Math Ext 1 help (1 Viewer)

Farhanthestudent005

Member

Need help with all of these questions...thanks

jimmysmith560

Le Phénix Trilingue
Moderator
Would the following working help for the third question?

$\bg_white y=8\ln \left(x-1\right)\:\rightarrow \:x=e^{\frac{y}{8}}+1$

$\bg_white V=\pi \int _0^6x^2\:dy$

$\bg_white =\pi \int _0^6\left(e^{\frac{y}{4}}+2e^{\frac{y}{8}}+1\right)dy$

$\bg_white =\pi \left[4e^{\frac{y}{4}}+16e^{\frac{y}{8}}+y\right]^6_0$

$\bg_white =\pi \left(4e^{1.5}+16^{0.75}+6-20\right)$

$\bg_white =\pi \left(4e^{1.5}+16^{0.75}-14\right)$

$\bg_white \approx 118.7\:\text{units}^3$

5uckerberg

Well-Known Member
Q20
$\bg_white \left(2-y\right)dy=\frac{dx}{\sqrt{2-x^{2}}}$ given that $\bg_white y\left(1\right)=0$.
That means when $\bg_white x=1, y=0$
$\bg_white \int_{0}^{y}2-ydy=\int_{1}^{x}\frac{dx}{\sqrt{2-x^{2}}}$
$\bg_white 2y-\frac{y^{2}}{2}=\left[\sin^{-1}\left(\frac{x}{\sqrt{2}}\right)\right]_{1}^{x}$
$\bg_white 2y-\frac{y^{2}}{2}=\sin^{-1}\left(\frac{x}{\sqrt{2}}\right)-\frac{\pi}{4}$
$\bg_white 4y-y^{2}=2\sin^{-1}\left(\frac{x}{\sqrt{2}}\right)-\frac{\pi}{2}$
$\bg_white -4+4y-y^{2}=2\sin^{-1}\left(\frac{x}{\sqrt{2}}\right)-\frac{\pi}{2}-4$
$\bg_white y^{2}-4y+4=-2\sin^{-1}\left(\frac{x}{\sqrt{2}}\right)+\frac{\pi}{2}+4$
$\bg_white \left(y-2\right)^{2}=\frac{\pi}{2}+4-2\sin^{-1}\left(\frac{x}{\sqrt{2}}\right)$
$\bg_white y-2=\pm{\sqrt{\frac{\pi}{2}+4-2\sin^{-1}\left(\frac{x}{\sqrt{2}}\right)}}$
$\bg_white y=2\pm{\sqrt{\frac{\pi}{2}+4-2\sin^{-1}\left(\frac{x}{\sqrt{2}}\right)}}$
Note that $\bg_white y\left(1\right)=0$ which is in the form $\bg_white y\left(x\right)=...$. Subbing it in we can see that it should be
$\bg_white y=2-\sqrt{\frac{\pi}{2}+4-2\sin^{-1}\left(\frac{x}{\sqrt{2}}\right)}$

Q21)
$\bg_white \frac{dP}{dt}=0.1P\left(\frac{C-P}{C}\right)$
$\bg_white \frac{C}{P\left(C-P\right)}=\frac{1}{P}+\frac{1}{C-P}$
$\bg_white \int\frac{CdP}{P\left(C-P\right)}=\int0.1dt$
$\bg_white \int\frac{1}{P}+\frac{1}{C-P}dP=\int0.1dt$
$\bg_white \ln{\left(P\right)}-\ln{\left(C-P\right)}=0.1t+D$
$\bg_white \ln{\left(\frac{P}{C-P}\right)}=0.1t+D$
When $\bg_white t=0, P=150,000$. Thus, $\bg_white D=\ln{\left(\frac{150000}{C-150000}\right)}$
$\bg_white \ln{\left(\frac{P}{C-P}\right)}=0.1t+\ln{\left(\frac{150000}{C-150000}\right)}$
$\bg_white \ln{\left(\frac{P\left(C-150000\right)}{\left(C-P\right)150000}\right)}=0.1t$
$\bg_white \frac{P\left(C-150000\right)}{\left(C-P\right)150000}=e^{0.1t}$
$\bg_white \frac{600000\left(C-150000\right)}{\left(C-600000\right)150000}=e^{2}$
$\bg_white \frac{C-150000}{C-600000}=\frac{e^{2}}{4}$
$\bg_white 1+\frac{450000}{C-600000}=\frac{e^{2}}{4}$
$\bg_white \frac{450000}{C-600000}=\frac{e^{2}}{4}-1$
$\bg_white \frac{450000}{\frac{e^{2}}{4}-1}=C-600000$
$\bg_white \frac{450000}{\frac{e^{2}}{4}-1}+600000=C$
Chuck the LHS in the calculator and you will see that $\bg_white C\approx{1130000}$

Q19) Show that $\bg_white \tan\left(\frac{5\pi}{12}\right)= \sqrt{3}+2$
To start off lets split $\bg_white \tan\left(\frac{5\pi}{12}\right)$ into $\bg_white \tan\left(\frac{2\pi}{12}+\frac{3\pi}{12}\right)$ which is just $\bg_white \tan\left(\frac{\pi}{6}+\frac{\pi}{4}\right)$
Now, $\bg_white \tan\left(A+B)\right)=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}=\frac{\frac{1}{\sqrt{3}}+1}{1-\frac{1}{\sqrt{3}}}$ and we multiply both the numberator and demnoominator by $\bg_white \sqrt{3}$ giving us $\bg_white \frac{{\sqrt{3}+1}}{\sqrt{3}-1}$.
Rationalise the denominator it will give us $\bg_white \frac{4+2\sqrt{3}}{2}=2+\sqrt{3}$ as required.

Part ii
Complete the square on the denominator which in turn will give you $\bg_white \left(x-2\right)^{2}+4$ and to find the area it is in the form of $\bg_white \int_{lower boundary}^{upper boundary}f\left(x\right)dx$

Knowing this we will now have $\bg_white \int_{0}^{2\sqrt{3}+6}\frac{2}{\left(x-2\right)^{2}+4}dx$.
Let $\bg_white x-2=2\tan{\theta}\rightarrow{dx=2\sec^{2}\theta{d\theta}}$
When $\bg_white x=0, \theta=\frac{-\pi}{4}$ and when $\bg_white x=2\sqrt{3}+6$ we have $\bg_white 2\sqrt{3}+6-2=2\tan{\theta}\rightarrow{2\sqrt{3}+4=2\tan{\theta}}\rightarrow{\sqrt{3}+2=\tan{\theta}$
Once we get here you should recognise that you have to find the inverse of tan which is $\bg_white \tan^{-1}\left(\sqrt{3}+2\right)=\theta$ is that just $\bg_white \theta=\frac{5\pi}{12}$ as discussed from part 1.

There, $\bg_white \int_{0}^{2\sqrt{3}+6}\frac{2}{\left(x-2\right)^{2}+4}dx=\int_{-\frac{\pi}{4}}^{\frac{5\pi}{12}}d\theta$
Then integrate once again $\bg_white \left[\theta\right]_{\frac{-\pi}{4}}^{\frac{5\pi}{12}}
We then end up finding out what is
and this gives us $\bg_white \frac{8\pi}{12}=\frac{2\pi}{3}$.

Q25

i) $\bg_white \frac{dx}{dt}=x\sin{t}$
$\bg_white \frac{dx}{x}=\sin{t}dt$
$\bg_white \ln\left(x\right)=-\cos{t}+C$
We are told initially the displacement is 1 cm so to interpret that we will say when $\bg_white t=0, x=1$ and using that $\bg_white C=1$.
$\bg_white \ln\left(x\right)=-\cos{t}+1$
$\bg_white x=e^{1-\cos{t]}$

ii)
$\bg_white x=e^{1-\cos{t}}$
$\bg_white \frac{dx}{dt}=\sin{t}e^{1-\cos{t}}$.
$\bg_white \frac{dx}{dt}=0$ when $\bg_white \sin{t}=0$, $\bg_white t=n\pi$ where n is an integer.
$\bg_white \frac{d^{2}x}{dt^{2}}=\left(\cos{t}+\sin^{2}t\right)e^{1-\cos{t}}$.
When $\bg_white t=0$ $\bg_white \frac{d^{2}x}{dt^{2}}=e$.
When $\bg_white t=\pi$ $\bg_white \frac{d^{2}x}{dt^{2}}=-e^{2}$.
Maximum displacement is $\bg_white e^{2}$ at intervals of $\bg_white \left(2n+1\right)\pi"/>"/>

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