Author Topic: 4U Maths Question Thread (Read 665489 times) Tweet Share

jakesilove · « **Reply #510 on:** September 06, 2016, 11:42:54 pm »

Quote from: katherine123 on September 06, 2016, 09:56:07 pm

how to do last part

Definitely a tough one. First, we know that if the particle is JUST at the point where it doesn't slide down the track, the force acting on it will be

$F=\mu N$

Now, from i)

$\frac{v_{min}^2}{Rg}=\frac{Nsin(\theta)-\mu Ncos(\theta)}{Ncos(\theta)+\mu Nsin(\theta)}$

From ii)

$\frac{v_{min}^2}{Rg}=\frac{Nsin(\theta)-\mu Ncos(\theta)}{Ncos(\theta)+\mu Nsin(\theta)}=\frac{tan(\theta)-\mu}{1+\mu tan(\theta)}$

Now, in the case that

$\mu \geqslant tan(\theta)$

$\frac{tan(\theta)-\mu}{1+\mu tan(\theta)} \leqslant 0$

Therefore, from above

$\frac{v_{min}^2}{Rg} \leqslant 0$

For all real velocities, this will only be true when the velocity equals zero! Therefore, the particle will not travel down the slant for any value of v

amandali · « **Reply #511 on:** September 08, 2016, 03:07:16 pm »

need help with this ques thanks

RuiAce · « **Reply #512 on:** September 08, 2016, 06:07:43 pm »

Quote from: amandali on September 08, 2016, 03:07:16 pm

(Image removed from quote.)

need help with this ques thanks

$\text{For the sake of convenience, let}Q\text{ be the point }(5,0)\\ \text{It doesn't matter what the exact coordinate of }Q\text{ is, it just has to be to the right of }A$
$\text{Let }\angle PAQ = \alpha\\ \text{Realise that we wish to find }\frac{d\alpha}{dt}$
$\text{We know that }\frac{d\theta}{dt}=1\\ \text{So we can use the chain rule:}\\ \frac{d\alpha}{dt}=\frac{d\alpha}{d\theta}\, \frac{d\theta}{dt} = \frac{d\alpha}{d\theta}\\ \text{So somehow we need to form a relationship between }\alpha\text{ and }\theta$
$\text{An awkward approach:}\\ \text{We know that }OP=5, \, OA = 3\\ \text{And that }\angle POA=\theta$
$\text{But using the exterior angle of a triangle, we find that }\angle OPA=\alpha-\theta$
$\text{So we now have all the ingredients to use the sine rule:}\\ \begin{align*}\frac{\sin (\alpha-\theta)}{3}&=\frac{\sin \theta}{5}\\ 5\sin (\alpha-\theta)&=3\sin \theta\end{align*}$
$\text{When }\theta=0,\, \alpha=0\text{ (unsurprising)}\\ \text{When }\theta = \frac{\pi}{2}\text{ we have}\\ 5\sin \left(\alpha-\frac{\pi}{2}\right)=3\sin \frac{\pi}{2}\implies -5\cos\alpha = 3 \implies \alpha = \cos^{-1}\left(-\frac{3}{5}\right)\\ \text{which is an obtuse angle, which sounds correct.}$
$\text{Implicitly differentiating the above equation, making careful use of the chain rule}\\ \begin{align*}5\sin(\alpha-\theta)&=3\sin \theta\\ 5\cos (\alpha-\theta)\left(\frac{d\alpha}{d\theta}-1\right)&=3\cos\theta\end{align*}$
$\text{Substitute in your values from above to ascertain an answer.}$
$\textit{Since the question asks for 'speed', you may also have to}\\\textit{take absolute value if your angular velocity is negative.}$
Let me know if something doesn't makes sense or I got a wrong answer. Plus I also don't do physics anymore and I'm using a brain from a university student's level. Implicit differentiation isn't an obvious approach and I feel it's unnecessarily overcomplicating for 4U. I wanted to use circle geometry but the sine rule was speaking more loudly to me.

massive · « **Reply #513 on:** September 10, 2016, 10:35:26 pm »

Hey guys how do you do this ??

RuiAce · « **Reply #514 on:** September 10, 2016, 11:26:08 pm »

Quote from: massive on September 10, 2016, 10:35:26 pm

Hey guys how do you do this ??

$\text{Force per unit mass is a fancy way of saying "acceleration".}\\ \text{Using what they gave us we have }\\ \begin{align*}F=m\ddot{x}&=-\frac{GmM}{x^2}\\ \ddot{x}&=-\frac{GM}{x^2}\end{align*}\\ \text{I use }x\text{ and not }r\text{ because I'm not a physicist.}$
$\text{When }x=c, \ddot{x}=g\\ \therefore \text{Upon subbing in, }GM=gc^2\\ \therefore \ddot{x}=-\frac{gc^2}{x^2}$
Note that the things here are negative as:
a) The centre of the planet is defined as x=0
b) Gravity brings things BACK to the centre of the planet
$\because\text{ we have acceleration as a function of }x\\ \begin{align*}\frac{d}{dx}\left(\frac{v^2}{2}\right)&=-\frac{gc^2}{x^2}\\ v^2&=\frac{2gc^2}{x}+k\end{align*}$
And this is where I got lost. I kinda had to use HSC physics here... so I reckon this question is pushing the boundaries of the 4U course.
$\text{According to HSC physics, when }x=a, \, v=\sqrt{\frac{GM}{a}}=\sqrt{\frac{gc^2}{a}}\\ \text{Proof to come.}$
$\text{So we find our constant of integration to be }\\ k=\frac{-gc^2}{a^2}\\ \therefore v^2=\frac{2gc^2}{x^2}-\frac{gc^2}{a^2}$
$\text{When }x=a, v=\frac{2\pi a}{T}\\ \therefore \left(\frac{2\pi a}{T}\right)^2=\frac{gc^2}{a}\\ T=2\pi \sqrt{\frac{a^3}{gc^2}}\text{ upon rearranging.}$
$\text{Note that }v=a\omega\text{ and }\omega = \frac{2\pi}{T}\\ \text{as }a\text{ is the radius of orbit.}$
$\text{Proof of the orbital velocity formula:}\\\text{The gravitational force is essentially what invokes the centripetal force.}\\ \begin{align*}F_C&=F_G\\ \frac{mv^2}{R}&=\frac{GmM}{R^2}\\ v&=\sqrt{\frac{GM}{R}}\end{align*}$
Can you please stop avoiding this question I've asked many times and tell us what is the source of these questions?

massive · « **Reply #515 on:** September 10, 2016, 11:36:58 pm »

Quote from: RuiAce on September 10, 2016, 11:26:08 pm

$\text{Force per unit mass is a fancy way of saying "acceleration".}\\ \text{Using what they gave us we have }\\ \begin{align*}F=m\ddot{x}&=-\frac{GmM}{x^2}\\ \ddot{x}&=-\frac{GM}{x^2}\end{align*}\\ \text{I use }x\text{ and not }r\text{ because I'm not a physicist.}$
$\text{When }x=c, \ddot{x}=g\\ \therefore \text{Upon subbing in, }GM=gc^2\\ \therefore \ddot{x}=-\frac{gc^2}{x^2}$
Note that the things here are negative as:
a) The centre of the planet is defined as x=0
b) Gravity brings things BACK to the centre of the planet
$\because\text{ we have acceleration as a function of }x\\ \begin{align*}\frac{d}{dx}\left(\frac{v^2}{2}\right)&=-\frac{gc^2}{x^2}\\ v^2&=\frac{2gc^2}{x}+k\end{align*}$
And this is where I got lost. I kinda had to use HSC physics here... so I reckon this question is pushing the boundaries of the 4U course.
$\text{According to HSC physics, when }x=a, \, v=\sqrt{\frac{GM}{a}}=\sqrt{\frac{gc^2}{a}}\\ \text{Proof to come.}$
$\text{So we find our constant of integration to be }\\ k=\frac{-gc^2}{a^2}\\ \therefore v^2=\frac{2gc^2}{x^2}-\frac{gc^2}{a^2}$
$\text{When }x=a, v=\frac{2\pi a}{T}\\ \therefore \left(\frac{2\pi a}{T}\right)^2=\frac{gc^2}{a}\\ T=2\pi \sqrt{\frac{a^3}{gc^2}}\text{ upon rearranging.}$
$\text{Note that }v=a\omega\text{ and }\omega = \frac{2\pi}{T}\\ \text{as }a\text{ is the radius of orbit.}$
$\text{Proof of the orbital velocity formula:}\\\text{The gravitational force is essentially what invokes the centripetal force.}\\ \begin{align*}F_C&=F_G\\ \frac{mv^2}{R}&=\frac{GmM}{R^2}\\ v&=\sqrt{\frac{GM}{R}}\end{align*}$
Can you please stop avoiding this question I've asked many times and tell us what is the source of these questions?

Hey thankyou so much. This particular question was from the catholic trial 1984...I don't usually do questions that are really old, it's just that i need to get good in maths HSC, theyre the only things pulling my atar up :L

RuiAce · « **Reply #516 on:** September 10, 2016, 11:45:47 pm »

Quote from: massive on September 10, 2016, 11:36:58 pm

Hey thankyou so much. This particular question was from the catholic trial 1984...I don't usually do questions that are really old, it's just that i need to get good in maths HSC, theyre the only things pulling my atar up :L

That was all you had to say though haha. That explains why these questions are extremely difficult - they're going beyond 1990.

Neutron · « **Reply #517 on:** September 12, 2016, 07:41:43 pm »

Hello y'all! I come with a complex numbers question (I have no bloody clue what's going on with this one):

Let the points A₁,A₂,...,A_n represent the nth roots of unity, w₁, w₂, ..., w_n, and suppose P represents any complex number z such that lzl=1.

a) Prove that w₁+w₂+...+w_n=0

b) Show that PA²_i=(z-w_i)(z (conjugate) - w_i (conjugate) for i=1,2,...,n

c)Prove that:
Σ (from i=1 to n) PA²_i=2n

Sorry it's really hard to type it out on this forum because of all the squares and subscripts and conjugate bars and sigmas and ah i hope it still makes sense because it doesn't to me! Thanks guys

Neutrons

RuiAce · « **Reply #518 on:** September 12, 2016, 08:01:05 pm »

Quote from: Neutron on September 12, 2016, 07:41:43 pm

Hello y'all! I come with a complex numbers question (I have no bloody clue what's going on with this one):

Let the points A₁,A₂,...,A_n represent the nth roots of unity, w₁, w₂, ..., w_n, and suppose P represents any complex number z such that lzl=1.

a) Prove that w₁+w₂+...+w_n=0

b) Show that PA²_i=(z-w_i)(z (conjugate) - w_i (conjugate) for i=1,2,...,n

c)Prove that:
Σ (from i=1 to n) PA²_i=2n

Sorry it's really hard to type it out on this forum because of all the squares and subscripts and conjugate bars and sigmas and ah i hope it still makes sense because it doesn't to me! Thanks guys

Neutrons

$\text{If }\omega\text{ is an }n\text{th root of unity, we have }\\ \omega^n = 1\\ \text{So we rewrite Qa) using powers and not subscripts.}$
$\omega+\omega^2+\dots+\omega^n\text{ is nothing more than a G.P.}\\ \text{So we have a formula for its sum:}\\ \frac{\omega(\omega^n-1)}{\omega-1} =\frac{0}{\omega-1}=0\\ \text{Notes: }\omega-1\text{ is not zero as }\omega \neq 1\\ \text{There are }n\text{ terms in the G.P.}\\ \textit{You should understand WHY }\omega_n=\omega^n$
___________________________________________
$\textbf{Draw a diagram, clearly showing the following points:}\\ \text{The origin }O(0,0)\\ \text{Any point }P\text{ on the unit circle}\\ \text{A point }A_i\text{ also on the unit circle.}$
$\text{By vector addition, we have the following result.}\\ \begin{align*}\overrightarrow{PA_i}&=\overrightarrow{PO}+\overrightarrow{OA_i}\\ &=-z+w_i\end{align*}\\ \text{Note that }P\text{ represents }z\\\text{and }A_i\text{ represents }\omega_i$
$\text{So upon taking moduli (lengths) and squaring:}\\ \begin{align*}|PA_i|&=|-z+w_i|\\ &=|z-w_i|\\ \implies PA_i^2 &= |z-w_i|^2\\ &= (z-w_i)\overline{(z-w_i)}\end{align*}\\ \text{Notes: Absolute values were discarded as:}\\ \text{On the LHS, }PA_i\text{ is just a length so it must be positive}\\ \text{On the RHS, as an exercise you should prove that }z\overline{z}=|z|^2$
___________________________________________
$\text{We EXPAND out the result from b) before summing}\\ \begin{align*}PA_i^2&=z\overline{z}-\omega_i\overline{z}-\overline{\omega_i}z+\omega_i\overline{\omega_i}\\ &= |z|^2-\omega_i\overline{z}-\overline{\omega_i}z+|\omega_i|^2\\ &=1-\omega_i\overline{z}-\overline{\omega_i}z+1\\ &=2-\omega_i\overline{z}-\overline{\omega_i}z\end{align*}$
$\text{And now we take the sum.}\\ \text{Note that what was used in part a) comes in.}\\ \text{Also note that the conjugate of the sum is the sum of the conjugates}\\ \text{i.e. }\overline{z+w}=\overline{z}+\overline{w}$
$\text{From a), }\sum_{i=1}^n\omega_i=0$
$\begin{align*}\sum_{i=1}^nPA_i^2&=\sum_{i=1}^n(2-\omega_i\overline{z}-\overline{\omega_i}z)\\ &= \sum_{i=1}^n2-\sum_{i=1}^n\omega_i\overline{z}-\sum_{i=1}^n\overline{\omega_i}z\\ &= 2n - \overline{z}\sum_{i=1}^n\omega_i - z\overline{\sum_{i=1}^n\omega_i}\\ &= 2n - 0 - \overline{0}\\ &= 2n\end{align*}$

massive · « **Reply #519 on:** September 16, 2016, 08:48:53 am »

How do you do this?

If abs(a)>2abs(b), prove that 2abs(a-b)>abs(a)

thanks!

RuiAce · « **Reply #520 on:** September 16, 2016, 09:30:22 am »

Quote from: massive on September 16, 2016, 08:48:53 am

How do you do this?

If abs(a)>2abs(b), prove that 2abs(a-b)>abs(a)

thanks!

$\text{Note that if }|a|>2|b|\\ \text{Then }-|b|>-\frac{1}{2}|a|$
$\text{We will use the reverse triangle inequality to help us out.}\\ \text{You may wish to prove this as an exercise.}\\ \text{Use a diagram if you do.}$
$\begin{align*}2|a-b|&\ge 2\left||a|-|b|\right|\\ &>2\left|a-\frac{1}{2}|a|\right|\tag{from above}\\ &= |a|\tag{QED}\end{align*}$

karenc. · « **Reply #521 on:** September 19, 2016, 04:44:56 pm »

Hi everyone can u please help me with this question, it was from my school's trial:

Ali_Abbas · « **Reply #522 on:** September 19, 2016, 06:06:30 pm »

Quote from: karenc. on September 19, 2016, 04:44:56 pm

Hi everyone can u please help me with this question, it was from my school's trial:

$\textrm{i) You are given the parametric equations of displacement for the} \\ x \space \textrm{and} \space y \space \textrm{directions and are required to find} \\ \textrm{the Cartesian equation of motion. This is achieved by eliminating the parameter,} \\ t, \textrm{from the given system. Conveniently, the equation of motion in the} \\ x \space \textrm{direction is linear in} \space t \space \textrm{and hence is easily} \\ \textrm{manipulated to get} \space t \space \textrm{as the subject. After doing this,} \\ \textrm{make a substitution into the equation for} \space y \space \textrm{and after some} \\ \textrm{simplifying you will obtain the desired result.}$

$\textrm{ii) The range over the ground will simply be the value of } \space x \space \textrm{when}\\ y = 0. \space \textrm{Substitute} \space y = 0 \space \textrm{into the Cartesian equation from part i) and} \\ \textrm{you will have a quadratic in} \space x. \space \textrm{Solve using the} \\ \textrm{quadratic formula and again, after some careful simplifying you} \\ \textrm{will obtain the required expression.}$

$\textrm{iii) For the keys to avoid the dam, we simply require the horizontal} \\ \textrm{range to be between} \space 0 \space \textrm{and} \frac{V^{2}(1 + \sqrt{3})}{4g} \space or\\ \textrm{the horizontal range is greater than} \frac{V^{2}(1 + \sqrt{3})}{4g} + \frac{V^{2}}{2g}. \\ \textrm{Set up these inequalities for the expression you found in part ii)} \\ \textrm{for the range and solve for} \space \theta, \space \textrm{noting that} \space 0 < \theta < 90 \space \textrm{degrees}.$

RuiAce · « **Reply #523 on:** September 26, 2016, 01:26:25 pm »

Here is a question I was asked during the 3U lecture break. I will address it in the next post.

Apologies for the poor quality photo

RuiAce · « **Reply #524 on:** September 26, 2016, 02:00:06 pm »

Quote from: RuiAce on September 26, 2016, 01:26:25 pm

Here is a question I was asked during the 3U lecture break. I will address it in the next post.
(Image removed from quote.)
Apologies for the poor quality photo

Unfortunately, I was unable to generate a better proof than what I offered up on the day for a 2 mark question. I will give a full statement of the proof I had then. It may be worth mentioning that I considered an approach that utilises stationary points, however felt it was a bit overkill for a 2 mark question. The stationary points do not have a tidy form: $ x=q+\frac{q-r}{-1\pm \sqrt{\frac{p-r}{p-q}}} $

(Note that this proof used a graph to get there, mostly to visualise the shape of the function, but doesn't properly involve the graph itself. I could not see how the partial fractions decomposition aided any further.)

So I'm basically welcoming a more elegant proof.

(Also, if you're the one who asked me this, sorry that I got the assumed inequality in the next line the wrong way around!)
$\text{Suppose, without loss of generality, } r < q< p$
$\text{The partial fractions decomposition in part (i) was}\\ f(x)=\frac{1}{q-r}\left(\frac{q-p}{x-q}+\frac{p-r}{x-r}\right)$
$\text{An easy approach to part (ii) is to take the derivative and deduce that it's strictly positive (or negative)}\\ \text{and to also observe that the range of }f(x)\text{ is all real }y\\ \text{Note, you would have something such as }r< p< q\text{ for this part.}$
$\text{A brief sketch-up for part (iii) exhibits a curious result. There exists a local min slightly below }y=0\\ \text{and a local max somewhere in the region }r< x< q\text{ well below }y=0$
$\text{To ensure we have a generalised case, try }f(x)=\frac{1}{q-r}\\ \text{Inspiration: It was factorised out too conveniently. That and it looked like it belonged in the gap}$
$\text{Then, we'd have}\\ \begin{align*}\frac{1}{q-r}&=\frac{1}{q-r}\left(\frac{q-p}{x-q}+\frac{p-r}{x-r}\right)\\ 1&=\frac{q-p}{x-q}+\frac{p-r}{x-r}\\ (x-q)(x-r)&=(q-p)(x-r)+(p-r)(x-q)\\ x^2-qx-rx+qr&= qx-px-qr+pr+px-rx-pq+qr\\ x^2-2qx+(qr+pq-pr)&=0\end{align*}$
$\text{The quadratic discriminant is}\\ \begin{align*}\Delta&=4q^2-4(1)(qr+pq-pr)\\ &= q^2-qr-pq+pr\\ &= (q-p)(q-r)\end{align*}$
$\text{Since we took }r< q< p\text{ we have}\\ q-p < 0\\ q-r > 0\\ \text{Hence their product is also negative so the discriminant is negative.}\\ \text{So the above equation must have no solution.}$
$\text{Hence working all the way back, the original equation must not have had a solution}\\ \text{for ANY choice of }q\text{ and }r\\ \text{thus satisfying the criteria.}$

Search the forums now!

We have moved!

Author Topic: 4U Maths Question Thread (Read 665489 times) Tweet Share

jakesilove

Re: 4U Maths Question Thread

amandali

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

massive

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

massive

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

Neutron

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

massive

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

karenc.

Re: 4U Maths Question Thread

Ali_Abbas

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

Recent Posts

User:
Password: