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

deswa1 · Feb 8, 2012

Re: 2012 HSC MX1 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=(x-3)^2@plus;y^2=1 \\ (x-3)^2=1-y^2 \\ (x-3)=\pm \sqrt(1-y^2) \\ x=3\pm \sqrt(1-y^2)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?(x-3)^2+y^2=1 \\ (x-3)^2=1-y^2 \\ (x-3)=\pm \sqrt(1-y^2) \\ x=3\pm \sqrt(1-y^2)" title="(x-3)^2+y^2=1 \\ (x-3)^2=1-y^2 \\ (x-3)=\pm \sqrt(1-y^2) \\ x=3\pm \sqrt(1-y^2)" /></a>

Examine · Feb 8, 2012

Re: 2012 HSC MX1 Marathon

$Tan^4\Theta =tan^2\Theta sec^2\Theta -sec^2\Theta +1$

How can you make the LHS equal the RHS? It's in my homework and I can't figure it out :/

SpiralFlex · Feb 8, 2012

Re: 2012 HSC MX1 Marathon

Examine said:
$Tan^4\Theta =tan^2\Theta sec^2\Theta -sec^2\Theta +1$

How can you make the LHS equal the RHS? It's in my homework and I can't figure it out :/

$RHS=\tan^2 x(1+\tan^2 x)-(1+\tan^2 x)+1$

$=\tan^2 x+\tan^4 x- 1 -\tan^2 x+1$

$=\tan^4 x$

$=LHS$

Used x instead of theta.

Carrotsticks · Feb 8, 2012

Re: 2012 HSC MX1 Marathon

Examine said:
$Tan^4\Theta =tan^2\Theta sec^2\Theta -sec^2\Theta +1$

How can you make the LHS equal the RHS? It's in my homework and I can't figure it out :/

$\\ \begin{align*} RHS &= tan^{2} \theta \sec ^{2} \theta - sec^{2} \theta + 1 \\\\ &= tan^{2} \theta \left ( tan^{2} \theta + 1 \right ) - \left ( tan^{2} \theta + 1 \right ) + 1 \\\\ &= tan^{4} \theta + tan^{2} \theta - tan^{2} \theta -1 + 1 \\\\ &= tan^{4} \theta \\\\ &= LHS \end{align*}$

Carrotsticks · Feb 8, 2012

Re: 2012 HSC MX1 Marathon

Is it just me, or is the 'Preview Post' for LaTeX very laggy?

math man · Feb 8, 2012

Re: 2012 HSC MX1 Marathon

yeh its bad

SpiralFlex · Feb 8, 2012

Re: 2012 HSC MX1 Marathon

Yes don't like it.

math man · Feb 8, 2012

Re: 2012 HSC MX1 Marathon

$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! LHS =tan^{4}x \\ = (tan^{2}x)^{2} \\ = (sec^{2}x -1)^{2} \\ = sec^{4}x- 2sec^{2}x + 1 \\ = sec^{2}x(sec^{2}x-2) +1 \\ = sec^{2}x(tan^{2}x - 1) +1 \\ = sec^{2}xtan^{2}x - sec^{2}x + 1 \\ = RHS$

Skeptyks · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

One question that's quite irrelevant, sorry, how do you make the latex "next line" thing so neat?
E.g. \\ for next line, but it looks really uneven for some reason. How do you do a double spacing?
Also, if I type something that is not latex and then latex underneath, how do I stop it from being so close to each other (lines 1 and 2 from http://community.boredofstudies.org/showthread.php?t=274161&p=5722682&viewfull=1#post5722682)

Carrotsticks · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

Skeptyks said:
One question that's quite irrelevant, sorry, how do you make the latex "next line" thing so neat?
E.g. \\ for next line, but it looks really uneven for some reason. How do you do a double spacing?
Also, if I type something that is not latex and then latex underneath, how do I stop it from being so close to each other (lines 1 and 2 from http://community.boredofstudies.org/showthread.php?t=274161&p=5722682&viewfull=1#post5722682)

$\\ $Here is line A$ \\ $Here is line B using only two slashes$ \\\\ $Here is line C using 4 slashes$$

Timske · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

Can someone explain the concept of subtraction of volumes when finding the volumes about the x-axis and y-axis on the same curve. I know for some of these questions you dont need to subtract any volume but why is that?

Nooblet94 · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

deswa1 said:
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

If it's not too much trouble, could somebody please post the working out for this question? I can evaluate the integral, but I can't get from the question to the integral itself.

bleakarcher · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

^ Don't worry lol, there is a whole 4U topic to explain that for you.

mnmaa · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

(x-3)^2 +y^2 =1
can be rearranged to x^2=1-y^2 +6(1-y^2)^(0.5) +9.
this in turn can be turned into (X^2)/6=10/6-(y^2)/6 + (1-Y^2)^0.5
from there it is self explanatory. I cant use latex because im retarded so sorry

Timske · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

$\int y\sqrt{y+1}\ dy, \ \textrm{using}\ \textrm{the}\ \textrm{substitution}\ u=x-1$

deswa1 · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

How do you use the substitution u=x-1 when there are no x's in the expression? I did the question though using the substitution u=y+1:
<a href="http://www.codecogs.com/eqnedit.php?latex=u=y@plus;1\\\frac{du}{dy}=1\\ \int y\sqrt(y@plus;1)=\int (u-1)\sqrt u\\ =\frac{2u^{\frac{5}{2}}}{5}- \frac{2u^{\frac{3}{2}}}{3}@plus;C \\=\frac{2(y@plus;1)^{\frac{5}{2}}}{5}-\frac{2(y@plus;1)^{\frac{3}{2}}}{3}@plus;C" target="_blank"><img src="http://latex.codecogs.com/gif.latex?u=y+1\\\frac{du}{dy}=1\\ \int y\sqrt(y+1)=\int (u-1)\sqrt u\\ =\frac{2u^{\frac{5}{2}}}{5}- \frac{2u^{\frac{3}{2}}}{3}+C \\=\frac{2(y+1)^{\frac{5}{2}}}{5}-\frac{2(y+1)^{\frac{3}{2}}}{3}+C" title="u=y+1\\\frac{du}{dy}=1\\ \int y\sqrt(y+1)=\int (u-1)\sqrt u\\ =\frac{2u^{\frac{5}{2}}}{5}- \frac{2u^{\frac{3}{2}}}{3}+C \\=\frac{2(y+1)^{\frac{5}{2}}}{5}-\frac{2(y+1)^{\frac{3}{2}}}{3}+C" /></a>

Nooblet94 · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

bleakarcher said:
^ Don't worry lol, there is a whole 4U topic to explain that for you.

Oh right, I'd assumed that since it was in the MX1 section you would be able to do it using MX1 methods. (Nevertheless, I still can't do it

)

deswa1 · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

Nooblet94 said:
Oh right, I'd assumed that since it was in the MX1 section you would be able to do it using MX1 methods. (Nevertheless, I still can't do it )

Its the last extension question in the Cambridge 3U book so you can do it, its just way harder than anything you'll ever see in the HSC. I won't do the whole thing for you, but I'll try and give you a hint and see if you can do the rest...

Because I can't draw a diagram, I'm going to call the side of the circle closer to the y axis CLOSE and the other side FAR. If you want to find the volume of the circle, what you're essentially doing is finding the volume when you rotate FAR around the y axis and then subtracting the volume you get when you rotate CLOSE around the y axis. Now, to evaluate this integral, you need to find an expression for FAR and for CLOSE. If you solve for x in the equation of the circle, one of the values will be FAR and the other CLOSE. You can work out which ones which. Then just apply the standard formula:

<a href="http://www.codecogs.com/eqnedit.php?latex=V=\pi\int_{a}^{b}(x_{2}^2-x_{1}^2)dy" target="_blank"><img src="http://latex.codecogs.com/gif.latex?V=\pi\int_{a}^{b}(x_{2}^2-x_{1}^2)dy" title="V=\pi\int_{a}^{b}(x_{2}^2-x_{1}^2)dy" /></a>

and you should get the answer... Tell me how you go

Timske · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

integral of e^x2 is 2xe^x2 right?

cutemouse · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

Timske said:
integral of e^x2 is 2xe^x2 right?

uh...

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