leehuan's All-Levels-Of-Maths SOS thread (11 Viewers)

leehuan · Feb 25, 2016

Only 3d) please. Missing the trick.

InteGrand · Feb 26, 2016

leehuan said:
Only 3d) please. Missing the trick.

$\noindent The pyramid $\mathcal{P}$ has volume $V_{1}=nV = \frac{r^2 h}{6} n\sin \frac{2\pi}{n}$. Since the volume of the cone is $V_2 = \frac{\pi r^2 h}{3}$, we have the ratio of the volumes $\rho $ being $\rho \equiv \rho_n :=\frac{V_1}{V_2}= \frac{\frac{r^2 h}{6} n\sin \frac{2\pi}{n}}{\frac{\pi r^2 h}{3}} =\frac{n\sin \frac{2\pi}{n}}{2\pi}$. Now, the sequence of $\rho_n$'s is monotonically increasing. This can be proved by showing that the function $f(x):= x\sin \frac{1}{x}$ is monotone increasing for $x\geq 1$, which we will do at the end. Now, we want $n$ such that $\rho _n \geq 0.99$. Since $\{\rho_n\}$ is montone increasing, we can just try values of $n$ in our calculator for the expression for $\rho _n$ until we get the first one that is at least $0.99$ (from memory, this is around about $n=26$). I think this (using the calculator) would suffice for HSC purposes.$

_______

$\noindent \textbf{Claim.} Let $f(x):= x\sin \frac{1}{x}$. Then $f$ is monotone increasing on $[1,\infty)$.$

$\noindent \textbf{Proof.} We have for $x\geq 1$$

$\begin{align*} f^\prime (x) &= \sin \frac{1}{x} + x\cos \frac{1}{x} \cdot \left(-\frac{1}{x^2}\right) \\ &= \sin \frac{1}{x} -\frac{1}{x} \cos \frac{1}{x} \\ &= \sin u -u \cos u, \quad \text{where } u\equiv \frac{1}{x}, \text{ so } 0<u \leq 1 \\ &> 0\Longleftrightarrow \sin u >u \cos u \Longleftrightarrow \tan u >u, \end{align*}$

$\noindent and this last inequality is well-known to be true (draw graph to see it, which is enough for HSC purposes) so our claim is true as $f$ has non-negative derivative on [1,\infty)$.$

$\noindent Edit: Sorry, this only shows that $\{\rho_n \}$ is monotone increasing after the $n$ such that $\frac{2\pi}{n} \leq 1$, which is $n=7$. So we can just show using our calculator that $\rho_n <0.99$ for all $n =1,2,3,4,5,6$ (in fact, these are also increasing as shown with a calculator), and then for $n\geq 7$, we have guaranteed a guaranteed monotone increase, so we can just search for the smallest $n\geq 7$ such that $\rho_n \geq 0.99$, without needing to test every single consecutive $n$.$

leehuan · Feb 26, 2016

Seems a bit rigorous though to test for values... even though yes this is an MX2 question...

InteGrand · Feb 26, 2016

leehuan said:
Seems a bit rigorous though to test for values... even though yes this is an MX2 question...

They probably wouldn't mind in MX2 if you just assumed it was monotonically increasing, unless that Q. was worth a lot of marks.

leehuan · Feb 26, 2016

InteGrand said:
They probably wouldn't mind in MX2 if you just assumed it was monotonically increasing, unless that Q. was worth a lot of marks.

What scares me is that it was worth 1mk.

Paradoxica · Feb 26, 2016

leehuan said:
What scares me is that it was worth 1mk.

They probably wanted you to exhaustively search for n. That's my best guess. Integrand's method is slightly more exact than what they wanted, but it is correct none the less.

leehuan · Feb 26, 2016

f(x)sin(x) = c for some constant c is not doable algebraically right? (for f(x) is some polynomial)

menodiks · Feb 26, 2016

help pls

leehuan · Feb 26, 2016

Another MX2 problem
$\\ $The point $P\left(a\sec{\theta},b\tan{\theta}\right)$ lies on the hyperbola $\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1$. The tangent and normal at $P$ meet the $y$-axis at $A$ and $B$ respectively. Show that the circle with diameter $AB$ passes through both foci of the hyperbola.$$

Progress:

$\\ $Easily enough, I have found that $A$ is $\left(0,-\frac{b}{\tan{\theta}}\right) $ and$ \\ B$ is $ \left( 0, \frac{\tan{\theta}}{b} \left( {a}^{2}{e}^{2} \right) \right)$

$\\ $This leads directly to$ \\ r=\frac{{a}^{2} \left( {e}^{2} \sec^{2}{\theta}-1 \right) }{2b\tan{\theta}} \\ \\ $and the centre of the circle being at$ \\ x=0 \\ y=-\frac{b}{\tan{\theta}} +\frac{{a}^{2} \left( {e}^{2} \sec^{2}{\theta}-1 \right) }{2b\tan{\theta}} \\ y=\frac{{a}^{2}+{a}^{2}{e}^{2}\left( \sec^{2}{\theta}-2 \right)}{2b\tan{\theta}}$

$\\ $Now, at this point I was about to apply the distance formula to the centre of the circle with $\left(ae,0\right)$ but I'm not prepared to. Substituting the centre of the circle does not yield a very nice perfect square to be expanded, requiring a ${4}^{$th$}$ degree. But with GeoGebra's aid I know my results are definitely not wrong. So is there a shortcut I missed at one point that can avoid this brute force?$

seanieg89 · Feb 26, 2016

leehuan said:
Another MX2 problem
$\\ $The point $P\left(a\sec{\theta},b\tan{\theta}\right)$ lies on the hyperbola $\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1$. The tangent and normal at $P$ meet the $y$-axis at $A$ and $B$ respectively. Show that the circle with diameter $AB$ passes through both foci of the hyperbola.$$

Progress:

$\\ $Easily enough, I have found that $A$ is $\left(0,-\frac{b}{\tan{\theta}}\right) $ and$ \\ B$ is $ \left( 0, \frac{\tan{\theta}}{b} \left( {a}^{2}{e}^{2} \right) \right)$

$\\ $This leads directly to$ \\ r=\frac{{a}^{2} \left( {e}^{2} \sec^{2}{\theta}-1 \right) }{2b\tan{\theta}} \\ \\ $and the centre of the circle being at$ \\ x=0 \\ y=-\frac{b}{\tan{\theta}} +\frac{{a}^{2} \left( {e}^{2} \sec^{2}{\theta}-1 \right) }{2b\tan{\theta}} \\ y=\frac{{a}^{2}+{a}^{2}{e}^{2}\left( \sec^{2}{\theta}-2 \right)}{2b\tan{\theta}}$

$\\ $Now, at this point I was about to apply the distance formula to the centre of the circle with $\left(ae,0\right)$ but I'm not prepared to. Substituting the centre of the circle does not yield a very nice perfect square to be expanded, requiring a ${4}^{$th$}$ degree. But with GeoGebra's aid I know my results are definitely not wrong. So is there a shortcut I missed at one point that can avoid this brute force?$

By symmetry about the y-axis, it suffices to show the circle just passes through S.

Since AB is the diameter of our circle, it suffices to show ASB is a right angle, which you can do just by multiplying gradients of AS and BS.

This does it v.quickly.

leehuan · Feb 26, 2016

seanieg89 said:
By symmetry about the y-axis, it suffices to show the circle just passes through S.

Since AB is the diameter of our circle, it suffices to show ASB is a right angle, which you can do just by multiplying gradients of AS and BS.

Of course! I was hoping that someone would find a circle geometry trick for me. Thanks!

seanieg89 · Feb 26, 2016

menodiks said:
help pls

$i)$\\ \\ $\phi(x):=\int_{\mathbb{R}^2}G(x,y)S(y)\, dy$\\ \\ solves this inhomogeneous problem, just from the definition of a Green's function.\\ \\ ii)\\ \\ We now need to construct a Green's function for the half-plane, which can be done via a reflection trick.\\ \\ We define \\ \\ $\tilde{G}(x,y):=G(x,y)-G(\tilde{x},y)$\\ \\ where $\tilde{x}$ simply denotes the reflection of $x$ about the $x_2$-axis.\\ \\ Then $\tilde{G}(x,y)=0$ when $x$ lies on the $x_2$-axis, and we have:\\ \\ $\Delta_x \tilde{G}(x,y)=\delta_y(x)-\delta_y(\tilde{x})=\delta_y(x)$ \\ \\ in the right half-plane. \\ \\ Hence the solution to this BVP is given by \\ \\ $u(x)=\int_{\mathcal{H}} (G(x_1,x_2,y_1,y_2)-G(-x_1,x_2,y_1,y_2))S(y_1,y_2)\, dy .$\\ \\ where $G$ denotes Green's function for the plane. $$

Edit: Note that we have skirted over the issue of existence and uniqueness of a solution, as we are not given enough information about S or boundary conditions at infinity for this PDE to be well posed. (Note that adding $x_1$ to our solution will give us another solution for instance. This integral formula is the natural approach though if S is nice enough for our problem to be well posed, as the resulting solution will also vanish at infinity, and it will be the unique such solution.)

leehuan · Mar 4, 2016

I can't quit BoS I need seanieg and InteGrand and co.

$\\ $Question: Find the greatest domain of $: \\ f\left(x\right)=\sqrt{1-2\sin{x}}$

$\\ $My thinking:$ \\ $Anything under a square root is positive. The equation therefore breaks down to$ \\ \sin{x}\le\frac{1}{2} \\ $ And then I got stuck for a bit. If there was an equal sign I would've jumped to:$ \\ x=n\pi+{(-1)}^{n}\left(\frac{\pi}{6}\right) \, \forall \, n\in \mathbb Z$

$\\ $So I drew the graph to visualise what was going on. I ended up with something messy like this...$$

$\\ \left[(2k-1)\pi + {(-1)}^{2k-1} \left(\frac{\pi}{6}\right) \, , \, 2k\pi+{(-1)}^{2k} \left(\frac{\pi}{6}\right)\right] \, \forall \, k \in \mathbb Z$

Is this correct? If so, is there a neater way to express this?

InteGrand · Mar 4, 2016

leehuan said:
I can't quit BoS I need seanieg and InteGrand and co.

$\\ $Question: Find the greatest domain of $: \\ f\left(x\right)=\sqrt{1-2\sin{x}}$

$\\ $My thinking:$ \\ $Anything under a square root is positive. The equation therefore breaks down to$ \\ \sin{x}\le\frac{1}{2} \\ $ And then I got stuck for a bit. If there was an equal sign I would've jumped to:$ \\ x=n\pi+{(-1)}^{n}\left(\frac{\pi}{6}\right) \, \forall \, n\in \mathbb Z$

$\\ $So I drew the graph to visualise what was going on. I ended up with something messy like this...$$

$\\ \left[(2k-1)\pi + {(-1)}^{2k-1} \left(\frac{\pi}{6}\right) \, , \, 2k\pi+{(-1)}^{2k} \left(\frac{\pi}{6}\right)\right] \, \forall \, k \in \mathbb Z$

Is this correct? If so, is there a neater way to express this?

$\noindent One way to express the domain is simply $D = \left \{ x \in \mathbb{R} : \sin x \leq \frac{1}{2} \right \}$.$

Carrotsticks · Mar 4, 2016

Paradoxica · Mar 4, 2016

InteGrand said:
$\noindent One way to express the domain is simply $D = \left \{ x \in \mathbb{R} : \sin x \leq \frac{1}{2} \right \}$.$

I mean yes that is technically the domain but... it doesn't directly tell you the range of x

Shuuya · Mar 4, 2016

leehuan said:
I can't quit BoS I need seanieg and InteGrand and co.

Welcome back haha

leehuan · Mar 4, 2016

InteGrand said:
$\noindent One way to express the domain is simply $D = \left \{ x \in \mathbb{R} : \sin x \leq \frac{1}{2} \right \}$.$

Well, I suppose, what Paradoxica said. But true you have a point

Carrotsticks said:
$\\ $If you prefer a more explicit answer, you may like to let $d=\left [ - \frac{7\pi}{6},\frac{\pi}{6} \right ]. $ Then the domain is $ D=d+2k \pi, k \in \mathbb{Z}.$

Ah I see. Thanks.

leehuan · Mar 4, 2016

Shuuya said:
Welcome back haha

Lol thanks. But I'm only residing in a few sections of the forum for the most part.

InteGrand · Mar 4, 2016

leehuan's All-Levels-Of-Maths SOS thread (11 Viewers)

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