{2-x} and use IBP with u = x^{n + 1/2} and \mathrm{d}v = \sqrt{2-x}. The uv portion clearly disappears and we are left with:

    \displaystyle 

\begin{align*}I_n &= \frac{(2n+1)}{3} \int_0^2 x^{n-1} \sqrt{2x-x^2} (2-x) \, \mathrm{d}x \\ \\ & \iff = 3I_n = 2(2n+1)I_{n-1} - (2n+1)I_n \\ \\ & \iff 2(n+2)I_n = 2(2n+1)I_{n-1} \\ \\ & \iff I_n = \frac{2n+1}{n+2}I_{n-1}\end{align*}

    as required.

    c) It follows straightforwardly that we end up with I_n = \frac{2(2n+1)!!}{(n+2)!} I_0 = \frac{(2n+1)!}{2^{n-1}n!(n+2)!} \times \frac{\pi}{2} = \frac{(2n+1)! \pi}{n! (n+2)! 2^n} using the well known double factorial identity.
    My method was about 7 steps longer even though it is fundamentally just like yours Wish I knew those tricks like shifting the limits and the double factorial thing.
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    (Original post by oShahpo)
    My method was about 7 steps longer even though it is fundamentally just like yours Wish I knew those tricks like shifting the limits and the double factorial thing.
    If it helps, "shifting the limits" is just me sneakily using the sub x \mapsto x-1 or u=x-1 in my head, nothing fancy there.

    And then, the double factorial thing isn't much work at all, it's just:

    \displaystyle (2n+1)!! = (2n+1)(2n-1)\cdots (1) = \frac{(2n+1)!}{(2n)(2n-2)(2n-4)\cdots(2)} = \frac{(2n+1)!}{2^n n!}

    So it's not as impressive as you think.
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    (Original post by Zacken)
    If it helps, "shifting the limits" is just me sneakily using the sub x \mapsto x-1 or u=x-1 in my head, nothing fancy there.

    And then, the double factorial thing isn't much work at all, it's just:

    \displaystyle (2n+1)!! = (2n+1)(2n-1)\cdots (1) = \frac{(2n+1)!}{(2n)(2n-2)(2n-4)\cdots(2)} = \frac{(2n+1)!}{2^n n!}

    So it's not as impressive as you think.
    What I did is that I multiplied both numerator and denominator by 2^n (n)!, it's just the double factorial thing looked more elegant
    Where does the factor of 2 in the numerator come from btw?
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    (Original post by Zacken)
    a) Write \int_0^2 \sqrt{1-(x-1)^2} \, \mathrm{d}x, then shift it over to get I_0 = \int_{-1}^{1} \sqrt{1-x^2} \, \mathrm{d}x where a straightforward u=\sin x sub sorts it out, gives \frac{\pi}{2}.

    b) Write x^n\sqrt{2x-x^2} = x^{n + 1/2} \sqrt{2-x} and use IBP with u = x^{n + 1/2} and \mathrm{d}v = \sqrt{2-x}. The uv portion clearly disappears and we are left with:

    \displaystyle 

\begin{align*}I_n &= \frac{(2n+1)}{3} \int_0^2 x^{n-1} \sqrt{2x-x^2} (2-x) \, \mathrm{d}x \\ \\ & \iff = 3I_n = 2(2n+1)I_{n-1} - (2n+1)I_n \\ \\ & \iff 2(n+2)I_n = 2(2n+1)I_{n-1} \\ \\ & \iff I_n = \frac{2n+1}{n+2}I_{n-1}\end{align*}

    as required.

    c) It follows straightforwardly that we end up with I_n = \frac{2(2n+1)!!}{(n+2)!} I_0 = \frac{(2n+1)!}{2^{n-1}n!(n+2)!} \times \frac{\pi}{2} = \frac{(2n+1)! \pi}{n! (n+2)! 2^n} using the well known double factorial identity.
    Hey Zack, where does the factor of 2 come from in C?
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    (Original post by A Slice of Pi)
    What method have you chosen for (b)?
    One second


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    (Original post by oShahpo)
    Hey Zack, where does the factor of 2 come from in C?
    We require n \geq 1 so the (n+2)! in the denominator doesn't contain the (0+2) = 2 terms. But I wrote it as a factorial and multiplied the numerator by 2 to account for the extra insertion of 2 in the denominator from the factorial.

    i.e: the denominator should really just be (n+2)(n+1) \cdots (3) but I write (n+2)! = (n+2)(n+1) \cdots (2)(1) so had to multiply the numerator by 2 to compensate for that.
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    (Original post by A Slice of Pi)
    What method have you chosen for (b)?
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    My way for this nice reduction question.



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    (Original post by Zacken)
    We require n \geq 1 so the (n+2)! in the denominator doesn't contain the (0+2) = 2 terms. But I wrote it as a factorial and multiplied the numerator by 2 to account for the extra insertion of 2 in the denominator from the factorial.

    i.e: the denominator should really just be (n+2)(n+1) \cdots (3) but I write (n+2)! = (n+2)(n+1) \cdots (2)(1) so had to multiply the numerator by 2 to compensate for that.
    Splendid Thanks a lot.
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    (Original post by physicsmaths)
    Name:  ImageUploadedByStudent Room1464821712.246474.jpg
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    My way for this nice reduction question.



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    Very nice! I'd rep but it won't let me
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    (Original post by Anon-)
    FP3 June 2009

    this paper, question 7 c). shortest distance between two skew vectors. mark scheme says you have to use a formula (link), but i've never seen this formula and i can't see it in the formula booklet?

    could someone please explain how you would find the solution?
    Just did this now, hope Im not too late
    Find the direction vector between the 2 position vectors and do the dot product of that with the unit normal vector (the normal of the 2 direction vectors)


    The absolute value of the dot product is the shortest distance
 
 
 
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