HSC 2012 MX2 Marathon (archive) (1 Viewer)

[Trebla]

Re: 2012 HSC MX2 Marathon

Hahahahah novice error by Trebla!

lol, I'm getting old...

 

[someth1ng]

Re: 2012 HSC MX2 Marathon

Ewwww

I'm lost. I know that it works but how did you do it like that?

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

I'm lost. I know that it works but how did you do it like that?

Substitution method u = 1-x^2.

After skipping a few steps too. Once you've done a fair few substitution method questions, you can jump from the question directly to the answer.

 

[someth1ng]

Re: 2012 HSC MX2 Marathon

Substitution method u = 1-x^2.

I see, how do you know which substitution to use? I probably don't have a chance at it since my school is doing some harder 3U topic first before integration.

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

Usually, it is let u = *inside of the square root* for the basic questions.

Alternatively, you could let u^2 = inside of square root.

However you cannot blindly do this. You have to see that it can lead to the correct answer, and this can only be done by carefully observing the integral and doing lots of questions to familiarise yourself with such questions.

 

[IamBread]

Re: 2012 HSC MX2 Marathon

You should recognise it's almost f'(x)/sqrt[f(x)], and with a little manipulation you can get it there, once it's like that you can just quickly integrate. But as Carrotsticks said, you need to do a lot of questions to get yourself familiar to this.

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

uhhhhh... how can this be true? the right hand series (the harmonic series) diverges

Some definite integrals can be divergent, one of the most trivial cases being:



It just so happens that the one Trebla provided was convergent that it was not true.

Trebla, your question looks very interesting. Will try.

 

[IamBread]

Re: 2012 HSC MX2 Marathon

Trebla's question didn't work as at pi/2 it is undefined.

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

Trebla's question didn't work as at pi/2 it is undefined.

Trebla's original question with limits 0 --> pi/2 was actually convergent. Just because a point is undefined, does not necessarily mean that the area is automatically infinite.

For example, consider this integral:



The above integral in fact converges for all p E (0,1)

These integrals (sometimes called the p-integrals) are perfect examples of this.

Even though the function may not be defined, the area may well be.

An interesting figure sorta related to this is Gabriel's Horn: http://en.wikipedia.org/wiki/Gabriel's_Horn

Oh and I forgot to put the dx in the first p integral I provided.

 

[IamBread]

Re: 2012 HSC MX2 Marathon

That does make sense, though you wouldn't be able to figure it out using conventional methods, (as in you can't just sub the number in), which what I was trying to point out lol.

 

[SpiralFlex]

Re: 2012 HSC MX2 Marathon

You should recognise it's almost f'(x)/sqrt[f(x)], and with a little manipulation you can get it there, once it's like that you can just quickly integrate. But as Carrotsticks said, you need to do a lot of questions to get yourself familiar to this.

This is the way.

 

[math man]

Re: 2012 HSC MX2 Marathon

I see, how do you know which substitution to use? I probably don't have a chance at it since my school is doing some harder 3U topic first before integration.

It was done using the reverse power
Rule, the best tool in 4u integration,
Substitution is hardly ever needed with
It

 

[math man]

Re: 2012 HSC MX2 Marathon

The reverse power rule is:

 

[math man]

Re: 2012 HSC MX2 Marathon

The skill to using the reverse chain rule
is to be very good at recognising powers of
functions and their derivatives.

 

[SpiralFlex]

Re: 2012 HSC MX2 Marathon

Some Conics...













.

For part alpha I can't work out why m<0. I can see it must be the case for the rest of the question because of the domain.

You can do it.

Other attempts pl0x. I like this question.

 

[Nooblet94]

Re: 2012 HSC MX2 Marathon

You can do it.

Other attempts pl0x. I like this question.

I enjoyed the question too - I did the rest of it last night after skipping the first part (which I still can't get )

 

[SpiralFlex]

Re: 2012 HSC MX2 Marathon

Use the force.

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

I did the rest of it last night after skipping the first part (which I still can't get )

Substitute y=mx+c into the equation of the semi-ellipse and make a quadratic in terms of x.

Let the discriminant be equal to zero and after a whole lot of algebra, you should get the solution.

However, I do not understand why m < 0 is a condition. I do not see why positive gradients are excluded as possible tangents.

This is only the case if we purely consider the first quadrant of the Ellipse in canonical form and the equation you provided covers the first 2 quadrants.

 

[Nooblet94]

Re: 2012 HSC MX2 Marathon

Substitute y=mx+c into the equation of the semi-ellipse and make a quadratic in terms of x.

Let the discriminant be equal to zero and after a whole lot of algebra, you should get the solution.

However, I do not understand why m < 0 is a condition. I do not see why positive gradients are excluded as possible tangents.

This is only the case if we purely consider the first quadrant of the Ellipse in canonical form and the equation you provided covers the first 2 quadrants.

That's exactly what I was struggling with (why m<0), I got the rest of it, I just assumed I was missing something since Spiral's almost always correct and nobody else had mentioned anything.

 

[IamBread]

Re: 2012 HSC MX2 Marathon

For part alpha I can't work out why m<0.

Lol I'm stuck at this point too, can't work that out.