HSC 2012 MX1 Marathon #2 (archive) (1 Viewer)

[Nooblet94]

Re: HSC 2012 Marathon

Might have to be a bit more specific about the equation of the cone. Best to use spherical coordinates but then that would make it out of syllabus.

So the best way to properly define the question is probably to give us the cross sectional angle of the cone.

y=|kx| rotated about the y axis

 

[barbernator]

Re: HSC 2012 Marathon

i will do the question tomorrow, but it will be intriguing, as you will have to find a relationship between the unit circle, and the line kx, so you will have to find the centre of the unit circle using perp. distance formula, and then integrate both and subtract appropriately. You could also use 4u cylindrical shells method which might be nice...

 

[barbernator]

Re: HSC 2012 Marathon

y=|kx| rotated about the y axis

that is a quality question!!!! did it, and i think the solution is correct as the lim k-->0 worked

For others, the question is find the volume of the area between the line y=kx rotated about the y axis, and a 1 unit radius sphere "dropped" in the cone formed. Find the general solution for all values k>0.

My answer was.

<a href="http://www.codecogs.com/eqnedit.php?latex=V=2\pi k\left ( \frac{k^2}{2\sqrt{1@plus;k^2}}@plus;\frac{1-k^4}{3\sqrt{(1@plus;k^2)^{\frac{3}{2}}}}-\frac{1}{3} \right )~units^{3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?V=2\pi k\left ( \frac{k^2}{2\sqrt{1+k^2}}+\frac{1-k^4}{3\sqrt{(1+k^2)^{\frac{3}{2}}}}-\frac{1}{3} \right )~units^{3}" title="V=2\pi k\left ( \frac{k^2}{2\sqrt{1+k^2}}+\frac{1-k^4}{3\sqrt{(1+k^2)^{\frac{3}{2}}}}-\frac{1}{3} \right )~units^{3}" /></a>

brilliant question even though u probably made it up off the cuff.

It is quite difficult, but rewarding. I used the method of cylindrical shells though, from the 4u course, but you could also subtract the volumes using 3u methods.

 

[Carrotsticks]

Re: HSC 2012 Marathon

I love these kinds of questions too.

Just think of 'random' scenarios and then derive cool results from it!

Then once you've got a result, generalise the problem!

 

[barbernator]

Re: HSC 2012 Marathon

The line y=mx-m intersects the curve sin(x) at n distinct points. For what values of m in the equation sin(x)=mx-m will n=9

 

[Nooblet94]

Re: HSC 2012 Marathon

I think I've got a vague idea of how to do it... Would i be correct in saying there's 4 possible values for m?

Wait, no there's not. Don't worry.

 

[barbernator]

Re: HSC 2012 Marathon

btw, i just made the question up from before, so while there is definitely a solution, it may be irrational and we may not be able to find it by 3 or 4u means. You can just give solutions to a few dp if you do get a solution. i dont have one myself yet.

 

[Nooblet94]

Re: HSC 2012 Marathon

I just did my question, and got a slightly different result to yours, barb. Can someone else give it a go and see what they get?

 

[barbernator]

Re: HSC 2012 Marathon

I just did my question, and got a slightly different result to yours, barb. Can someone else give it a go and see what they get?

post up your result and compare

 

[Nooblet94]

Re: HSC 2012 Marathon

post up your result and compare

<a href="http://www.codecogs.com/eqnedit.php?latex=V=2\pi k \left (\frac{k^2}{2\sqrt{1@plus;k^2}} @plus;\frac{1}{3(1@plus;k^2)^\frac{3}{2}}-\frac{1}{3}\right )" target="_blank"><img src="http://latex.codecogs.com/gif.latex?V=2\pi k \left (\frac{k^2}{2\sqrt{1+k^2}} +\frac{1}{3(1+k^2)^\frac{3}{2}}-\frac{1}{3}\right )" title="V=2\pi k \left (\frac{k^2}{2\sqrt{1+k^2}} +\frac{1}{3(1+k^2)^\frac{3}{2}}-\frac{1}{3}\right )" /></a>

From what I can see, one of us has messed up subbing in for x.

 

[barbernator]

Re: HSC 2012 Marathon

if you did cylindrical shells method, did u get

<a href="http://www.codecogs.com/eqnedit.php?latex=h=\sqrt{1@plus;k^2}-\sqrt{1-x^2}-kx\\ r=kx\\ V=2\pi k\int_{0}^{\frac{k}{\sqrt{1@plus;k^2}}}()x\sqrt{1@plus;k^2}-x\sqrt{1-x^2}-kx^2)dx" target="_blank"><img src="http://latex.codecogs.com/gif.latex?h=\sqrt{1+k^2}-\sqrt{1-x^2}-kx\\ r=kx\\ V=2\pi k\int_{0}^{\frac{k}{\sqrt{1+k^2}}}()x\sqrt{1+k^2}-x\sqrt{1-x^2}-kx^2)dx" title="h=\sqrt{1+k^2}-\sqrt{1-x^2}-kx\\ r=kx\\ V=2\pi k\int_{0}^{\frac{k}{\sqrt{1+k^2}}}()x\sqrt{1+k^2}-x\sqrt{1-x^2}-kx^2)dx" /></a>

 

[barbernator]

Re: HSC 2012 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=V=2\pi k \left (\frac{k^2}{2\sqrt{1@plus;k^2}} @plus;\frac{1}{3(1@plus;k^2)^\frac{3}{2}}-\frac{1}{3}\right )" target="_blank"><img src="http://latex.codecogs.com/gif.latex?V=2\pi k \left (\frac{k^2}{2\sqrt{1+k^2}} +\frac{1}{3(1+k^2)^\frac{3}{2}}-\frac{1}{3}\right )" title="V=2\pi k \left (\frac{k^2}{2\sqrt{1+k^2}} +\frac{1}{3(1+k^2)^\frac{3}{2}}-\frac{1}{3}\right )" /></a>

From what I can see, one of us has messed up subbing in for x.

also, this doesn't work for lim k-->0 = 0.

 

[Timske]

Re: HSC 2012 Marathon

What topic is this under

 

[Nooblet94]

Re: HSC 2012 Marathon

if you did cylindrical shells method, did u get

<a href="http://www.codecogs.com/eqnedit.php?latex=h=\sqrt{1@plus;k^2}-\sqrt{1-x^2}-kx\\ r=kx\\ V=2\pi k\int_{0}^{\frac{k}{\sqrt{1@plus;k^2}}}()x\sqrt{1@plus;k^2}-x\sqrt{1-x^2}-kx^2)dx" target="_blank"><img src="http://latex.codecogs.com/gif.latex?h=\sqrt{1+k^2}-\sqrt{1-x^2}-kx\\ r=kx\\ V=2\pi k\int_{0}^{\frac{k}{\sqrt{1+k^2}}}()x\sqrt{1+k^2}-x\sqrt{1-x^2}-kx^2)dx" title="h=\sqrt{1+k^2}-\sqrt{1-x^2}-kx\\ r=kx\\ V=2\pi k\int_{0}^{\frac{k}{\sqrt{1+k^2}}}()x\sqrt{1+k^2}-x\sqrt{1-x^2}-kx^2)dx" /></a>

Ahh shit. Left out kx on the height. Amateur mistake.

What topic is this under

4 unit volumes, although you could do it with 3 unit knowledge, it'd just be a massive pain.

 

[barbernator]

Re: HSC 2012 Marathon

What topic is this under

volumes. If u do 4u, do it by the cylindrical shells method. If you do 3u, subtract a couple of the volumes created from each other.

 

[Timske]

Re: HSC 2012 Marathon

Ahh i see it looked complicated

 

[barbernator]

Re: HSC 2012 Marathon

basically if you can do that question, you should be able to get most HSC volumes questions.

 

[barbernator]

Re: HSC 2012 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=solve~for~x: x^2@plus;2x-4@plus;\frac{3}{x^2@plus;2x}=0" target="_blank"><img src="http://latex.codecogs.com/gif.latex?solve~for~x: x^2+2x-4+\frac{3}{x^2+2x}=0" title="solve~for~x: x^2+2x-4+\frac{3}{x^2+2x}=0" /></a>

this is the very first question in the 1986 CSSA 3u paper. Quite difficult for Q1.

 

[Sy123]

Re: HSC 2012 Marathon

let u=x^2+2x, solve from there.

Some probably will miss that but yeah.

My solutions are

x=-3, 1, -1+2^(1/2) , -1 - 2^(1/2)

 

[Fus Ro Dah]

Re: HSC 2012 Marathon

let u=x^2+2x, solve from there.

Some probably will miss that but yeah.

My solutions are

x=-3, 1, -1+2^(1/2) , -1 - 2^(1/2)

Solution can be found more easily by realising that we are solving two even functions simultaneously, and we can take advantage of that because they both share x=-1 as their axes of symmetry.