Author Topic: 3U Maths Question Thread (Read 1239048 times) Tweet Share

RuiAce · « **Reply #3390 on:** April 12, 2018, 08:11:33 pm »

Quote from: KT Nyunt on April 12, 2018, 10:12:17 am

Hi, could someone please show me the working for this? The answer is 110km/h.

Thanks !

$\text{The question deceptively makes it look like we're minimising}\\ \text{the velocity, but that's just a redundancy.}\\ \text{Because if we want to minimise the cost, the simple solution is to not drive.}$
(i.e. $v=0$).
$\text{Note that we want to travel a fixed distance of 500.}\\ \text{Using }v_{av} = \frac{D}{t}, \text{ where }D=500\text{, we see that}\\ \boxed{v = \frac{500}{t}}$
$\text{Therefore the cost per hour is}\\ \boxed{150 + \frac{3125}{t^2}}\text{ cents per hour.}$
$\text{This means, that the total cost, is}\\ \boxed{C = 150t + \frac{3125}{t}}.$
(You should figure out why multiplying by $t$ turns our "rate" into a "quantity".)
_______________________________________________
$\text{With the hard part out of the way, we follow the usual process.}\\ \frac{dC}{dt} = 150 - \frac{3125}{t^2}\text{, which equals 0 when }t=\pm 5 \sqrt{\frac56}.$
$\text{As usual, we reject the negative solution because of negative time, so }t = 5\sqrt{\frac{5}{6}}.\\ \text{Now using the velocity formula again, }v = \frac{500}{5\sqrt{\frac56}}\\ \text{which is approximately 110.}$

RuiAce · « **Reply #3391 on:** April 12, 2018, 08:28:14 pm »

Quote from: beeangkah on April 12, 2018, 03:32:02 pm

Could someone help with 19&20 please x

$u=t+2\implies du = dt.\\ \text{So we have}\\ \begin{align*}\dot{x} &= \int \frac{t}{\sqrt{t+2}}\,dt\\ &= \int \frac{u-2}{\sqrt{u}}\,du\\ &= \int \left( u^{1/2} - 2u^{-1/2}\right)\,du\\ &= \frac{2u^{3/2}}{3}- 4u^{1/2}+C\\ &= \frac23 \sqrt{(t+2)^3} - 4\sqrt{t+2}+C\end{align*}$
$\dot{x} = \frac43\text{ when }t=2\text{ so}\\ \frac43 = -\frac83 + C \implies C = 4.\\ \text{So when }t=0,\\ \begin{align*}\dot{x} &= \frac23\sqrt{8} - 4\sqrt2 - 4\\ &= \frac43\sqrt2 - 4\sqrt2 + 4\\ &= -\frac83\sqrt2+4\end{align*}$
__________________________________________________
$\text{The main hint I'll give you for Q20 is this.}\\ \begin{align*}\ddot{x} &= \cos^2 \left(t+\frac\pi4 \right) - \sin^2 \left(t+\frac\pi4 \right)\\ &= \cos \left[2\left(t+\frac\pi4\right)\right]\\ &= \cos \left( 2t+\frac\pi2\right)\\ &= \cos 2t \cos \frac\pi2 - \sin 2t \sin \frac\pi2\\ &= -\sin2t.\end{align*}\\ \text{I will let you figure out the remainder from here.}$

bdobrin · « **Reply #3392 on:** April 16, 2018, 01:05:55 pm »

Hi,
Can someone please help me with this question:

Prove tan^-1 (3/4) = 2tan^-1 (1/3)

Thanks

RuiAce · « **Reply #3393 on:** April 16, 2018, 01:15:31 pm »

Quote from: bdobrin on April 16, 2018, 01:05:55 pm

Hi,
Can someone please help me with this question:

Prove tan^-1 (3/4) = 2tan^-1 (1/3)

Thanks

$\text{Let }\theta = \tan^{-1} \frac13\text{, so that }\tan \theta = \frac13$
Note that $ -\frac\pi2 < \theta < \frac\pi3$ is assured from the range of the $ \tan^{-1}$ function.
$\text{Then,}\\ \begin{align*}\tan \left(2\tan^{-1} \frac13\right)& = \tan 2\theta\\ &= \frac{2\tan\theta}{1-\tan^2\theta}\\ &= \frac{\frac23}{1-\frac19} \\ &= \frac34.\end{align*}$
$\text{So taking }\tan^{-1}\text{ on both sides, we have}\\ 2\tan^{-1}\frac13 = \tan^{-1}\frac34.$

beeangkah · « **Reply #3394 on:** April 16, 2018, 05:38:20 pm »

Could I have help for this please:

the period of a particle moving in shm is 6s and its amplitude is 8cm. calculate its velocity and acceleration when the displacement is 5cm from the centre of motion

RuiAce · « **Reply #3395 on:** April 16, 2018, 05:50:27 pm »

Quote from: beeangkah on April 16, 2018, 05:38:20 pm

Could I have help for this please:

the period of a particle moving in shm is 6s and its amplitude is 8cm. calculate its velocity and acceleration when the displacement is 5cm from the centre of motion

$\text{We know that }n = \frac{2\pi}{6} = \frac\pi3\text{ and }a = 8.$
From here, it's just plugging into the formulas $ \ddot{x} = -n^2 x$ and $v^2 = n^2(a^2-x^2)$, where $x=5$.

arii · « **Reply #3396 on:** April 19, 2018, 09:44:40 am »

Can I get some help for this question?

RuiAce · « **Reply #3397 on:** April 19, 2018, 12:04:30 pm »

Quote from: arii on April 19, 2018, 09:44:40 am

Can I get some help for this question?

In the future, for long multi-part questions please include relevant progress/ideas you thought of.

Some thoughts for the meantime with part a:
a) i) You should be able to set $y = x$ in the Cartesian equation of motion
a) ii) Set $y = 0$. It might not be immediately obvious as to how $T_1$ comes into play.
a) iii) Hint: If $ 0 < \alpha < \frac\pi4$, then $T_2$ will be negative... and we can't have negative time...

Edit: Also, b) ii) looks like it might involve double angles.

arii · « **Reply #3398 on:** April 19, 2018, 09:14:10 pm »

I managed to get that previous one after trying various things... Anyway, this one is a slightly different type of a project compared to the classic. It continues on so the range is not just 2x, x being the "distance" to the amplitude (highest point), so I'm having some trouble working that part out as I feel like it'll help me get the tan theta stuff.

RuiAce · « **Reply #3399 on:** April 19, 2018, 10:07:24 pm »

Quote from: arii on April 19, 2018, 09:14:10 pm

I managed to get that previous one after trying various things... Anyway, this one is a slightly different type of a project compared to the classic. It continues on so the range is not just 2x, x being the "distance" to the amplitude (highest point), so I'm having some trouble working that part out as I feel like it'll help me get the tan theta stuff.

When I tried doing it, I couldn't force a $\cos\theta$ term to appear. So I'm not sure where the mistake in my working is, but it would follow in a similar line to this. When you typed $\cos$, are you sure you didn't mean $ \cot$?

(Also, I later realised that less working out was needed. The same Cartesian equation of motion works, i.e. $ y = x - \frac{gx^2}{V^2} $, so long as we assume that it passes through $ \left( \frac{30}{\tan \theta}, 30 \right) $ and $ \left( \frac{60}{\tan \theta}, -60\right) $. This is because we can always change the orientation of the problem.

006896 · « **Reply #3400 on:** April 22, 2018, 05:35:59 pm »

Hello friends,
Could you please help me with part b of the attached question?
Thanks in advance!
(I tried to post this question earlier but I'm not sure if it posted because of computer problems, so if I have already posted it, just ignore this post.)

Dragomistress · « **Reply #3401 on:** April 22, 2018, 06:52:25 pm »

I do not quite understand how I am meant to approach this question.

RuiAce · « **Reply #3402 on:** April 22, 2018, 06:54:12 pm »

Quote from: 006896 on April 22, 2018, 05:35:59 pm

Hello friends,
Could you please help me with part b of the attached question?
Thanks in advance!
(I tried to post this question earlier but I'm not sure if it posted because of computer problems, so if I have already posted it, just ignore this post.)

$\text{Since }x^4-x^3+x^2-x+1 = (x^2+4)(x^2-x-3) + 3x+13\\ \text{upon moving the remainder }3x+13\text{ to the other side we have}\\ \boxed{x^4-x^3+x^2 - 4x -12 = (x^2+4)(x^2-x-3)}$
Therefore $a =-4$ and $b=-12$.

RuiAce · « **Reply #3403 on:** April 22, 2018, 07:00:14 pm »

Quote from: Dragomistress on April 22, 2018, 06:52:25 pm

I do not quite understand how I am meant to approach this question.

$\text{We know that }\frac{dS}{dt} = 1.9$
$\text{The surface area and volume formulae are }S = 4\pi^2\text{ and }V = \frac43 \pi r^3.\\ \text{This implies that }\frac{dS}{dr} = 8\pi\text{ and }\frac{dV}{dr} = 4\pi r^2.$
$\text{From the chain rule, used twice,}\\ \begin{align*}\frac{dV}{dt} &= \frac{dV}{dr} \, \frac{dr}{dS} \, \frac{dS}{dt}\\ &= 4\pi r^2 \times \frac{1}{8\pi} \times 1.9\end{align*}\\ \text{Now put in }r=0.6$
Note: You learnt from inverse functions that $ \frac{dx}{dy} = \frac{1}{dy/dx}$.

vikasarkalgud · « **Reply #3404 on:** April 30, 2018, 02:55:06 pm »

Hi, I want help for this motion in a straight line question. I don't seem to understand when u take the positive/negative case for velocity so I end up with +- not knowing which to take

The acceleration of a particle is 2x-5/ms^2, where x is the distance in metres from the origin
a) Find an expression for the velocity of this particle in terms of x, given that the particle is at rest one metre to the left of the origin initially

Answer: -sqrt(2x^2-10x-12). (Why negative???)

ty for help

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