Hey there! Sign in to join this conversationNew here? Join for free
    Offline

    22
    ReputationRep:
    (Original post by EnglishMuon)
    Wait whaaat? I asked my teacher and he said there are only old ones which dont cover the new spec. I just went straight into past papers and try to work it out as I go along. If you feel that books useful Ill definitely get one.
    I can send you a free PDF copy of it, if you'd like. (M5 too, if you want).
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by Zacken)
    I can send you a free PDF copy of it, if you'd like. (M5 too, if you want).
    Wow that would be great! Thanks very much.
    Offline

    2
    ReputationRep:
    (Original post by physicsmaths)
    Lol damped harmonic motion always ****s me up. One of my constants in the 2ODE is always opposite sign since i always write it the wrong way. Although Im on the review exercises so I shud be able to 95+ with some practice.
    Relative morion is a treat for geomters
    M5 is kinda hard since i dont know what In doing with forces sometimes lol!


    Posted from TSR Mobile
    I find damped simple harmonic motion just an application of 2nd order differential equation, so I find it pretty easy. EXCEPT, that the biggest questions involve finding extensions, like the questions in M3 SHM and I currently have trouble with that. I think I just need a bit more practice since I usually skip review exercises.

    Wait, so are you currently doing review exercises on M4 to get some practice going? Did you do M4/M5 last year? If so, what did you get on them?

    I don't classify myself as a geometer but I do find it an easy chapter if you have a bit of practice with it first.
    Offline

    2
    ReputationRep:
    (Original post by EnglishMuon)
    Wait whaaat? I asked my teacher and he said there are only old ones which dont cover the new spec. I just went straight into past papers and try to work it out as I go along. If you feel that books useful Ill definitely get one.
    Wait what? You have the old specification textbook?
    Offline

    3
    ReputationRep:
    (Original post by EnglishMuon)
    Maths-(C1-4,M1-3,S1-2)
    You're pretty badass for doing M3 in Y12, damn! Good luck meeting your offer, English-\mu ^{-}.
    Offline

    3
    ReputationRep:
    (Original post by Zacken)
    I can send you a free PDF copy of it, if you'd like. (M5 too, if you want).
    Mate, Kenya hook me up with S4?
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by aymanzayedmannan)
    You're pretty badass for doing M3 in Y12, damn! Good luck meeting your offer, English-\mu ^{-}.
    lol thanks Im currently year 13, but I did m3 last year
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by aymanzayedmannan)
    Mate, Kenya hook me up with S4?
    this is gonna turn into some underground maths book sharing thread soon
    Offline

    2
    ReputationRep:
    (Original post by physicsmaths)
    M5 is kinda hard since i dont know what In doing with forces sometimes lol!
    Posted from TSR Mobile
    Please don't say that, you make me even more worried.
    Offline

    3
    ReputationRep:
    (Original post by EnglishMuon)
    lol thanks Im currently year 13, but I did m3 last year
    Yeah, that's what I meant - most people sit M3 in Y13. Well done so far
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by Insight314)
    Wait what? You have the old specification textbook?
    I dont myself but the school have a copy I think.
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by aymanzayedmannan)
    Yeah, that's what I meant - most people sit M3 in Y13. Well done so far
    Ah ok! Thanks and good luck to you too (sorry for my bad interpretation of english )
    Offline

    2
    ReputationRep:
    (Original post by EnglishMuon)
    Ah ok! Thanks and good luck to you too (sorry for my bad interpretation of english )
    I guess we could say that.. you forgot the English in EnglishMuon.
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by Insight314)
    I guess we could say that.. you forgot the English in EnglishMuon.
    XD I have a Russian friend who comes up with terrible jokes. I should recommend this one to him
    • Thread Starter
    Offline

    15
    ReputationRep:
    Day 5 Summary
    It seems like large amount of both Mechanics A level modules and Physics are based heavily on the assumptions of Newton's laws acting in either one of two ways- in the intertial frame  F=m\ddot{x} form or the circular motion  F=\dfrac{mv^{2}}{r} form. I find it quite strange that they don't even mention where this comes from when there is such an easy derivation for all 2 dimensional cases of rotation! I thought I'd write out the proof here as it helps me appreciate some of the nicer maths behind it
    Name:  FullSizeRender (1).jpg
Views: 125
Size:  504.1 KB
    Equations of Motion in Polar Coordinates
    In our polar system, we can see
      \hat{r}=cos{\theta}i+sin{\theta}  j, \hat{\theta}=-sin{\theta}i+cos{\theta}j
    Differentiating, we have
     \dfrac{d\hat{r}}{dt}=-\hat{\theta}sin{\theta}i+ \hat{\theta}cos{\theta}j= \dot{\theta}\hat{\theta}

\dfrac{d\hat{\theta}}{dt}=-\hat{\theta}cos{\theta}i-\hat{\theta}sin{\theta}j= -\dot{\theta}{\hat{r}}
    We can therefore see (by subbing in the expressions from the previous part) that
     \dfrac{d}{dt} (r\hat{r})=\dot{r}\hat{r}+r \dot{\theta} \hat{\theta}

\dfrac{d^2}{dt^2}(r\hat{r})=\dot  {r}r+ 2\dot{r} \dot{\theta} \hat{\theta}+ r\ddot{\theta} \hat{\theta}- r\dot{\theta^2}\hat{r}
    Let  F_{r} be the force due to the r vector and  F_{\theta} be the force to the theta vector. Now we have an expression for the acceleration of our 'r vector', we can apply Newton's 2nd law (with m as the mass of our object) to get
    F_{r}+F_{\theta}=m(\ddot{r}- r\dot{\theta^2}) \hat{r} + (2\dot{r}\dot{\theta}+r \ddot{\theta}) \hat{\theta} .
    In our classic example of circular motion,  \dot{r}=0 as the radius of the circle is constant. Our above equation with  \dot{r}=0 gives us our usual  F=-m\dot{\theta^2}r form.
    (And thanks Insight314 for the mech. help )
    • Thread Starter
    Offline

    15
    ReputationRep:
    Day 6 Summary
    So one of the main points in STEP recently is to state the obvious whether you think they are asking for it or not!
    A good example is Q7 STEP III 2003. At the end of the question we are asked for a geometrical interpretation. In the previous part I had already mentioned the product of these =-1 can be shown by the perpendicular lines my diagram, therefore I assumed there was some deeper result. In fact there wasn't and it wasted lots of time!

    A technique that I have seen multiple times now is differentiating 'expansions' to evaluate new series. Consider
     (1-x)^{- \frac{1}{2}}= \displaystyle\sum_{r=0}^{\infty} \frac{^{2r} \mathrm{C}_{r}x^{r}}{4^{r}} (assuming we know this from previous workings). We can then differentiate both sides with respect to x to obtain more complex expressions. A common technique is adding multiples of the differentiated expansion to other expansions in the question.
    Offline

    22
    ReputationRep:
    (Original post by EnglishMuon)
    Day 6 Summary
    Don't forget integrating the series as well!
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by Zacken)
    Don't forget integrating the series as well!
    Oh yes of course! Thanks for reminding me
    • Thread Starter
    Offline

    15
    ReputationRep:
    Day 7 Summary
    One of the most satisfying ideas used in mathematical problem solving I find is coming up with a general solution! It is nice to answer the question, but why answer just this one situation when you can solve all of them at once?

    Consider
    (Original post by joostan)
    Compute the exact value of:x=13+\dfrac{1}{2+\dfrac{1}{26+ \dfrac{1}{2+\frac{1}{26+\frac{1}  {2+...}}}}}Where this fraction continues in this recurring way ad infinitum.Find an expression of this type (known as a continued fraction) for x=\dfrac{1+\sqrt{5}}{2}, and x=\sqrt{2}.
    It may well just be me being strange, but here Id love to find a way to write any number in the form of a continued fraction (of this type).
    We can then consider the general case of
     x=a+ \dfrac{1}{b+\dfrac{1}{c+ \dfrac{1}{b+\frac{1}{c+\frac{1}{  b+...}}}}}
    So
     \dfrac{1}{x-a} = b+ \dfrac{1}{c+ \dfrac{1}{b+...}}}
     \Rightarrow \dfrac{(ba+1)-bx}{x-a}= \dfrac{1}{c+ \dfrac{1}{b+...}}}
     \Rightarrow \dfrac{x-a}{(ba+1)-bx} = c-a+x
    Leading to
     bx^{2}+b(c-2a)x+(a^{2}b-c(1+ba))=0
    This may seem unnecessary at first, but this can actually save lots of time if a similar (the same) method is to be undertaken more than once. Now we can work out  x when given a continued fraction in this form by quadratic formula (or easier) and we can pick any  a,b,c to represent a number in this nested form (so long as the discrim of our quadratic >0!).
    A useful application of this is in proving integration techniques in STEP.
    The classic problem of evaluating
     \displaystyle\int^{ \frac{\pi}{2}}_0 \dfrac{sinx}{cosx+sinx} \dx can be extended by looking at
     \displaystyle \begin{equation*}\int_a^b f(x) \, \mathrm{d}x = \int_a^b f(a+b-x) \, \mathrm{d}x\end{equation*} (Thanks to Zacken for that last line- all this typing's given me RSI )
    Offline

    11
    ReputationRep:
    (Original post by EnglishMuon)
    It may well just be me being strange, but here Id love to find a way to write any number in the form of a continued fraction (of this type).
    It is indeed possible to write any real number as a continued fraction.
    Sadly however there's not always a nice recurrence. Even in seemingly simple cases.
    For example, \sqrt{190} has 14 recurring integers rather than just the two in the example I gave.
    There is an algorithm for computing the continued fraction of a given number, any rational's continued fraction will necessarily terminate. For a general real number you can compute the values of the integers known as partial quotients, the corresponding fraction formed by the partial quotients will tend to the real value you're computing from.
    This gives a way to find rational approximations to real values, such as \pi. Where many of the well known approximations to \pi arise from this technique.

    There are more general cases without requirements of integers and so forth, but simple continued fractions are easier to deal with.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Has a teacher ever helped you cheat?
    Useful revision links

    Articles

    Writing revision notes

    Our top revision articles

    Tips and advice on making the most of your study time.

    Boomarked book

    Superpowered study

    Take the hard work out of revising with our masterplan.

    Rosette

    Essay expert

    Learn to write like a pro with our ultimate essay guide.

    Can you help? Study Help unanswered threadsStudy Help rules and posting guidelines

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • 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.