Untitled (Urgently needs an idea)

AQA S2B June 2006

Question 7
The continuous random variable $X$ has probability density function defined by
\[\mathrm{f}(x)=\begin{cases}\frac{1}{5}(2x+1)&0\le x\le 1\\ \frac{1}{15}(4-x)^2&1\lt x\le4\\[0.75ex]0&\text{otherwise}\end{cases}\]

Part a) (2) Sketch the graph of $\mathrm{f}$
Moving on.

Part b) (i) (3) Show the cumulative distribution function, $\mathrm{F}(x)$, for $0\le x\le1$ is
\[\mathrm{F}(x)=\frac15x(x+1)\]

Solution
\begin{align*}
\mathrm{F}(x_1)&=\int_0^{x_1}\frac{1}{5}(2x+1)\mathrm{d}x\\
&=\frac15\left[x^2+x\right]_0^{x_1}\\
&=\frac15(x_1^2+x_1)\\
&=\frac15x_1(x_1+1)\\
\Rightarrow\mathrm{F}(x)&=\frac15x(x+1)
\end{align*}

Part ii) (1) Hence write down the value of $\mathrm{P}(X\le 1)$

Solution
\[\mathrm{P}(X\le 1)=\mathrm{F}(1)=\frac151(1+1)=\frac25\]

Part iii) (5) Find the value of $x$ for which $\mathrm{P}(X\ge x)=\frac{17}{20}$

Solution
\[\mathrm{P}(X\ge x)=\frac{17}{20}\Rightarrow\mathrm{P}(X\le x)=\frac{3}{20}\]
\begin{align*}
\frac{3}{20}&=\frac15x(x+1)\\
\frac34&=x(x+1)\\
x^2+x-\frac34&=0\\
\left(x+\frac32\right)\left(x-\frac12\right)&=0\\
\therefore x=-\frac32, x=\frac12
\end{align*}
As $x=-\frac32$ is outside domain of function $\mathrm{f}(x)$, $x=\frac12$

Part iv) (4) Find the lower quartile of the distribution

Solution
Let the lower quartile = $q$
\begin{align*}
\frac14&=\frac15q(q+1)\\
5&=4q^2+4q\\
4q^2+4q-5&=0\\
4(q+\frac12)^2-1-5&=0\\
q+\frac12&=\pm\sqrt{\frac64}\\
q&=-\frac12\pm\sqrt\frac32\\
\end{align*}
As $q=-\frac12-\sqrt\frac32$ is outside domain of function $\mathrm{f}(x)$, $q=-\frac12+\sqrt\frac32=\frac12(\sqrt6-1)$

AQA M2B January 2006

Question 8

A particle, of mass 10 kg, is attached to one end of a light elastic string of natural length 0.4 metres and modulus of elasticity 100 N. The other end of the string is fixed to the point $O$.

Part a) (2) Find the length of the elastic string when the particle hangs in equilibrium directly below $O$.
Solution
Equilibrium implies $mg=\frac{\lambda}{l}x$
\[10g=\frac{100}{0.4}x\]
\[\therefore x=\frac{10g\times0.4}{100}=0.392\mathrm{m}\]
Therefore total length below $O=0.392+0.4=0.792$

Part b) (2) The particle is pulled down and held at a point $P$, which is 1 metre vertically below $0$. Show that the elastic potential energy of the string when the particle is in this position is 45 J.
Solution
\[E_P=\frac12\frac{\lambda}{l} x^2=0.5\frac{100}{0.4}\times\left(1-0.4\right)^2\]
\[E_P=0.5\times250\times0.36=45\]

Part c) The particle is released from rest at the point $P$. In the subsequent motion, the particle
has speed $v$ ms$^{-1}$ when it is $x$ metres below $O$.
i) (7) Show that, while the string is taut, \[v^2=39.6x-25x^2-14.6\]
Solution
Assuming point $O$ is taken as having most potential energy, and when string is $x$ metres from $O$, $E_P$ is at a minimum.
At $P$, $E_K=0$, $E_P=mg\Delta h=10g(x-1)=10gx-10g$, $E_{PE}=45$
At a point $x$ metres below $O$, $E_K=\frac12mv^2=5v^2$, $E_P=0$, $E_{PE}=125\times(x-0.4)^2$
\[0+10g(x-1)+45=5v^2+125(x-0.4)^2\]
\[10gx-98+45=5v^2+125(x^2-0.8x+0.16)\]
\[10gx-53=5v^2+125x^2-100x+20\]
\[5v^2=(10g+100)x-125x^2-53-20\]
\[v^2=\frac{198x-125x^2-73}{5}=39.6x-25x^2-14.6\]
Part ii) (3) Find the value of $x$ when the particle comes to rest for the first time after being released, given that the string is still taut.
Solution
When the particle comes to rest, $v$=0
\[0=39.6x-25x^2-14.6\]
\[25 (x-0.792)^2-1.0816 = 0\]
\[(x-0.792)^2=\frac{1.0816}{25}\]
\[x=0.792\pm\sqrt{\frac{1.0816}{25}}\]
\[x=0.792\pm0.208\]
\[x=1,x=0.584\]
As the particle was released when $x=1$, that solution can be eliminated, and the first point at which the particle comes to rest can be seen to be 0.584m below $O$.

AQA M2B January 2006
\(\require{AMSmath}\)
Question 7
A particle $P$ of mass $m$ kg, is placed at the point $Q$ on the top of a smooth upturned hemisphere of radius 3 metres and centre $O$. The plane face of the hemisphere is fixed to a horizontal table. The particle is set into motion with an initial horizontal velocity of 2 ms$^{-1}$. When the particle is on the surface of the hemisphere, the angle between $OP$ and $OQ$ is $\theta$ and the particle has speed $v$ ms$^{-1}$.

Part a) (4) Show that $v^2=4+6g(1-\cos\theta)$
Solution
At $Q$, $E_K=\frac12mv^2=\frac12m\times2^2=2m$, $E_P=mg\Delta h=3mg$
At $P$, $E_K=\frac12mv^2$, $E_P=mg\Delta h=mg 3\cos\theta$
As no work done on $P$, energy at $P=$ energy at $Q$
\[\Rightarrow 2m+3mg=\frac12mv^2+3mg\cos\theta\]
\[\therefore 2+3g=\frac12v^2+3g\cos\theta\]
\[\therefore v^2=2\times\left(2+3g-3g\cos\theta\right)=4+6g(1-\cos\theta)\]

Part b) (5) Find the value of $\theta$ when the particle leaves the hemisphere.
Solution
Particle leaves the hemisphere when reaction force equals 0.
\[\Rightarrow mg\cos\theta=\frac{mv^2}r\]
\[\overset{v^2=4+6g(1-\cos\theta)}{\Rightarrow}mg\cos\theta=m\frac{4+6g(1-\cos\theta)}r\]
\[3g\cos\theta=4+6g(1-\cos\theta)\]
\[4+6g=9g\cos\theta\]
\[\frac{4+6g}{9g}=\cos\theta\]
\[\therefore\theta=\cos^{-1}\left(\frac{4+6g}{9g}\right)=44.6^{\circ}\]

AQA MFP3 Jan 06
\(\require{AMSmath}\)
Question 6
Part a) (4) A circle \(C_1\) has cartesian equation \[x^2+(y-6)^2=36\]Show the polar equation of \(C_1\) is\[r=12\sin\theta\]
Solution
\[x^2+y^2-12y+36=36\]
Given that \(x^2+y^2=r^2\) and \(y=r\sin\theta\)
\[r^2-12r\sin\theta=0\]
\[r^2=12r\sin\theta\]
\[r=12\sin\theta\]

Part b) (6) Find the area bound by the curve \(C_2\) with polar equation \[r=2\sin\theta+5\]for \(0\le\theta\le2\pi\)
Solution
\[A=\frac{1}{2} \int_{0}^{2\pi}r^2\mathrm{d}\theta\]
\[ =\frac{1}{2} \int_0^{2\pi}(2\sin\theta+5)^2\mathrm{d}\theta\]
\[ =\frac{1}{2} \int_0^{2\pi}(4\sin^2\theta+20\sin\theta+25)\mathrm{d}\theta\]
Given that \(\frac{1}{2}\left(1-\cos2\theta\right)=\sin^2\theta\)
\[ =\frac{1}{2} \int_0^{2\pi}(2(1-\cos2\theta)+20\sin\theta+25)\mathrm{d}\theta\]
\[ =\frac{1}{2}\left[2\theta-\sin2\theta-20\cos\theta+25\theta\right]_0^{2\pi}\]
\[ =\frac{1}{2}\left[27\theta-\sin2\theta-20\cos\theta\right]_0^{2\pi}\]
\[ =27\pi\]

Part c) (6) The circle \(C_1\) intersects the curve \(C_2\) at the points \(P\) and \(Q\). Find, in surd form, the area of the quadrilateral \(OPMQ\), where \(M\) is the centre of the circle and \(O\) is the pole.
Solution
\[C_1=C_2 \therefore 12\sin\theta=2\sin\theta+5\]\[\therefore 10\sin\theta=5 \therefore \sin\theta=\frac12\therefore \theta=\frac\pi6,\frac{5\pi}6\]
\[\therefore P = (6,\frac{\pi}6), Q=(6,\frac{5\pi}6)\]
\(OPMQ\) is a rhombus of side 6 
\[Area=2\times\frac12 6^2\sin\frac{2\pi}{3}\]
\[=36\frac{\sqrt{3}}{2}\]

\[\text{Mind}=\text{Blown}\]
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
\[\sin^2 x + \cos^2 x \equiv 1\]
\[\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}+\ldots=\sum\limits_{r=0}^{\infty}(-1)^{r}\frac{x^{2r}}{(2r)!}\]
Edit: But not if you view it from the dashboard
Edit2: But if you view it in Chrome with this

Re-installed Windows and all the programs in a weekend.
Loving the clean windows smell

Just love the rain, lol, jk, just absorbed enough to prevent a rise in sea levels

When I first started this, I’d hoped that I’d be able to write a regularly updated technology blog. In all fairness, this has been less successful than I’d hoped. I have managed to put out more posts than I thought I ever would, but that isn’t setting a particularly high target. Literally three or four posts in, and the depressing stuff started to set in. At that point though, I’d just found that someone I’d considered to be a good friend, wasn’t. In all honesty, they were just about the total opposite. At that point, this was about the best outlet I had for what I felt, and became something of a crutch. The problem is, I can feel that becoming close to happening again. So, instead, I’m going to actually try and post some technology related stuff, and details of any projects I’m working on. Also, any tips on design would be greatly appreciated, particularly for the LaTeX document I’m producing for my computing coursework, particularly for stuff like the title page. Thank you to everyone who hasn’t abandoned this in the silence.

51 Notes. I really ought to look at old posts more often.

Breaking news: the poor mite’s had an awful sleep, and is still sulking, he’s also missed breakfast to further reinforce his “woe is me” act. On the bright side, he’s been more useful this morning than most others, so, mustn’t grumble. Back to you in the studio.