Q20
![](https://latex.codecogs.com/png.latex?\bg_white \left(2-y\right)dy=\frac{dx}{\sqrt{2-x^{2}}})
given that
![](https://latex.codecogs.com/png.latex?\bg_white y\left(1\right)=0)
.
That means when
![](https://latex.codecogs.com/png.latex?\bg_white y=2\pm{\sqrt{\frac{\pi}{2}+4-2\sin^{-1}\left(\frac{x}{\sqrt{2}}\right)}})
Note that
![](https://latex.codecogs.com/png.latex?\bg_white y\left(1\right)=0)
which is in the form
![](https://latex.codecogs.com/png.latex?\bg_white y\left(x\right)=...)
. Subbing it in we can see that it should be
Q21)
![](https://latex.codecogs.com/png.latex?\bg_white \ln{\left(\frac{P}{C-P}\right)}=0.1t+D)
When
![](https://latex.codecogs.com/png.latex?\bg_white t=0, P=150,000)
. Thus,
![](https://latex.codecogs.com/png.latex?\bg_white \frac{450000}{\frac{e^{2}}{4}-1}+600000=C)
Chuck the LHS in the calculator and you will see that
Q19) Show that
![](https://latex.codecogs.com/png.latex?\bg_white \tan\left(\frac{5\pi}{12}\right)= \sqrt{3}+2)
To start off lets split
![](https://latex.codecogs.com/png.latex?\bg_white \tan\left(\frac{5\pi}{12}\right))
into
![](https://latex.codecogs.com/png.latex?\bg_white \tan\left(\frac{2\pi}{12}+\frac{3\pi}{12}\right))
which is just
![](https://latex.codecogs.com/png.latex?\bg_white \tan\left(\frac{\pi}{6}+\frac{\pi}{4}\right))
Now,
![](https://latex.codecogs.com/png.latex?\bg_white \tan\left(A+B)\right)=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}=\frac{\frac{1}{\sqrt{3}}+1}{1-\frac{1}{\sqrt{3}}})
and we multiply both the numberator and demnoominator by
![](https://latex.codecogs.com/png.latex?\bg_white \sqrt{3})
giving us
![](https://latex.codecogs.com/png.latex?\bg_white \frac{{\sqrt{3}+1}}{\sqrt{3}-1})
.
Rationalise the denominator it will give us
![](https://latex.codecogs.com/png.latex?\bg_white \frac{4+2\sqrt{3}}{2}=2+\sqrt{3})
as required.
Part ii
Complete the square on the denominator which in turn will give you
![](https://latex.codecogs.com/png.latex?\bg_white \left(x-2\right)^{2}+4)
and to find the area it is in the form of
Knowing this we will now have
![](https://latex.codecogs.com/png.latex?\bg_white \int_{0}^{2\sqrt{3}+6}\frac{2}{\left(x-2\right)^{2}+4}dx)
.
Let
![](https://latex.codecogs.com/png.latex?\bg_white x-2=2\tan{\theta}\rightarrow{dx=2\sec^{2}\theta{d\theta}})
When
![](https://latex.codecogs.com/png.latex?\bg_white x=0, \theta=\frac{-\pi}{4})
and when
![](https://latex.codecogs.com/png.latex?\bg_white x=2\sqrt{3}+6)
we have
![](https://latex.codecogs.com/png.latex?\bg_white 2\sqrt{3}+6-2=2\tan{\theta}\rightarrow{2\sqrt{3}+4=2\tan{\theta}}\rightarrow{\sqrt{3}+2=\tan{\theta})
Once we get here you should recognise that you have to find the inverse of tan which is
![](https://latex.codecogs.com/png.latex?\bg_white \tan^{-1}\left(\sqrt{3}+2\right)=\theta)
is that just
![](https://latex.codecogs.com/png.latex?\bg_white \theta=\frac{5\pi}{12})
as discussed from part 1.
There,
![](https://latex.codecogs.com/png.latex?\bg_white \int_{0}^{2\sqrt{3}+6}\frac{2}{\left(x-2\right)^{2}+4}dx=\int_{-\frac{\pi}{4}}^{\frac{5\pi}{12}}d\theta)
Then integrate once again
![](https://latex.codecogs.com/png.latex?\bg_white \left[\theta\right]_{\frac{-\pi}{4}}^{\frac{5\pi}{12}}
We then end up finding out what is [TEX]\frac{5\pi}{12}+\frac{\pi}{4})
and this gives us
![](https://latex.codecogs.com/png.latex?\bg_white \frac{8\pi}{12}=\frac{2\pi}{3})
.
Q25
i)
![](https://latex.codecogs.com/png.latex?\bg_white \ln\left(x\right)=-\cos{t}+C)
We are told initially the displacement is 1 cm so to interpret that we will say when
![](https://latex.codecogs.com/png.latex?\bg_white t=0, x=1)
and using that
![](https://latex.codecogs.com/png.latex?\bg_white C=1)
.
ii)
![](https://latex.codecogs.com/png.latex?\bg_white \frac{dx}{dt}=\sin{t}e^{1-\cos{t}})
.
![](https://latex.codecogs.com/png.latex?\bg_white \frac{dx}{dt}=0)
when
![](https://latex.codecogs.com/png.latex?\bg_white \sin{t}=0)
,
![](https://latex.codecogs.com/png.latex?\bg_white t=n\pi)
where n is an integer.
![](https://latex.codecogs.com/png.latex?\bg_white \frac{d^{2}x}{dt^{2}}=\left(\cos{t}+\sin^{2}t\right)e^{1-\cos{t}})
.
When
![](https://latex.codecogs.com/png.latex?\bg_white \frac{d^{2}x}{dt^{2}}=e)
.
When
![](https://latex.codecogs.com/png.latex?\bg_white \frac{d^{2}x}{dt^{2}}=-e^{2})
.
Maximum displacement is
![](https://latex.codecogs.com/png.latex?\bg_white e^{2})
at intervals of
![](https://latex.codecogs.com/png.latex?\bg_white \left(2n+1\right)\pi[TEX] where n is a natural number.)
[/TEX][/TEX]