Math Ext 1 help (1 Viewer)

Farhanthestudent005 · Jun 22, 2022

Need help with all of these questions...thanks

jimmysmith560 · Jun 23, 2022

Would the following working help for the third question?

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

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

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

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

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

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

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

5uckerberg · Jun 23, 2022

Farhanthestudent005 said:
View attachment 35852
View attachment 35853

View attachment 35854

Need help with all of these questions...thanks

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

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

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

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

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

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

Q25

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

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

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Math Ext 1 help (1 Viewer)

Farhanthestudent005

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jimmysmith560

Le Phénix Trilingue

5uckerberg

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