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HSC 2012 MX1 Marathon #2 (archive) (2 Viewers)

Timske

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Re: HSC 2012 Marathon :)

<a href="http://www.codecogs.com/eqnedit.php?latex=\int \frac{x^4}{1@plus;x^{10}} \textup{dx} ~\\ Let ~u = x^5 \\ \frac{du}{dx}= 5x^4 \\dx=\frac{du}{5x^4} \\ \frac{1}{5}\int \frac{1}{1@plus;u^2}~ \textup{dx} = \frac{1}{5}\tan^{-1}u @plus; \textup{C} , ~since~ u ~= x^5 \\\\ \frac{1}{5}\tan^{-1}x^5 @plus; \textup{C}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int \frac{x^4}{1+x^{10}} \textup{dx} ~\\ Let ~u = x^5 \\ \frac{du}{dx}= 5x^4 \\dx=\frac{du}{5x^4} \\ \frac{1}{5}\int \frac{1}{1+u^2}~ \textup{dx} = \frac{1}{5}\tan^{-1}u + \textup{C} , ~since~ u ~= x^5 \\\\ \frac{1}{5}\tan^{-1}x^5 + \textup{C}" title="\int \frac{x^4}{1+x^{10}} \textup{dx} ~\\ Let ~u = x^5 \\ \frac{du}{dx}= 5x^4 \\dx=\frac{du}{5x^4} \\ \frac{1}{5}\int \frac{1}{1+u^2}~ \textup{dx} = \frac{1}{5}\tan^{-1}u + \textup{C} , ~since~ u ~= x^5 \\\\ \frac{1}{5}\tan^{-1}x^5 + \textup{C}" /></a>
 

Carrotsticks

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Re: HSC 2012 Marathon :)

<a href="http://www.codecogs.com/eqnedit.php?latex=\int \frac{x^4}{1@plus;x^{10}} \textup{dx} ~\\ Let ~u = x^5 \\ \frac{du}{dx}= 5x^4 \\dx=\frac{du}{5x^4} \\ \frac{1}{5}\int \frac{1}{1@plus;u^2}~ \textup{dx} = \frac{1}{5}\tan^{-1}u @plus; \textup{C} , ~since~ u ~= x^5 \\\\ \frac{1}{5}\tan^{-1}x^5 @plus; \textup{C}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int \frac{x^4}{1+x^{10}} \textup{dx} ~\\ Let ~u = x^5 \\ \frac{du}{dx}= 5x^4 \\dx=\frac{du}{5x^4} \\ \frac{1}{5}\int \frac{1}{1+u^2}~ \textup{dx} = \frac{1}{5}\tan^{-1}u + \textup{C} , ~since~ u ~= x^5 \\\\ \frac{1}{5}\tan^{-1}x^5 + \textup{C}" title="\int \frac{x^4}{1+x^{10}} \textup{dx} ~\\ Let ~u = x^5 \\ \frac{du}{dx}= 5x^4 \\dx=\frac{du}{5x^4} \\ \frac{1}{5}\int \frac{1}{1+u^2}~ \textup{dx} = \frac{1}{5}\tan^{-1}u + \textup{C} , ~since~ u ~= x^5 \\\\ \frac{1}{5}\tan^{-1}x^5 + \textup{C}" /></a>
Don't really like how you had dx = du/5x^4.... mixing X and U...
 

Kingportable

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Re: HSC 2012 Marathon :)

John Fitzpatrick 3U Mathematics 25(c) Question 3

A particle movies in a straight line. At time t seconds, its displacement x cm from a fixed poin O in the line is given by x=5sin((pi/2)T + pi/6). Express the acceleration in terms of x only and hence show that the motion is simple harmonic. Find:

iii) The speed when x=-2+1/2
iv) the acceleration when x = -2 + 1/2
Stuck on 3 and 4, i tried using v^2=n^2(a^2+x^2) and -n^2 . x
 

Carrotsticks

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Re: HSC 2012 Marathon :)

I've always done it like that :S
Best to avoid it.

du = 5x^4 dx would have sufficed.

Also, I remember I made a nice integration question a while ago and I posted it up somewhere...

It basically was a way of evaluating:



Without actually knowing the integral of ln(x).
 

deswa1

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Re: HSC 2012 Marathon :)

Best to avoid it.

du = 5x^4 dx would have sufficed.

Also, I remember I made a nice integration question a while ago and I posted it up somewhere...

It basically was a way of evaluating:



Without actually knowing the integral of ln(x).
Oooh, if you could find it, can you post it again. A nice question that I remember you posting was how to integrate 1/1+tan^n(x) between 0 and pi/2 but that needs 4U techniques...
 

nightweaver066

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Re: HSC 2012 Marathon :)

Oooh, if you could find it, can you post it again. A nice question that I remember you posting was how to integrate 1/1+tan^n(x) between 0 and pi/2 but that needs 4U techniques...
Not necessarily.

Could be asked in a 3U paper as a challenging question but given the substitution u = pi/2 - x and being broken up in to parts lol.
 

deswa1

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Re: HSC 2012 Marathon :)

Just looking at that wouldnt you make u = x^5 + 5x^2-5x ?
Yep. If you can though, try and learn to do these sorts of things without having to let u=whatever. Something like this you can just recognise after you do a lot of them which saves a lot of time in tests.
 

nightweaver066

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Re: HSC 2012 Marathon :)

Just looking at that wouldnt you make u = x^5 + 5x^2-5x ?
Yep.

lol reason i'm posting these questions is to highlight the fact that questions may look intimidating need not be so hard as you just need to stick to basics.
 

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