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    (Original post by joostan)
    Lol, I tried doing that but they never seem interesting enough, but go for it
    Haha I guess so, let me try using my skills :cyber:
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    so i was thinking.... How hard is AFM??

    Ryan
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    (Original post by tigerz)
    Haha I guess so, let me try using my skills :cyber:
    Like a :ninja:
    (Original post by ryanb97)
    so i was thinking.... How hard is AFM??

    Ryan
    It's largely mechanics and stats. Apparently M4/5 are hard but the rest are ok.
    DJ or Jkn can tell you more, or you can take a look at this thread.

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    (Original post by ryanb97)
    so i was thinking.... How hard is AFM??

    Ryan
    I think the difficulty depends on the module
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    (Original post by Boy_wonder_95)
    I think the difficulty depends on the module
    By that time, I'd expected you'd have completed most of the modules so you'll probably be doing the same modules as the other's dong AFM.
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    (Original post by Boy_wonder_95)
    ...
    (Original post by joostan)
    ,,,
    hehe,, thanks for pointing that out..il post it on that thread...

    ryan
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    (Original post by joostan)
    Like a :ninja:


    It's largely mechanics and stats. Apparently M4/5 are hard but the rest are ok.
    DJ or Jkn can tell you more, or you can take a look at this thread.

    LOOL, this is probably a failure of a question but...
    Differentiate  y=7x(cosx)^\frac{x}{2} If its too easy i'll pick a better one
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    (Original post by tigerz)
    LOOL, this is probably a failure of a question but...
    Differentiate  y=7x(cosx)^\frac{x}{2} If its too easy i'll pick a better one
    Sorry, I was filling out an open day form.
    \dfrac{d}{dx}\left(7x\cos^{\frac  {x}{2}}(x)\right) = 7\cos^{\frac{x}{2}}(x) + 7x\dfrac{d}{dx}\left(\cos^{\frac  {x}{2}}(x)\right)
    Consider:
    \dfrac{d}{dx}\left(\cos^{\frac{x  }{2}}(x)\right) = \dfrac{d}{dx}\left(e^{\frac{x}{2  }\ln(\cos(x))}\right)

    Let:
    e^u=e^{\frac{x}{2}\ln(\cos(x))}

\therefore \dfrac{d}{dx}\left(e^{\frac{x}{2  }}\ln(\cos(x))\right) = e^{\frac{x}{2}\ln(\cos(x))} \times \frac{1}{2}(\ln(\cos(x)) -x\tan(x))

    \therefore \dfrac{d}{dx}\left(7x\cos^{\frac  {x}{2}}(x)\right) = 7\cos^{\frac{x}{2}}(x) + \frac{7x}{2}\cos^{\frac{x}{2}}(x  )\times (\ln(\cos(x)) -x\tan(x))

= 7\cos^{\frac{x}{2}}(x)\left(1 + \frac{7x}{2}\times (\ln(\cos(x)) -x\tan(x))\right)
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    (Original post by joostan)
    Sorry, I was filling out an open day form.
    \dfrac{d}{dx}\left(7x\cos^{\frac  {x}{2}}(x)\right) = 7\cos^{\frac{x}{2}}(x) + 7x\dfrac{d}{dx}\left(\cos^{\frac  {x}{2}}(x)\right)
    Consider:
    \dfrac{d}{dx}\left(\cos^{\frac{x  }{2}}(x)\right) = \dfrac{d}{dx}\left(e^{\frac{x}{2  }\ln(\cos(x))}\right)

    Let:
    e^u=e^{\frac{x}{2}\ln(\cos(x))}

\therefore \dfrac{d}{dx}\left(e^{\frac{x}{2  }}\ln(\cos(x))\right) = e^{\frac{x}{2}\ln(\cos(x))} \times (\frac{1}{2}\ln(\cos(x)) -x\tan(x))

    \therefore \dfrac{d}{dx}\left(7x\cos^{\frac  {x}{2}}(x)\right) = 7\cos^{\frac{x}{2}}(x) + \frac{7x}{2}\cos^{\frac{x}{2}}(x  )\times (\frac{1}{2}\ln(\cos(x)) -x\tan(x))

= 7\cos^{\frac{x}{2}}(x)\left(1 + \frac{7x}{2}\times (\frac{1}{2}\ln(\cos(x)) -x\tan(x))
    Haha no problemo, I think I need to find a more interesting question lols >.<
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    (Original post by tigerz)
    Haha no problemo, I think I need to find a more interesting question lols >.<
    That one was ok, it wasn't so much difficult as tedious
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    (Original post by joostan)
    That one was ok, it wasn't so much difficult as tedious
    LOOOL, haha that was a warm up, one sec lemme search for a better one
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    (Original post by joostan)
    Sorry, I was filling out an open day form.
    \dfrac{d}{dx}\left(7x\cos^{\frac  {x}{2}}(x)\right) = 7\cos^{\frac{x}{2}}(x) + 7x\dfrac{d}{dx}\left(\cos^{\frac  {x}{2}}(x)\right)
    ....
    \therefore \dfrac{d}{dx}\left(7x\cos^{\frac  {x}{2}}(x)\right) = 7\cos^{\frac{x}{2}}(x) + \frac{7x}{2}\cos^{\frac{x}{2}}(x  )\times (\frac{1}{2}\ln(\cos(x)) -x\tan(x))

= 7\cos^{\frac{x}{2}}(x)\left(1 + \frac{7x}{2}\times (\frac{1}{2}\ln(\cos(x)) -x\tan(x))\right)
    Hmmm, I did it differently. Can you tell me if mine's equivalent or just wrong?
    \dfrac{dy}{dx}=y  \left(\dfrac{-xtanx+lncosx}{2}\right)+\dfrac{y  }{x}
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    (Original post by reubenkinara)
    Hmmm, I did it differently. Can you tell me if mine's equivalent or just wrong?
    \dfrac{dy}{dx}=y  \left(\dfrac{-xtanx+lncosx}{2}\right)+\dfrac{y  }{x}
    Assuming y is the original function it looks pretty much the same, I considered that but went down the exponential route, your way may well have been quicker
    EDIT: In fact a sneaky half has crept its way into my solution :eek:
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    (Original post by joostan)
    Assuming y is the original function it looks pretty much the same, I considered that but went down the exponential route, your way may well have been quicker
    EDIT: In fact a sneaky half has crept its way into my solution :eek:
    Yep. y=7x \left(cos[x] \right)^\frac{x}{2}
    Latex error?
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    (Original post by reubenkinara)
    Yep. y=7x \left(cos[x] \right)^\dfrac{x}{2}
    Yeah, I tidied up my tex a little.
    I also factored out y
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    (Original post by joostan)
    Assuming y is the original function it looks pretty much the same, I considered that but went down the exponential route, your way may well have been quicker
    EDIT: In fact a sneaky half has crept its way into my solution :eek:
    You let your guard down tut tut
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    pssht my excuse of floppiness


    Right this is a weird one haha:

    There is a rabbit that runs in a perfect circle of radius r with a constant speed v.
    A fox chases the rabbit, starting from the center of the circle and also moves with a constant speed v such that it is always between the center of the circle and the rabbit. How long will it take for the fox to catch the rabbit?
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    (Original post by joostan)
    Yeah, I tidied up my tex a little.
    I also factored out y
    I am slightly worried about my solution as it lacks 7s!
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    (Original post by reubenkinara)
    I am slightly worried about my solution as it lacks 7s!
    The 7s are in the y
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    (Original post by tigerz)
    You let your guard down tut tut
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    pssht my excuse of floppiness


    Right this is a weird one haha:

    There is a rabbit that runs in a perfect circle of radius r with a constant speed v.
    A fox chases the rabbit, starting from the center of the circle and also moves with a constant speed v such that it is always between the center of the circle and the rabbit. How long will it take for the fox to catch the rabbit?
    I'm gonna go eat, I'll sort that in a bit
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    (Original post by joostan)
    I'm gonna go eat, I'll sort that in a bit
    Okays, I look forward to seeing what method you decide use
 
 
 
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