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    (Original post by und)
    They're on different sides of an equals sign...
    Could you quote the exact bits of my solution you're unhappy with? You've essentially got  - kx = -c , so x has to be positive.
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    (Original post by DJMayes)
    Could you quote the exact bits of my solution you're unhappy with? You've essentially got  - kx = -c , so x has to be positive.
    Your solution is fine. I just for some reason thought that we should have been adding the constant rather than subtracting it. :rolleyes:
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    (Original post by und)
    Your solution is fine. I just for some reason thought that we should have been adding the constant rather than subtracting it. :rolleyes:
    Ah, OK then. What are your thoughts on the question?
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    (Original post by DJMayes)
    Ah, OK then. What are your thoughts on the question?
    Damnit! http://www.thestudentroom.co.uk/show...&postcount=374
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    (Original post by bananarama2)
    I would, but I still have a stray minus Care to assist?

    Spoiler:
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    It seems that you haven't completed the square properly and are missing a -\dfrac{\mu ^2l^2}{(1+\mu )^2 } on the RHS in that line...
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    (Original post by Star-girl)
    It seems that you haven't completed the square properly and are missing a -\dfrac{\mu ^2l^2}{(1+\mu )^2 } on the RHS in that line...
    Oh dear, of course :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: I still can't see why you have an equality rather than inequality.
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    (Original post by Blazy)
    Problem 55*

    The problem is concerned with constructing lines of length root  n , where  n is not a perfect square. Suppose you only have a ruler which can only measure integer lengths and assume you can draw perfect right-angled triangles. Show that unless  n is of the form \displaystyle 4k-2 , then you only ever need one right-angled triangle to construct a line of length root  n , e.g. To construct  \sqrt 2 , you create a  \displaystyle \left \{ 1, 1, \sqrt2 \right \} triangle, to construct  \sqrt3 you create a  \displaystyle \left \{ 1, \sqrt3, 2 \right \} triangle.
    I quite like this

    Consider the set of perfect squares: 1, 4, 9, 16, 25, 36, .... these are the numbers we must use, by addition or subtraction to find n (we can only use two).

    Note that the difference between adjacent primes is of the form 2k+1 (e.g. 1 and 4 is 3, 4 and 9 is 5, etc.). So for all odd n, we only have to construct sides of a and a+1 where 2a+1 = n and a+1 is the hypotenuse of the triangle. e.g. for n= 17 we need sides of length 8 and 9, and because 81-64 = 17, the side of length root n is root 17. And so all odd n are possible.

    Now note that the sum of two prime numbers which are two apart in the sequence is a^2 + (a+2)^2 = 2a^2 + 4a + 4, which is divisible by 4. So for all n where n = 4m, we can construct a triangle with rootn as the hypotenuse and side length a and a+2 where (a+2)^2 - a^2 = n. e.g. for root 12 we would need sides of 2 and 4. And so all multiples of 4 are possible.

    This only leaves even n which are not multiples of 4, i.e. n of the form 4k-2.

    Edit: I may come back and do this more rigorously to prove that the difference of summation of two perfect squares cannot be of the form 4k-2
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    (Original post by bananarama2)
    Oh dear, of course :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: :facepalm: I still can't see why you have an equality rather than inequality.
    Don't worry about it, I made numerous silly mistakes before finally getting to the right answer. :rofl:

    Sorted! http://www.thestudentroom.co.uk/show...3#post42214543
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    (Original post by und)
    Using u as the initial velocity.

    Since the function of v is continuous, if v \to 0^{+} for some x, then at that instant the frictional force must be greater than the weight, thus the rope will remain stationery. It remains to find a range of values of  u for which v will become  0 .

    Creating and solving the differential equation, we obtain v^2=\frac{g(\mu +1)x^2}{l}-2\mu x+u^2, so if v is to be  0 for some positive x, then there must exist a positive root x=\frac{\mu l \pm \sqrt{(\mu l)^2-u^{2}gl(\mu +1)}}{g(\mu +1)}, so a necessary and sufficient condition is u^2 \leq \frac{{\mu}^{2}l}{g(\mu +1)} as (almost ) required.
    Just so that I can better follow what you did: for v^2=\frac{g(\mu +1)x^2}{l}-2\mu x+u^2, did you take  v=0 \Leftrightarrow v^2=0 and use the quadratic formula on the quadratic on the RHS to get x=\frac{\mu l \pm \sqrt{(\mu l)^2-u^{2}gl(\mu +1)}}{g(\mu +1)}?
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    (Original post by Star-girl)
    Just so that I can better follow what you did: for v^2=\frac{g(\mu +1)x^2}{l}-2\mu x+u^2, did you take  v=0 \Leftrightarrow v^2=0 and use the quadratic formula on the quadratic on the RHS to get x=\frac{\mu l \pm \sqrt{(\mu l)^2-u^{2}gl(\mu +1)}}{g(\mu +1)}?
    Yes, but I've now fixed it so that I get the correct answer. It turns out I missed a g when I was solving the DE.

    DJ, I think this question is very STEP-like in that you can approach it in plenty of ways, but some solutions are certainly quicker than others.
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    (Original post by und)
    Yes, but I've now fixed it so that I get the correct answer. It turns out I missed a g when I was solving the DE.

    DJ, I think this question is very STEP-like in that you can approach it in plenty of ways, but some solutions are certainly quicker than others.
    Awesome! Onwards to part (ii)!

    Do you reckon that something like this could be in STEP III (I rarely look at III mechanics because I don't think I'll be able to do the qs)?
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    (Original post by Star-girl)
    Awesome! Onwards to part (ii)!

    Do you reckon that something like this could be in STEP III (I rarely look at III mechanics because I don't think I'll be able to do the qs)?
    I've done very few STEP III mechanics questions, but this one certainly seems somewhat similar to the style you can expect.
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    (Original post by und)
    I've done very few STEP III mechanics questions, but this one certainly seems somewhat similar to the style you can expect.
    In that case, I might try to have a go at some more (I always thought before that they were out of my reach).
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    (Original post by Llewellyn)
    I quite like this
    Ty, it was inspired by an interview question for a non-maths course - draw root 2
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    (Original post by DJMayes)
    ...
    (Original post by und)
    ...
    (Original post by bananarama2)
    ...
    If you're interested, my completed solution is here: http://www.thestudentroom.co.uk/show...3#post42214543.
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    (Original post by metaltron)
     \frac{1}{2}absin\theta we can see the area will be maximised by maximising two sides and making the angle between them a right-angle. This is because sin theta is greatest when theta = 90 degrees when sintheta is one. Since the third side is not involved in the area formula, the area will be maximised by maximising the two shortest sides.
    Sorry to be blunt but this is all rather unconvincing. Consider what happens if we decide to make theta smaller. We can then increase ab.
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    (Original post by Star-girl)
    If you're interested, my completed solution is here: http://www.thestudentroom.co.uk/show...3#post42214543.
    I like it, especially as energy considerations are a route I don't tend to go down.
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    (Original post by DJMayes)
    I like it, especially as energy considerations are a route I don't tend to go down.
    Thanks. As for me, energy consideration tends to be my first route to go down because then I can visualise a problem better. :cute:

    The differential equations route is cool too.
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    I just lost the will to live with that question after I completed a load of school work (binomial expansion). I generally go for energy, but just didn't in that instance, nice one Star-girl (I had to think about that name.)
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    (Original post by Star-girl)
    Thanks. As for me, energy consideration tends to be my first route to go down because then I can visualise a problem better. :cute:

    The differential equations route is cool too.
    No, that's fair enough. I tend to opt for Differential Equations because I like transforming a Mechanics question into a Pure question (And one of the nicer Pure topics in my opinion) but some of the energy solutions are really nice.
 
 
 
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