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    1. #1
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      Maths question lul

      Why is 0.25sin^4(0.5pie) - 0.25sin^4(0) = 0.25 :S

    2. #2
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      It's easier with fractions. Note that sin(pi/2) = 1, and sin(0) = 0.
      Rewrite it as:

      = (1/4) [sin(pi/2)]^4 - (1/4) [sin(0)]^4
      = (1/4) (1^4) - (1/4) (0^4)
      = (1/4)(1) - (1/4)(0)
      = 1/4 - 0
      = 1/4
      = 0.25

      Did you really need a thread for this!?
      Last edited by Dianeva; 12-24-2011 at 01:57 PM.

    3. #3
      Xei
      UnitedKingdom Xei is offline
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      Yeah... because if you input the values for sin that's what you get, lul.

    4. #4
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      I'm stuck on a maths question as well. Might as well just post it here to get an answer.

      Find the general solution of the differential equation: tan(x)(dy/dx) +y = e^(2x)*tanx

      I found that the integrating factor is e^(1/tanx) = e^(lnsinx) = sinx

      Multiplying every term by sinx:
      sinx (dy/dx) + y (sinx/tanx) = e^(2x)*sinx

      But now when I do integration by parts it just goes in a loop..
      Last edited by ThePieMan; 12-24-2011 at 06:36 PM.

    5. #5
      Xei
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      Check out the homework guideline sticky...

      If you've studied DEs I don't see why you shouldn't be able to do that. What's your specific confusion?

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      Flaw in my math knowledge soz, didnt realise that sin^4 is the same as the 0.5pie^4 if that makes sense.

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      Quote Originally Posted by Xei View Post
      If you've studied DEs I don't see why you shouldn't be able to do that. What's your specific confusion?
      Sorry I put up the wrong question. I actually managed to do the previous one (just a standard 2nd order homogeneous DE). It is in fact this trigonometric function (above) that I'm stuck on right now; reasons for confusion stated.

    8. #8
      Xei
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      You'll be trying to integrate 1/tanx, yesh?

      Write down the definition of tan. You should be able to integrate by substitution.

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      sinx (dy/dx) + y (sinx/tanx) = e^(2x)*sinx

      =>I started by using some basic trig sinx/tanx = sinx/(sinx/cosx) = cosx

      =>So now the equation looks like: sinx y'' + ycosx = e^(2x)*sinx

      I then tried to do integration by parts by letting u = e^(2x) , u'=2e^(2x) , v = -cosx , v' = sinx

      Was this the wrong step to take?
      Last edited by ThePieMan; 12-25-2011 at 01:02 AM.

    10. #10
      Xei
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      Yeah, to use that method you first have to get it into the form y' + fy = g; in other words, divide the whole thing by sinx.

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      Oh I see. Thanks man

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      Out of curiosity, what subjects did you take before university? Further maths I presume?

    13. #13
      Xei
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      Yeah, and phys, chem, bio. They're not very demanding. The only exam I had to spend a long time training for was STEP, although the syllabus is just maths and further maths.

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      Wow that's impressive. I wouldn't be able to cope with that many subjects.
      And how is it that you don't find further maths demanding?!

    15. #15
      Xei
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      Meh I found it to just be learning set methods. I don't think there's much conceptually hard stuff... they just give you a list of different things you have to be able to do, and the set of instructions you follow to do them. There's no thinking involved, just memorisation of a list of things. Science A-Levels are exactly the same, it gets boring once you get used to it.

      Check out some of these which have the same syllabus as A-Level but the opposite emphasis:

      STEP Test Preparation | University Admissions Tests from Cambridge Assessment

      I had to pass II and III.

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      That stuff looks intimidating... But I suppose it ought to be if its the admissions test to one of the best universities in the world.

      Now I know who to ask when I'm stuck on a question
      Last edited by ThePieMan; 12-25-2011 at 06:52 PM.

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      A mechanics problem:

      A particle is projected from a point P on an inclined plane, up the line of greatest slope through P, with initial speed V. The angle of the plane to the horizontal is θ.

      (i) If the plane is smooth, and the particle travels for a time 2Vcosθ/g before coming instantaneously to rest, show that θ=pi/4

      (ii) If the same plane is given a roughened surface, with a coefficient of friction 0.5, find the distance travelled before the particle comes to rest.

      ---
      I managed to do (i) by considering the components of gravity on an inclined plane, then using v=u+at. V, g terms cancel and you get θ=pi/4.

      Not sure how to do part (ii)...

      Help is appreciated.

    18. #18
      Xei
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      The key with pretty much all of those mechanics modules is just to draw a diagram... it's basically the same deal in this question, except you have a frictional force. I can't really remember how those things work, but isn't it just that a frictional force of mu times the reaction force of the surface opposes the motion?

      In this case it's pretty much the same question as before, except you have to work out the reaction force (which is done in pretty much the same way as you found the decelerating component of the weight), and add that to the deceleration. Then it's just equations of motion again.

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      I'm guessing you'd have to find the mass, right? Then maybe use the work-energy theorem?

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      Hmm not quite - I think the correct approach is to actually consider the components of gravity (gsinθ, gcosθ etc), but I have no idea how to deduce the reaction force from that..

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      I just figured the coefficient of friction wouldn't be of use without knowledge of the normal force on the particle.
      Last edited by Wayfaerer; 01-31-2012 at 11:00 PM.

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      Meh I'll try to solve it at some later date. I'll be sure to post detailed solutions to satiate your curiosity.

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      Sounds good, I could probably find it helpful to post some of my own homework problems here later

    24. #24
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      Cool. You might see some complex number problems here in the near future.

    25. #25
      Xei
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      Bro I answered your question lol...



      Random helpful image from Google. The green line makes an angle @ with the horizontal by definition, and also by simple geometry the red and black lines also make an angle @. Fg is just the weight force of the particle, and from that you can calculate Fperp which is the reaction force just like you calculated Fpara which is the retarding gravitational component.

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