Author Topic: 3U Maths Question Thread (Read 1239931 times) Tweet Share

winstondarmawan · « **Reply #2760 on:** September 16, 2017, 05:58:22 pm »

Would appreciate help with the following:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21624159_1326480187477501_1176035748_n.png?oh=92550c1f11eede0a4835be2c7ff544d8&oe=59BF8C80
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21763755_1326495504142636_437795940_n.png?oh=b3afa5de0caef916a1815852a78e7462&oe=59BE7304
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21763778_1326613077464212_1046725160_n.png?oh=a459ec486a972af6b1d111364a39b35b&oe=59BF7E3D
Part (iii): Was able to get out most of the proof, but not sure where the sigma sign came from and why n is at the top.
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/21682576_1326617404130446_514269388_o.png?oh=e0be0e3e2cced7dbb6c0fadac9f708e1&oe=59BF8D93
Part (ii).
Thanks in advance!

RuiAce · « **Reply #2761 on:** September 16, 2017, 06:00:01 pm »

Quote from: winstondarmawan on September 16, 2017, 05:58:22 pm

Would appreciate help with the following:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21624159_1326480187477501_1176035748_n.png?oh=92550c1f11eede0a4835be2c7ff544d8&oe=59BF8C80
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21763755_1326495504142636_437795940_n.png?oh=b3afa5de0caef916a1815852a78e7462&oe=59BE7304
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21763778_1326613077464212_1046725160_n.png?oh=a459ec486a972af6b1d111364a39b35b&oe=59BF7E3D
Part (iii): Was able to get out most of the proof, but not sure where the sigma sign came from and why n is at the top.

The second and third are already in the compilation thread.
____________________________________________
$\text{The first is nothing more than the general solution formula}\\ \begin{align*}\sin 2x &= a\\ 2x &= n\pi \pm (-1)^n \sin^{-1}a\\ x &= \frac{n\pi \pm (-1)^n \sin^{-1}a}2\end{align*}$

RuiAce · « **Reply #2762 on:** September 17, 2017, 01:20:26 pm »

Quote from: winstondarmawan on September 16, 2017, 05:58:22 pm

https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/21682576_1326617404130446_514269388_o.png?oh=e0be0e3e2cced7dbb6c0fadac9f708e1&oe=59BF8D93
Part (ii).
Thanks in advance!

$\text{Here is one possible approach}$
Will be added to the compilation thread after completion.
$\text{We have the curves }y=e^{rx}\text{ and }y=\log_e x\\ \text{Their respective derivatives are }\frac{dy}{dx} = re^{rx} \text{ and }\frac{dy}{dx} = \frac1x$
__________________________________________________________
$\text{We seek a common tangent specifically at a point of intersection}\\ \text{For this, we have two necessary and sufficient conditions}\\ \text{1. At a certain point }x\text{, their }y\text{ values are equal}\\ \text{2. At the same point, their }\frac{dy}{dx}\text{ values are equal}$
Intuitively, the first criteria is just guaranteeing that we have a point of intersection. But the only way we can have a common tangent there is if the gradient of the tangent is ALSO the same there, hence criteria #2.
$\text{We consider condition 2 first}\\ \text{Equating derivatives, }re^{rx} = \frac1x \implies \boxed{e^{rx} = \frac{1}{rx}}$
$\text{Let }\boxed{t = rx}\text{, so the equation becomes }\boxed{e^t = \frac1t}\\ \text{We know, from part i), this has a solution at }\boxed{t\approx 0.56}\\ \text{Hence, subbing back }t=rx\text{, our equation has a solution at}\\ rx \approx 0.56 \implies \boxed{x \approx \frac{0.56}{r}}$
__________________________________________________________
$\text{So we've now found an }x\text{-coordinate to ensure a common tangent}\\ \text{To ensure a point of intersection, we now sub these in for y}\\ \text{and then equate the relevant ordinates.}$
$\begin{align*}e^{rx} &= \log_e x\\ \therefore e^{0.56} &\approx \log \frac{0.56}{r}\\ e^{e^{0.56}}&\approx \frac{0.56}{r}\\ r&\approx \frac{0.56}{e^{e^{0.56}}}\\ &\approx 0.97 \end{align*}$

J.B · « **Reply #2763 on:** September 17, 2017, 04:30:01 pm »

Hi,
I was just wondering if someone could explain how you know that f'(1) > 1, as I couldn't tell if it was a) or b)?
Thank you

RuiAce · « **Reply #2764 on:** September 17, 2017, 04:33:10 pm »

Quote from: J.B on September 17, 2017, 04:30:01 pm

Hi,
I was just wondering if someone could explain how you know that f'(1) > 1, as I couldn't tell if it was a) or b)?
Thank you

Already addressed.

This has received a surprisingly large amount of requests and hence will be added to the compilation.

winstondarmawan · « **Reply #2765 on:** September 17, 2017, 06:57:45 pm »

Hello!
Would appreciate help with the following (2008 HSC):
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21291833_1327536244038562_855053827_n.png?oh=0ec3b8867b9326259a4d0fb22af65f7c&oe=59BFB548
Part (iii). Also for part (ii), my reasoning was that by inspection the largest change in theta will occur when x=0. Is this sufficient?
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21698076_1327536934038493_934609861_n.png?oh=d6833ec00283d7f84645b0c2dfc11af3&oe=59C0E23E
Part (iii).
The following links are for the same Q:
{ https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21767659_1327537074038479_1743018830_n.png?oh=ada6b1cd4cb38b63c97243e5575664ba&oe=59C0D387
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21691073_1327537157371804_1410072627_n.png?oh=bad31e219c4fcb6e255a3716e5ee24b2&oe=59C0E60A }
Parts (d-f).
Thanks in advance.

RuiAce · « **Reply #2766 on:** September 17, 2017, 07:08:46 pm »

Quote from: winstondarmawan on September 17, 2017, 06:57:45 pm

Hello!
Would appreciate help with the following (2008 HSC):
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21291833_1327536244038562_855053827_n.png?oh=0ec3b8867b9326259a4d0fb22af65f7c&oe=59BFB548
Part (iii). Also for part (ii), my reasoning was that by inspection the largest change in theta will occur when x=0. Is this sufficient?
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21698076_1327536934038493_934609861_n.png?oh=d6833ec00283d7f84645b0c2dfc11af3&oe=59C0E23E
Part (iii).
The following links are for the same Q:
{ https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21767659_1327537074038479_1743018830_n.png?oh=ada6b1cd4cb38b63c97243e5575664ba&oe=59C0D387
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21691073_1327537157371804_1410072627_n.png?oh=bad31e219c4fcb6e255a3716e5ee24b2&oe=59C0E60A }
Parts (d-f).
Thanks in advance.

Found the 2008 one here; do be mindful of a few things in the solution. E.g. with part iii, go further than just x
$\text{Note that since }\angle PLQ = \angle PKQ\quad (= 90^\circ)\\ \text{the points }P, K, L, Q\text{ are concyclic}\\ \text{from the theorem: if an interval subtends angles of}\\ \text{equal size on the same side, then all points involved are concyclic}$
Which is also a theorem taught in the Cambridge textbook but not MIF.
$\text{Furthermore, as }\angle PKQ = 90^\circ\text{ it is the angle in a semicircle}\\ \text{Hence }PQ\text{ is a diameter of the circle }PKLQ$
$\text{Since }M\text{ is the midpoint of a diameter}\\ \text{it follows that }M\text{ is the centre of the circle}\\ \therefore MK = ML\text{ (equal radii)}$

J.B · « **Reply #2767 on:** September 18, 2017, 04:55:15 pm »

Quote from: RuiAce on September 17, 2017, 04:33:10 pm

Already addressed.

This has received a surprisingly large amount of requests and hence will be added to the compilation.

Thank you

RuiAce · « **Reply #2768 on:** September 18, 2017, 06:56:25 pm »

Quote from: winstondarmawan on September 17, 2017, 06:57:45 pm

Hello!
Would appreciate help with the following (2008 HSC):
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21291833_1327536244038562_855053827_n.png?oh=0ec3b8867b9326259a4d0fb22af65f7c&oe=59BFB548
Part (iii). Also for part (ii), my reasoning was that by inspection the largest change in theta will occur when x=0. Is this sufficient?
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21698076_1327536934038493_934609861_n.png?oh=d6833ec00283d7f84645b0c2dfc11af3&oe=59C0E23E
Part (iii).
The following links are for the same Q:
{ https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21767659_1327537074038479_1743018830_n.png?oh=ada6b1cd4cb38b63c97243e5575664ba&oe=59C0D387
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21691073_1327537157371804_1410072627_n.png?oh=bad31e219c4fcb6e255a3716e5ee24b2&oe=59C0E60A }
Parts (d-f).
Thanks in advance.

$\text{d) Applying trigonometry in }\triangle OLP\text{ and }\triangle NLP\\ OP = h \cot \alpha\text{ and }PN = h\cot \beta\\ \text{So since }\boxed{OP+PN= r}\\\boxed{ r = h\cot\alpha + h\cot\beta}$
$\text{Also from the symmetry of the parabola examined in the}\\ \text{previous parts, we have }OP=QN\\ \text{So we see that }w = r - 2OP\implies\boxed{r - 2h\cot\alpha}\\ \text{Subbing the result from above gives}\\ \boxed{w = h\cot\beta - h\cot\alpha}$
_____________________________
$\text{f) can be done quite easily by working backwards on scrap paper}\\ \text{simply assuming the result of e)}$
$\begin{align*}\frac{r}{w}&=\frac{h(\cot \beta+\cot \alpha)}{h(\cot\beta-\cot\alpha)}\\ &= \frac{\cot\beta + \cot \alpha}{\cot \beta - \cot \alpha}\end{align*}$
$\text{Multiply top and bottom by }\tan\alpha\tan\beta\text{ to obtain}\\ \frac{r}{w} = \frac{\tan\alpha+\tan\beta}{\tan\alpha-\tan\beta} = \frac{\tan\theta}{\tan\phi}$
$\therefore \frac{w}{\tan \phi}= \frac{r}{\tan\theta}$

winstondarmawan · « **Reply #2769 on:** September 18, 2017, 09:02:30 pm »

Quote from: RuiAce on September 18, 2017, 06:56:25 pm

$\text{d) Applying trigonometry in }\triangle OLP\text{ and }\triangle NLP\\ OP = h \cot \alpha\text{ and }PN = h\cot \beta\\ \text{So since }\boxed{OP+PN= r}\\\boxed{ r = h\cot\alpha + h\cot\beta}$
$\text{Also from the symmetry of the parabola examined in the}\\ \text{previous parts, we have }OP=QN\\ \text{So we see that }w = r - 2OP\implies\boxed{r - 2h\cot\alpha}\\ \text{Subbing the result from above gives}\\ \boxed{w = h\cot\beta - h\cot\alpha}$
_____________________________
$\text{f) can be done quite easily by working backwards on scrap paper}\\ \text{simply assuming the result of e)}$
$\begin{align*}\frac{r}{w}&=\frac{h(\cot \beta+\cot \alpha)}{h(\cot\beta-\cot\alpha)}\\ &= \frac{\cot\beta + \cot \alpha}{\cot \beta - \cot \alpha}\end{align*}$
$\text{Multiply top and bottom by }\tan\alpha\tan\beta\text{ to obtain}\\ \frac{r}{w} = \frac{\tan\alpha+\tan\beta}{\tan\alpha-\tan\beta} = \frac{\tan\theta}{\tan\phi}$
$\therefore \frac{w}{\tan \phi}= \frac{r}{\tan\theta}$

How about part (e)? Sorry should've made it more clearer - parts d to f.

Also would appreciate help with this one:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21641678_1328712140587639_1334058776_n.png?oh=b919267c38a3104e619ed90af4346867&oe=59C23D44

Sukakadonkadonk · « **Reply #2770 on:** September 18, 2017, 11:02:53 pm »

Hi, could I get some hints with this one?

Thanks.

RuiAce · « **Reply #2771 on:** September 19, 2017, 12:11:08 am »

Quote from: winstondarmawan on September 18, 2017, 09:02:30 pm

How about part (e)? Sorry should've made it more clearer - parts d to f.

Also would appreciate help with this one:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21641678_1328712140587639_1334058776_n.png?oh=b919267c38a3104e619ed90af4346867&oe=59C23D44

I had left part e at the time because I wanted to spend longer thinking about it (I had my mind focused on other things). It was clear that you needed help with it - I just didn't want to do it at the time. The main issue was, that 3 marker is most likely the one that requires more thought, and whilst I figured out some methods on doing it I wasn't convinced at which was the better/faster/my recommended choice, so I wanted to look at some other solutions beforehand.
$\text{In general, when it comes to invoking an angle}\\ \text{we must rely on the relationships }\boxed{\tan \alpha = m = \frac{dy}{dx} = \frac{dy/dt}{dx/dt}}$
where alpha is the angle made with the horizontal, m is the gradient, and dy/dx is the derivative. Note that the derivative IS the gradient of the tangent, and that the fourth expression is obtained via the chain rule, and the identity $ \frac{dy}{dx}\, \frac{dx}{dy} = 1 $
$\text{Now, in our case, first note}\\ \dot{x} = V\cos \theta\, \quad \dot{y} = V\sin\theta - gt\\ \text{Hence, we have}\\ \begin{align*}\tan \phi &= \frac{V\sin \theta-gt_1}{V\cos\theta}\\ &= \tan\theta - \frac{gt_1}{V\cos\theta}\end{align*}\\ \text{noting that when }t=t_1, \text{ the particle is at }L$
_________________________________________
$\text{But what is }t_1?$
$\text{Recall, from part a), that you obtained a quadratic}\\ gt^2 - 2Vt\sin\theta + 2h = 0\\ \text{Observe that }t=t_1\text{ is the first solution of this quadratic}$
$\text{From the quadratic formula, it then follows that}\\ t_1 = \frac{V\sin\theta - \sqrt{V^2\sin^2\theta -2gh } }{g}$
$\text{So substituting back into the above}\\ \begin{align*} \tan \phi &= \tan \theta - \frac{V\sin\theta - \sqrt{V^2\sin^2\theta - 2gh}}{V\cos\theta} \end{align*}$
_________________________________________
$\text{It now falls to just rearrange, with step 1 using }\tan\theta = \frac{\sin\theta}{\cos\theta}\\ \begin{align*}\tan \phi &= \frac{\sqrt{V^2\sin^2\theta-2gh}}{V\cos\theta}\\ V^2\tan^2\phi &= \frac{V^2\sin^2\theta}{\cos^2\theta} - \frac{2gh}{\cos^2\theta}\\ \tan^2\phi &= \tan^2\theta - 4\tan\alpha\tan\beta\\&= (\tan\alpha+\tan\beta)^2 - 4\tan\alpha\tan\beta\\ &= (\tan\alpha-\tan\beta)^2\end{align*}$
$\text{Taking positive roots (careful why we do this)}\\ \tan \phi = \tan\alpha - \tan\beta$

The 2009 question is already in the compilation

RuiAce · « **Reply #2772 on:** September 19, 2017, 12:22:48 am »

Quote from: Sukakadonkadonk on September 18, 2017, 11:02:53 pm

Hi, could I get some hints with this one?

Thanks.

$\text{The question starts in a similar procedure to the one above}\\ \text{Using }\tan \theta = \frac{\dot{y}}{\dot{x}}\text{ we have }\boxed{\tan\theta = \frac{-10t}{V}}\\ \text{but because the angle of landing is }-45^\circ \text{ (Careful about the minus!)}\\ -1 = \frac{-10t}{V} \implies \boxed{V = 10t}$
$\text{It falls to determine }t\\ \text{Here, }t\text{ is the time of flight, which is when }x=4000\\ x =4000\implies Vt = 4000 \implies \boxed{t = \frac{4000}{V}}$
$\text{So substituting it in}\\ \begin{align*}V &= \frac{10\times 4000}{V}\\ V^2&=40000\\ V&=200\end{align*}$
(tagged: 2001)

Sukakadonkadonk · « **Reply #2773 on:** September 19, 2017, 07:46:55 am »

Quote from: RuiAce on September 19, 2017, 12:22:48 am

$\text{The question starts in a similar procedure to the one above}\\ \text{Using }\tan \theta = \frac{\dot{y}}{\dot{x}}\text{ we have }\boxed{\tan\theta = \frac{-10t}{V}}\\ \text{but because the angle of landing is }-45^\circ \text{ (Careful about the minus!)}\\ -1 = \frac{-10t}{V} \implies \boxed{V = 10t}$
$\text{It falls to determine }t\\ \text{Here, }t\text{ is the time of flight, which is when }x=4000\\ x =4000\implies Vt = 4000 \implies \boxed{t = \frac{4000}{V}}$
$\text{So substituting it in}\\ \begin{align*}V &= \frac{10\times 4000}{V}\\ V^2&=40000\\ V&=200\end{align*}$
(tagged: 2001)

Oh ok, thank you.

bsdfjnlkasn · « **Reply #2774 on:** September 19, 2017, 07:54:38 am »

Hey there!

I'm not sure how they moved from the second line to the last (given in (ii) the screenshot) and would love some help...

I also don't get how they performed the actual derivative after the substitution of r here in b)

Thank you!

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Author Topic: 3U Maths Question Thread (Read 1239931 times) Tweet Share

winstondarmawan

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