HSC 2012 MX1 Marathon #2 (archive) (8 Viewers)

[nightweaver066]

Re: HSC 2012 Marathon

How about..

 

[barbernator]

Re: HSC 2012 Marathon

Mind if I extend the question a bit?

k! - 1

 

[Carrotsticks]

Re: HSC 2012 Marathon

How about..

Herp derp let z=9sintheta because obviously there's no geometric way of doing it =p

 

[SpiralFlex]

Re: HSC 2012 Marathon

herp derp let z=9sintheta because obviously there's no geometric way of doing it =p

spoiler alert noob!

 

[Timske]

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>

 

[barbernator]

Re: HSC 2012 Marathon

Herp derp let z=9sintheta because obviously there's no geometric way of doing it =p

yeh quadrants of circles dont exist

 

[Carrotsticks]

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...

 

[Timske]

Re: HSC 2012 Marathon

Don't really like how you had dx = du/5x^4.... mixing X and U...

I've always done it like that :S

 

[Kingportable]

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]

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).

 

[nightweaver066]

Re: HSC 2012 Marathon

yeh quadrants of circles dont exist

Who said anything about a circle? And quadrants?

 

[Timske]

Re: HSC 2012 Marathon

Stuck on 3 and 4, i tried using v^2=n^2(a^2+x^2) and -n^2 . x

It's v^2=n^2(a^2-x^2)

 

[nightweaver066]

Re: HSC 2012 Marathon

Here's a really tricky one,

 

[deswa1]

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...

 

[barbernator]

Re: HSC 2012 Marathon

Who said anything about a circle? And quadrants?

not sure if srs -_-

 

[deswa1]

Re: HSC 2012 Marathon

Here's a really tricky one,

This isn't that hard is it? The left of the e is just the derivative of the exponent divided by 5.

 

[Timske]

Re: HSC 2012 Marathon

Here's a really tricky one,

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

 

[nightweaver066]

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]

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]

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.