Year 13 Maths Help Thread

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    (Original post by RDKGames)
    If you can do it with numbers then im sure you can figure out the letters...


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    R u the new zacken btw. U seem to know everything bout maths


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    (Original post by physicsmaths)
    R u the new zacken btw. U seem to know everything bout maths


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    Nah fam I'll wait for zacken to help 2 im stuck on ur question now
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    http://www.physicsandmathstutor.com/...02.pdf#page=36

    Question and worked solution I'm stuck on in the link; I can't figure out how to do parts b and c.

    My working:
    a) as in worked solutions found that R (apq, a(p+q) )

    b) gradient = 2/(p+q), insert (a, 0) into y=mx+c to get y(p+q) = 2x-2a ; which is equivalent to the equation found in the worked solutions when considering that pq=-1. Not sure how to get the locus of R being x=-a from that.

    c) No working done as not sure how to find a locus as in b.

    What I'm specifically stuck on:
    - How to find the locus equation in b and c.
    - Why inserting the coordinates of P into y=mx+c seems to work in the worked solutions, but inserting (a, 0), which the chord PQ passes through, doesn't.
    - Most importantly: why finding the equation of the chord PQ has anything to do with finding the locus of R. R doesn't lie upon PQ, so what is the link? This is the part I need to understand most to be able to complete anything in this chapter.

    Any help would be very much appreciated, I've been stuck on this for a couple of days and not been able to move on with the chapter, and exhausted every possible avenue to try to understand it. This is in the Edexcel FP3 book btw, Chapter 2 Ex F.
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    (Original post by karim_elo)
    http://www.physicsandmathstutor.com/...02.pdf#page=36

    Question and worked solution I'm stuck on in the link; I can't figure out how to do parts b and c.

    My working:
    a) as in worked solutions found that R (apq, a(p+q) )

    b) gradient = 2/(p+q), insert (a, 0) into y=mx+c to get y(p+q) = 2x-2a ; which is equivalent to the equation found in the worked solutions when considering that pq=-1. Not sure how to get the locus of R being x=-a from that.

    c) No working done as not sure how to find a locus as in b.

    What I'm specifically stuck on:
    - How to find the locus equation in b and c.
    - Why inserting the coordinates of P into y=mx+c seems to work in the worked solutions, but inserting (a, 0), which the chord PQ passes through, doesn't.
    - Most importantly: why finding the equation of the chord PQ has anything to do with finding the locus of R. R doesn't lie upon PQ, so what is the link? This is the part I need to understand most to be able to complete anything in this chapter.

    Any help would be very much appreciated, I've been stuck on this for a couple of days and not been able to move on with the chapter, and exhausted every possible avenue to try to understand it. This is in the Edexcel FP3 book btw, Chapter 2 Ex F.
    So you got pq = -1, so that means if you put that into your R co-ordinates, you now get  R : ( -a , a(p + q)) so as p and q vary, only the y-coordinate of R will change, and it's x-coordinate stays at -a.

    For c, if the gradient is  \dfrac{2}{p+q} and we are given that in this case for a chosen p and q that the gradient is 2, then p + q = 1, Now you can see that in the R co-ordinates something happens rather similar to part b. Let me know if you get it or not.
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    (Original post by AMarques)
    So you got pq = -1, so that means if you put that into your R co-ordinates, you now get  R : ( -a , a(p + q)) so as p and q vary, only the y-coordinate of R will change, and it's x-coordinate stays at -a.

    For c, if the gradient is  \dfrac{2}{p+q} and we are given that in this case for a chosen p and q that the gradient is 2, then p + q = 1, Now you can see that in the R co-ordinates something happens rather similar to part b. Let me know if you get it or not.
    Ah yes, if p+q=1 then the y coordinate = a. So it seems the key thing here in finding loci is to find a relationship between p and q and then substitute that in to your general R coordinates, and finding the equation of the chord PQ is simply a vehicle to do that - just as using the gradient in part c is sufficient to do the same? I think that's what was puzzling me the most since I couldn't understand the relationship between PQ and R.

    And one final question does that mean this is the 'standard' way to find loci (other than when the question involves parametric equations)? i.e. Find the general coordinates of the variable point in terms of the two variables, find a relationship/equation between the two variables, and so substitute that into your coordinates to find the locus? Just asking because I want to understand how to solve all locus problems individually rather than coming back on here to ask for each slightly different question.

    Thanks for the help by the way!
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    (Original post by karim_elo)
    Ah yes, if p+q=1 then the y coordinate = a. So it seems the key thing here in finding loci is to find a relationship between p and q and then substitute that in to your general R coordinates, and finding the equation of the chord PQ is simply a vehicle to do that - just as using the gradient in part c is sufficient to do the same? I think that's what was puzzling me the most since I couldn't understand the relationship between PQ and R.

    And one final question does that mean this is the 'standard' way to find loci (other than when the question involves parametric equations)? i.e. Find the general coordinates of the variable point in terms of the two variables, find a relationship/equation between the two variables, and so substitute that into your coordinates to find the locus? Just asking because I want to understand how to solve all locus problems individually rather than coming back on here to ask for each slightly different question.

    Thanks for the help by the way!
    This is the easiest kind of loci questions because either x or y is constant, and so the x and y variables of the co-ordinates of R do not depend on eachother. Later on in the loci chapter you should find much harder questions where y depends on x as the loci.

    In the FP3 chapter, you will need to familiarize yourself with simple trig identities (normally  \sin^{2}{x} + \cos^{2}{x} = 1 ). Sometimes the answer may be given, and it is sufficient to simply verify the result by plugging in the values of x and y (this is normally really hard loci questions, but they are still do-able with trig identities).

    For instance, say we found the coordinates of a point  M to be  (\dfrac{\sin{\theta}}{4}, \dfrac{\cos{\theta}}{3}) . The way I do it is I let  X = \dfrac{\sin{\theta}}{4}, Y = \dfrac{\cos{\theta}}{3} . If we make  \sin{\theta} and  \cos{\theta} the subject, and take the square, we get  \sin^{2}{\theta} = 16X^{2} and  \cos^{2}{\theta} =  9Y^{2} . We can then imply, using the trig identity, that  16X^{2} + 9Y^{2} = 1 .

    I hope this made sense, and no problem.
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    what c4 topics do i need in order to comlete fp2 as i have only done c3 and was wondering if that is neough to do fp2 or which topics will i need from c4
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    (Original post by youreanutter)
    what c4 topics do i need in order to comlete fp2 as i have only done c3 and was wondering if that is neough to do fp2 or which topics will i need from c4
    What board are you doing? If edexcel, then Partial Fractions and Integration. Implicit differentiation will come in handy for more complicated questions later on when dealing with substitutions in 2nd order differential equations.
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    (Original post by youreanutter)
    what c4 topics do i need in order to comlete fp2 as i have only done c3 and was wondering if that is neough to do fp2 or which topics will i need from c4
    I would say learn all of C4 first.
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    which applied module is quickest for most ppl to selfteach s3 d2 or m3
    also does anyone have any good resources for s3 as wanted to take a better look
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    (Original post by youreanutter)
    which applied module is quickest for most ppl to selfteach s3 d2 or m3
    also does anyone have any good resources for s3 as wanted to take a better look
    S3 and M3 for sure. D2 is a fat module but is really straightforward.
    Fo for S3 its super easy and quick.


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    Would you recommend that I finish off C4 vectors (the only topic which I haven't fully learnt) when I recap C4 over the rest of the summer in preparation for MAT? I know only C1 and C2 can come up in the MAT but the person who runs the MAT thread recommended learning C3/C4 as it improves mathematical dexterity.

    I'm done recapping C3 and I'm up to recapping C4 binomial at the moment.
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    Q2. I'm not sure about the last part, I've successfully done everything before it. These numerical methods with Euler are all over the place.

    I've gotten:

    y(1+h)=y(1)+\frac{1}{2}h[f(1,2)+f(1+h,-2(1-h^2))] but I'm pretty sure the last part of that is wrong. I'm confused.

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    (Original post by RDKGames)
    Q2. I'm not sure about the last part, I've successfully done everything before it. These numerical methods with Euler are all over the place.

    I've gotten:

    y(1+h)=y(1)+\frac{1}{2}h[f(1,2)+f(1+h,-2(1-h^2))] but I'm pretty sure the last part of that is wrong. I'm confused.

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    It's not  -2(1-h^2) , it should be  f(1,2) + f(1+h,2(1-h)) as that is the y-coordinate estimate that we obtained from Euler's formula. Then you just plug both of them into f(x,y). Let me know what you get for a and b since I've done it.

    Note: we have found what  f(1+h,2(1-h)) was from the previous part.
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    (Original post by AMarques)
    It's not  -2(1-h^2) , it should be  f(1,2) + f(1+h,2(1-h)) as that is the y-coordinate estimate that we obtained from Euler's formula. Then you just plug both of them into f(x,y). Let me know what you get for a and b since I've done it.

    Note: we have found what  f(1+h,2(1-h)) was from the previous part.
    Ah of course! Got too confused between all the y's there.

    a=2, b=1?

    Also: when carrying out Euler's improved method, is there any efficient way to do it? Drawing tables or working them out 1-by-1 is a bit tedious.
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    (Original post by RDKGames)
    Ah of course! Got too confused between all the y's there.

    a=2, b=1?

    Also: when carrying out Euler's improved method, is there any efficient way to do it? Drawing tables or working them out 1-by-1 is a bit tedious.
    Precisely.

    I've never had this topic in my syllabus but knew it from doing other exam boards to gain more practice for step, I just do it the tedious way. I'm not quite sure what you mean by tables.
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    (Original post by AMarques)
    Precisely.

    I've never had this topic in my syllabus but knew it from doing other exam boards to gain more practice for step, I just do it the tedious way. I'm not quite sure what you mean by tables.
    Ah okay. For table I'm referring to the y_{r+1}=y_r+\frac{1}{2}(k_1+k_2) version where k_1=hf(x_r,y_r) and k_2=hf(x_r+h,y_r+k_1)

    You can tabulate this with variables r, x_r, y_r, k_1, k_2,... and work your way through the table. Take quite a while though but ensures accuracy.
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    (Original post by RDKGames)
    Ah okay. For table I'm referring to the y_{r+1}=y_r+\frac{1}{2}(k_1+k_2) version where k_1=hf(x_r,y_r) and k_2=hf(x_r+h,y_r+k_1)

    You can tabulate this with variables r, x_r, y_r, k_1, k_2,... and work your way through the table. Take quite a while though but ensures accuracy.
    Oh yes, I understand now.
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    Im not sure where to begin with this question as I don't understand what they're asking for, any hints? Q7

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    (Original post by RDKGames)
    Im not sure where to begin with this question as I don't understand what they're asking for, any hints? Q7

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    Differentiate it twice and sub it into the differential equation.
 
 
 
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