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Mega A Level Maths Thread MK II

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Reply 4000

tigerz

Original post by joostan

Lol, I tried doing that but they never seem interesting enough, but go for it :biggrin:

Haha I guess so, let me try using my skills :cyber:

Reply 4001

ryanb97

so i was thinking.... How hard is AFM??

Ryan

Reply 4002

joostan

Original post by tigerz

Haha I guess so, let me try using my skills :cyber:

Like a

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.

(edited 10 years ago)

Reply 4003

Boy_wonder_95

Original post by ryanb97

so i was thinking.... How hard is AFM??

Ryan

I think the difficulty depends on the module

Reply 4004

Kvothe the Arcane

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.

Reply 4005

ryanb97

Original post by Boy_wonder_95

...

Original post by joostan

,,,

hehe,, thanks for pointing that out..il post it on that thread...

ryan

Reply 4006

tigerz

Original post by joostan

Like a

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

Reply 4007

joostan

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))}[br]\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))[br]= 7\cos^{\frac{x}{2}}(x)\left(1 + \frac{7x}{2}\times (\ln(\cos(x)) -x\tan(x))\right)

(edited 10 years ago)

Reply 4008

tigerz

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))}[br]\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))

Unparseable latex formula:

\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))[br]= 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 >.<

Reply 4009

joostan

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 :s-smilie:

Reply 4010

tigerz

Original post by joostan

That one was ok, it wasn't so much difficult as tedious :s-smilie:

LOOOL, haha that was a warm up, one sec lemme search for a better one

Reply 4011

Kvothe the Arcane

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))[br]= 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}

(edited 10 years ago)

Reply 4012

joostan

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 :smile:

EDIT: In fact a sneaky half has crept its way into my solution :eek:

Reply 4013

Kvothe the Arcane

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 :smile:

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?

Reply 4014

joostan

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 :smile:

Reply 4015

tigerz

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 :smile:

EDIT: In fact a sneaky half has crept its way into my solution :eek:

You let your guard down :wink:

tut tut

Spoiler

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?