HSC 2012 MX1 Marathon #2 (archive) (4 Viewers)

SpiralFlex · Oct 20, 2012

Re: HSC 2012 Marathon

I'll pop by to make some questions later.

twinklegal19 · Oct 20, 2012

Re: HSC 2012 Marathon

$By considering the gradient function of $f(x)=\sin{x}+x-1$, show that $f(x)=0$ has exactly one real root and that this root lies between 0.4 and 0.6. Find this root, to an accuracy of two decimal places, by applying the method of halving the interval.$

GoldyOrNugget · Oct 20, 2012

Re: HSC 2012 Marathon

the derivative cos(x) + 1 is always >= 0 so f(x) is always increasing, so it has at most one root. f(0.4) < 0, f(0.6) > 0, so f(x) = 0 for some 0.4<x<0.6. Then just halve the interval.

How do you do math notation?

Sy123 · Oct 20, 2012

Re: HSC 2012 Marathon

GoldyOrNugget said:
the derivative cos(x) + 1 is always >= 0 so f(x) is always increasing, so it has at most one root. f(0.4) < 0, f(0.6) > 0, so f(x) = 0 for some 0.4<x<0.6. Then just halve the interval.

How do you do math notation?

To do math notation you type normal latex code within the tags [.tex] and [./tex] (without the dots)

Also for the hexagon question I havent quite got the answer yet but Im quite close, I feel like I need to somehow prove that BP=BC since everything falls nicely from there. I havent done it yet but you could consider constructing a circle around the regular hexagon (its justified since it is regular), and then applying circle properties or something.

Sy123 · Oct 20, 2012

Re: HSC 2012 Marathon

Another question I made up:

$$A projectile is fired from the origin with velocity V, under the conditions of gravity g and no air resistance. The projectile is fired such that the maximum height is h, and the horizontal distance from the origin to the maximum height is h.$ \\ \\ $Show that $ V=\sqrt{\frac{5gh}{2}}$

kenkap · Oct 20, 2012

Re: HSC 2012 Marathon

CAN somebody please check my aproach for Sy123's question part (v)

i subbed the equation of normal in equation of the semicircle (function) making a quadratic equation in x...since it only touches the curve, i made the discriminant=0 but it gave me answers of theta=0,90,180 degrees which seems weird??

where am i going wrong?? is my method correct??

Sy123 · Oct 20, 2012

Re: HSC 2012 Marathon

kenkap said:
CAN somebody please check my aproach for Sy123's question part (v)

i subbed the equation of normal in equation of the semicircle (function) making a quadratic equation in x...since it only touches the curve, i made the discriminant=0 but it gave me answers of theta=0,90,180 degrees which seems weird??

where am i going wrong?? is my method correct??

Nope, your method is correct, I did it and it yielded my results (45, 135)

Did you end up with:

$\frac{8\sin^2 \theta-4}{\sin^2 \theta \cos^2 \theta}=0 \Rightarrow \sin^2 \theta=\frac{1}{2} \\ \\ sin \theta=\frac{1}{\sqrt{2}}$

GoldyOrNugget · Oct 20, 2012

Re: HSC 2012 Marathon

For Sy's projectile problem (this is quite a convoluted method):

$The concave-down parabola representing the path of the projectile has a root at x=0 and a maximum at $x=h$. Symmetry of parabolae dictates its roots are 0, 2h. Hence it is of the form $y=-ax(x-2h)$. Furthermore, $x=h$ when $y=h$, so $h=-ah(h-2h)$ \implies $a = \frac{1}{h}$ \implies $y = -\frac{x}{h}(x-2h)$ \\ \\ Now, the standard form for the path of a projectile is $y = x\tan \theta - \frac{gx^2}{2V^2}\sec^2 \theta$. $\tan \theta$ is given by $\frac{\mathrm{d}y}{\mathrm{d}x}$ at $x=0$ so $\tan \theta = 2$. Equating coefficients of $x^2$, $-\frac{g}{2V^2}\sec^2 \theta = -\frac{1}{h}$, so $V^2 = \frac{hg \sec^2 \theta}{2}$, where $\sec^2(\tan^{-1} 2) = 5$. So $V = \sqrt{\frac{5gh}{2}}$. \blacksquare$

Sy123 · Oct 20, 2012

Re: HSC 2012 Marathon

GoldyOrNugget said:
For Sly's projectile problem (this is quite a convoluted method):

$The concave-down parabola representing the path of the projectile has a root at x=0 and a maximum at $x=h$. Symmetry of parabolae dictates its roots are 0, 2h. Hence it is of the form $y=-ax(x-2h)$. Furthermore, $x=h$ when $y=h$, so $h=-ah(h-2h)$ \implies $a = \frac{1}{h}$ \implies $y = -\frac{x}{h}(x-2h)$ \\ \\ Now, the standard form for the path of a projectile is $y = x\tan \theta - \frac{gx^2}{2V^2}\sec^2 \theta$. $\tan \theta$ is given by $\frac{\mathrm{d}y}{\mathrm{d}x}$ at $x=0$ so $\tan \theta = 2$. Equating coefficients of $x^2$, $-\frac{g}{2V^2}\sec^2 \theta = -\frac{1}{h}$, so $V^2 = \frac{hg \sec^2 \theta}{2}$, where $\sec^2(\tan^{-1} 2) = 5$. So $V = \sqrt{\frac{5gh}{2}}$. \blacksquare$

Very nice, new question will come up soon.

Sy123 · Oct 20, 2012

Re: HSC 2012 Marathon

$$A projectile is fired from the origin with velocity V at an angle $ \theta $ to the horizontal, the projectile is fired such that the maximum height is h. T seconds later a projectile is dropped h metres above the point where the first projectile is to reach its maximum height. The particles then collide when the first projectile is at its maximum height$ \\ \\ $Take g as acceleration due to gravity and disregard air-resistance$ \\ \\ $Show that $ T=\sqrt{\frac{2h}{g}}-\frac{V\sin \theta}{g}$

Yes I cant think of any other good questions from other topics, Id rather not have easy questions but sharing questions that are nice and moderate to hard.

GoldyOrNugget · Oct 20, 2012

Re: HSC 2012 Marathon

I think that question is missing information. Right now it's just a scenario with two unrelated projectiles. Do they hit each other or something?

Sy123 · Oct 20, 2012

Re: HSC 2012 Marathon

GoldyOrNugget said:
I think that question is missing information. Right now it's just a scenario with two unrelated projectiles. Do they hit each other or something?

Oh of course, sorry for that, yes the particles collide at maximum height. Sorry for the inconvinience

GoldyOrNugget · Oct 20, 2012

Re: HSC 2012 Marathon

I found that one much easier. Now back to this stupid hexagon >.<

$$Time taken for projectile B to fall h metres is given by $h = \frac{1}{2}gt^2$ \implies $t = \sqrt{\frac{2h}{g}}$. \\ \\ Time taken for projectile A to reach max height is given by $\.{y}=0$ \implies $V \sin \theta - gt = 0$ \implies $t = \frac{V \sin \theta}{g}$. $\therefore$ for collision, projectile B must be dropped $\sqrt{\frac{2h}{g}}$ seconds before $\frac{V \sin \theta}{g}$ i.e. at $t = \frac{V \sin \theta}{g} - \sqrt{\frac{2h}{g}}$. So the time \textbf{after} the first particle is launched is given by the negation of this, i.e. $T = \sqrt{\frac{2h}{g}} - \frac{V \sin \theta}{g}$. \blacksquare$

EDIT: derp, got the terms the wrong way around. fixed now.

Sy123 · Oct 20, 2012

Re: HSC 2012 Marathon

GoldyOrNugget said:
I found that one much easier. Now back to this stupid hexagon >.<

$$Time taken for projectile B to fall h metres is given by $h = \frac{1}{2}gt^2$ \implies $t = \sqrt{\frac{2h}{g}}$. \\ \\ Time taken for projectile A to reach max height is given by $\.{y}=0$ \implies $V \sin \theta - gt = 0$ \implies $t = \frac{V \sin \theta}{g}$. $\therefore$ for collision, projectile B must be dropped $\sqrt{\frac{2h}{g}}$ seconds before $\frac{V \sin \theta}{g}$ i.e. at $t = \frac{V \sin \theta}{g} - \sqrt{\frac{2h}{g}}$. So the time \textbf{after} the first particle is launched is given by the negation of this, i.e. $T = \sqrt{\frac{2h}{g}} - \frac{V \sin \theta}{g}$. \blacksquare$

EDIT: derp, got the terms the wrong way around. fixed now.

Very nice way of looking at it, I did it by treating (t+T) as a single parameter and found equation of projectile B then solved from there.

And I too must get back to the hexagon, post a solution if you find it out

GoldyOrNugget · Oct 21, 2012

Re: HSC 2012 Marathon

If you represent the hexagon's vertices as points on a plane and do lots of coordinate geometry, you obtain the following expression for the angle:

$\theta = \tan^{-1}(\frac{\frac{3}{2}\tan10^{\circ}\tan50^{\circ} + \frac{\sqrt{3}}{2}\tan50^{\circ}}{\tan 50^{\circ} - \frac{\sqrt{3}}{2} - \frac{1}{2} \tan 10^{\circ}}})$

And if you type that into a calculator, it does indeed give 80^o. I don't know how to simplify it manually though.

Sy123 · Oct 21, 2012

Re: HSC 2012 Marathon

$$Using the definition of e$ \\ \\ e=\lim_{n \to \infty}(1+\frac{1}{n})^n \\ \\ $Prove from first principles $ \frac{d}{dx}(\ln x)=\frac{1}{x} \\ \\ $Use the substitution $ u=\frac{h}{x} \ \ \ $for $ f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$

Carrotsticks · Oct 21, 2012

Re: HSC 2012 Marathon

Okay worked out the hexagon Q.

HINT: Construct AC and it follows that angle BAC = angle BCA = 30 degrees (isosceles triangle). Work from there and focus purely on ABCP.

seanieg89 · Oct 21, 2012

Re: HSC 2012 Marathon

First part of this question is a 3U induction if you replace the word complex with real

.

http://community.boredofstudies.org/showthread.php?t=292775

Sy123 · Oct 21, 2012

Re: HSC 2012 Marathon

seanieg89 said:
First part of this question is a 3U induction if you replace the word complex with real .

http://community.boredofstudies.org/showthread.php?t=292775

When you define u' does it mean this:

$u=(u_1,u_2,u_3...) \\ \\ u'=(u_1',u_2',u_3'...)$

If it means this then I will have a go at the question first thing tomorrow.

seanieg89 · Oct 21, 2012

Re: HSC 2012 Marathon

Yep, exactly. The question itself is a pretty standard induction...but it raises other interesting questions about this construction.

HSC 2012 MX1 Marathon #2 (archive) (4 Viewers)

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