2014 Extension 2 BOS Trial Exam Discussion Thread (3 Viewers)

[alficio]

how does the (1-x)S(x) way work?

 

[Trebla]

Did you differentiate the geometric series formula or did you use some other trick for accounting for the linearly increasing coefficients? (The terms in the series being kz^{k-1}).

To get the closed form initially I used the standard differentiation trick, I think in Trebla's solutions he also used the sum of the sum of gp method. There's a third way I know of using (1-x)S(x) trick that handles it nicely.

I deliberately avoided the differentiation of GP for the solution because that would involve differentiation of complex numbers which is outside the scope of the syllabus (though the result would still work out if you did it this way)

 

[aDimitri]

The last option was supposed to be - 1-n.

This one was my fault. In the construction of the question, I summed two geometric series, one including w and another with w bar. I forgot though that I had to subtract 1 from both sides.

i noticed this just by testing. same goes for Q6, none of the options were correct when i just subbed a random value for theta.
does this mean you just remove the question when marking?

 

[egress]

It was a circle.

Holy moly I just realised an almost identical question came up in my half yearlies and I lost a couple of marks over it. Thought I'd learnt from that mistake, but apparently not. (for anyone's that wondering, it's just using the basic angle subtended by arc at centre = twice the angle subtended at circumference circle geometry property…)

Is it weird that I still have no idea what's happening in 12d? If a car turns along a flat road, and r stays constant, doesn't F just keep rising alongside V?

 

[Carrotsticks]

Holy moly I just realised an almost identical question came up in my half yearlies and I lost a couple of marks over it. Thought I'd learnt from that mistake, but apparently not. (for anyone's that wondering, it's just using the basic angle subtended by arc at centre = twice the angle subtended at circumference circle geometry property…)

Is it weird that I still have no idea what's happening in 12d? If a car turns along a flat road, and r stays constant, doesn't F just keep rising alongside V?

That would be true assuming that F is not bounded, but remember the question stated that F acquires a maximum value at those values of V.

Intuitively, it doesn't make sense for F to keep rising alongside V unless the tyres are ultra grippy, and even so at some point the tyres will give. Because you can imagine at some point, if you make the turn too quickly, the car will start to slip.

As you drive faster and faster around the bend, your value of F increases accordingly. However, your tyres can only provide an output of at most say K newtons of frictional force. If you increase your velocity, but your F has already attained a maximum, the only thing that CAN possibly change afterwards (to accommodate for the increasing values of V) is your radius R, which is why you slip in an outwards manner (the radius is increasing).

 

[aDimitri]

There's no problem with Q6, the answer is cos4θ + isin4θ. Quickest way of doing it is letting z=cisθ and multiplying the contents of the brackets by z/z, you'll find that it simplifies to z^4.

unfortunately it's not. it's cos4θ - isin4θ. sub theta = 1 and just test it. glittergal got the same in her solutions.

 

[aDimitri]

That would be true assuming that F is not bounded, but remember the question stated that F acquires a maximum value at those values of V.

Intuitively, it doesn't make sense for F to keep rising alongside V unless the tyres are ultra grippy, and even so at some point the tyres will give. Because you can imagine at some point, if you make the turn too quickly, the car will start to slip.

As you drive faster and faster around the bend, your value of F increases accordingly. However, your tyres can only provide an output of at most say K newtons of frictional force. If you increase your velocity, but your F has already attained a maximum, the only thing that CAN possibly change afterwards (to accommodate for the increasing values of V) is your radius R, which is why you slip in an outwards manner (the radius is increasing).

it took me a little to understand what the question meant by maximum friction actually. i interpreted it as maximum lateral force that the car can take while still saying in uniform circular motion, the same way that when a question says "maximum tension" it means the most force it can take before it snaps. but i still got the question wrong

 

[Carrotsticks]

First and second place currently separated by one mark. Still got question 12 and the multiple choice to mark.

 

[Ixuvia]

Are marks still being released later today?

 

[Carrotsticks]

Yes, they will be released tonight.

What I will do is release the table containing the marks (no names) and then ask people to PM me if they wish to acquire their marks.

 

[miamiheat]

Yes, they will be released tonight.

What I will do is release the table containing the marks (no names) and then ask people to PM me if they wish to acquire their marks.

Will worked solutions be released? if so when?

 

[Ixuvia]

Awesome! Feel free to PM me my mark whenever that happens, I'm pretty keen to find out how I went.

 

[aDimitri]

i would also like my mark PM'd to me

 

[Trebla]

Solutions will be released a little bit later. They require a bit of time to finalise

 

[mirachael]

PM me as well when they're released

 

[egress]

That would be true assuming that F is not bounded, but remember the question stated that F acquires a maximum value at those values of V.

Intuitively, it doesn't make sense for F to keep rising alongside V unless the tyres are ultra grippy, and even so at some point the tyres will give. Because you can imagine at some point, if you make the turn too quickly, the car will start to slip.

As you drive faster and faster around the bend, your value of F increases accordingly. However, your tyres can only provide an output of at most say K newtons of frictional force. If you increase your velocity, but your F has already attained a maximum, the only thing that CAN possibly change afterwards (to accommodate for the increasing values of V) is your radius R, which is why you slip in an outwards manner (the radius is increasing).

Ahh I misinterpreted the question quite badly then :L

Yep, okay, with this, I managed to solve the problem. Got stuck with rgV₀²cosθ = rg(V₁² - V₂²) - (V₀V₂)²sinθ for a while, though. That was rather tricky.

 

[Carrotsticks]

My apologies all, but the results will need to be released tomorrow along with 2U and 3U.
I was ACTUALLY ready to make the thread and clicked on the button to release it, then I got logged out for some reason and I got kicked out of the office for staying back too late. Now I have to go somewhere else to tend to other things.

My sincerest apologies all!

 

[mreditor16]

My apologies all, but the results will need to be released tomorrow along with 2U and 3U.
I was ACTUALLY ready to make the thread and clicked on the button to release it, then I got logged out for some reason and I got kicked out of the office for staying back too late. Now I have to go somewhere else to tend to other things.

My sincerest apologies all!

haha no worries carret.

 

[Kurosaki]

Well that was anticlimactic LOL.

 

[tomnomnom696]

ahahah no worries