HSC 2012 MX1 Marathon #1 (archive) (3 Viewers)

[deswa1]

Re: 2012 HSC MX1 Marathon

Just a quick Q. If a question in a three unit paper says solve this integral using the substitution... or otherwise, can you use integration by parts or are you limited to 3U methods?

 

[math man]

Re: 2012 HSC MX1 Marathon

if the question adds otherwise then you are free to do it any other method, but keep in
mind it is still 3 unit so the question is designed to be done in 3u ways and is most likely
easier that way.

 

[SpiralFlex]

Re: 2012 HSC MX1 Marathon

There was one question for 2U in the HSC dealing with exponentials and the marking notes said it was okay to use integration by parts.

"On rare occasions students demonstrated their knowledge of integration by parts."

 

[deswa1]

Re: 2012 HSC MX1 Marathon

There was one question for 2U in the HSC dealing with exponentials and the marking notes said it was okay to use integration by parts.

"On rare occasions students demonstrated their knowledge of integration by parts."

Which paper was that? I can imagine a sizable number of 3U candidates doing IBP as they might have learnt it in 4U but how many 2U students have come across it (unless they were accelerated and planned to do 4U)?

 

[SpiralFlex]

Re: 2012 HSC MX1 Marathon

Which paper was that? I can imagine a sizable number of 3U candidates doing IBP as they might have learnt it in 4U but how many 2U students have come across it (unless they were accelerated and planned to do 4U)?

I would assume accelerants.

I don't remember, but I helped someone on BOS with that question. I will go dig it up.

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

Which paper was that? I can imagine a sizable number of 3U candidates doing IBP as they might have learnt it in 4U but how many 2U students have come across it (unless they were accelerated and planned to do 4U)?

One of my students is in Year 11, and has already learnt IBP.

 

[cutemouse]

Re: 2012 HSC MX1 Marathon

Which paper was that? I can imagine a sizable number of 3U candidates doing IBP as they might have learnt it in 4U but how many 2U students have come across it (unless they were accelerated and planned to do 4U)?

Like 2007 or something. It was the last question (Q10) that asked to find the area underneath a log graph.

 

[SpiralFlex]

Re: 2012 HSC MX1 Marathon

Like 2007 or something. It was the last question (Q10) that asked to find the area underneath a log graph.

YES THAT ONE!

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

 

[Nooblet94]

Re: 2012 HSC MX1 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\\\textrm{Equation of the tangent:}~y=tx-at^2\\~\\ \textrm{Finding points of intersection:}\\ x^2=-4a(tx-at^2)\\ x^2@plus;4atx-4a^2t^2=0\\ \begin{align*} x&=\frac{-4at \pm \sqrt{(4at)^2-4(-4a^2t^2)}}{2}\\ &=\frac{-4at \pm 4\sqrt{2}at}{2}\\ &=2at(-1 \pm 2\sqrt{2}) \end{align*}\\ \therefore \textrm{the x-coordinate of the midpoint of PQ is:}\\ \frac{2at(-1@plus;\sqrt{2})@plus;2at(-1-\sqrt{2})}{2}=-2at\\ ~\\ y=\frac{x^2}{4a}=\frac{4a^2t^2(-1 \pm \sqrt{2})^2}{4a}=at^2(-1 \pm \sqrt{2})^2\\ ~\\ \therefore\textrm{the y-coordinate of the midpoint of PQ is:}\\ \frac{at^2(-1@plus;\sqrt{2})^2@plus;at^2(-1-\sqrt{2})^2}{2}=3at^2\\ ~\\ t=\frac{x}{-2a}\\ \therefore y=3a(\frac{x}{-2a})^2=\frac{3x^2}{4a}\\ x^2=\frac{4a}{3}y ~\\ \textrm{The locus of the midpoint of PQ is a parabola with vertex (0,0) and focal length}~\frac{a}{3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\\textrm{Equation of the tangent:}~y=tx-at^2\\~\\ \textrm{Finding points of intersection:}\\ x^2=-4a(tx-at^2)\\ x^2+4atx-4a^2t^2=0\\ \begin{align*} x&=\frac{-4at \pm \sqrt{(4at)^2-4(-4a^2t^2)}}{2}\\ &=\frac{-4at \pm 4\sqrt{2}at}{2}\\ &=2at(-1 \pm 2\sqrt{2}) \end{align*}\\ \therefore \textrm{the x-coordinate of the midpoint of PQ is:}\\ \frac{2at(-1+\sqrt{2})+2at(-1-\sqrt{2})}{2}=-2at\\ ~\\ y=\frac{x^2}{4a}=\frac{4a^2t^2(-1 \pm \sqrt{2})^2}{4a}=at^2(-1 \pm \sqrt{2})^2\\ ~\\ \therefore\textrm{the y-coordinate of the midpoint of PQ is:}\\ \frac{at^2(-1+\sqrt{2})^2+at^2(-1-\sqrt{2})^2}{2}=3at^2\\ ~\\ t=\frac{x}{-2a}\\ \therefore y=3a(\frac{x}{-2a})^2=\frac{3x^2}{4a}\\ x^2=\frac{4a}{3}y ~\\ \textrm{The locus of the midpoint of PQ is a parabola with vertex (0,0) and focal length}~\frac{a}{3}" title="\\\textrm{Equation of the tangent:}~y=tx-at^2\\~\\ \textrm{Finding points of intersection:}\\ x^2=-4a(tx-at^2)\\ x^2+4atx-4a^2t^2=0\\ \begin{align*} x&=\frac{-4at \pm \sqrt{(4at)^2-4(-4a^2t^2)}}{2}\\ &=\frac{-4at \pm 4\sqrt{2}at}{2}\\ &=2at(-1 \pm 2\sqrt{2}) \end{align*}\\ \therefore \textrm{the x-coordinate of the midpoint of PQ is:}\\ \frac{2at(-1+\sqrt{2})+2at(-1-\sqrt{2})}{2}=-2at\\ ~\\ y=\frac{x^2}{4a}=\frac{4a^2t^2(-1 \pm \sqrt{2})^2}{4a}=at^2(-1 \pm \sqrt{2})^2\\ ~\\ \therefore\textrm{the y-coordinate of the midpoint of PQ is:}\\ \frac{at^2(-1+\sqrt{2})^2+at^2(-1-\sqrt{2})^2}{2}=3at^2\\ ~\\ t=\frac{x}{-2a}\\ \therefore y=3a(\frac{x}{-2a})^2=\frac{3x^2}{4a}\\ x^2=\frac{4a}{3}y ~\\ \textrm{The locus of the midpoint of PQ is a parabola with vertex (0,0) and focal length}~\frac{a}{3}" /></a>

I think that's correct, although I haven't checked it for mistakes.

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

Nooblet, might I suggest something to you. This is a very useful trick that will save you a lot of time during exams.

You wanted to find the midpoint of PQ. To do this, you combined the equations for the parabola and the line simultaneously. Afterwards, you used the quadratic formula, added the two solutions, then divided by two.

However if you didn't notice, the two x co-ordinates of the points of intersections are the roots of the quadratic that you acquired by combining the two equations.

What you can recognise is that the midpoint of the x coordinates is the 'average' of the x coordinates.

The 'average' as taught in Year 7 (or earlier) can be found by adding the two numbers, then dividing them by 2.

Instead of actually finding both the points of intersections, then adding them, you could have used the 'Sum of Roots' method on that quadratic, which would have acquired the 'Adding of the two numbers' as I mentioned above.

Now, all you need to do is divide by two!

Allow me to demonstrate... Maths is better shown than said:

 

[AAEldar]

Re: 2012 HSC MX1 Marathon

That is genius... I have never thought of doing that before :\

 

[deswa1]

Re: 2012 HSC MX1 Marathon

I like .

Don't you still have to solve the quadratic though to find the y values or is there another hack that I'm not aware of?

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

Don't you still have to solve the quadratic though to find the y values or is there another hack that I'm not aware of?

Once you find the midpoint of the interval, all you need is the y value of the same interval, which can easily be found by substituting that x coordinate -2at into the equation of the tangent (since the midpoint of PQ lies on the tangent).

Alternatively, you could make another quadratic as Nooblet94 did. But instead of it being in terms of x, you could make it in terms of y, then use that sum of roots method again to find the y coordinate of the midpoint.

 

[deswa1]

Re: 2012 HSC MX1 Marathon

Mate, you are a god! What did you get in the HSC for maths?

 

[Nooblet94]

Re: 2012 HSC MX1 Marathon

Mate, you are a god!

I concur

 

[deswa1]

Re: 2012 HSC MX1 Marathon

New question:

The region inside the circle <a href="http://www.codecogs.com/eqnedit.php?latex=(x-3)^2@plus;y^2=1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?(x-3)^2+y^2=1" title="(x-3)^2+y^2=1" /></a> is rotated about the y-axis. Show that the volume of the solid formed is given by <a href="http://www.codecogs.com/eqnedit.php?latex=V=24\pi\int_{0}^{1}(1-y^2)^{1/2}dy" target="_blank"><img src="http://latex.codecogs.com/gif.latex?V=24\pi\int_{0}^{1}(1-y^2)^{1/2}dy" title="V=24\pi\int_{0}^{1}(1-y^2)^{1/2}dy" /></a> and evaluate the integral.

Sorry about the way the latex appeared... I still haven't worked out how to make it neat, or even type in latex

 

[Timske]

Re: 2012 HSC MX1 Marathon

is the integral 12pi^2

 

[bleakarcher]

Re: 2012 HSC MX1 Marathon

naa its pi^2/2

 

[largarithmic]

Re: 2012 HSC MX1 Marathon

naa its pi^2/2

Timske's answer is correct