VCE Specialist 3/4 Question Thread!

[ahat]

technically b = 2a is fine. b and a without the tilde means magnitude.

Thanks I just noticed that they did remove it in their solution (the tilde that is) Thanks man!

The main problem I see is that PM is a factor of PR, but OM means little.
I would have went PR.QM=0           insert vector arrows over all capitals

Btw Nilu, why PR.QM opposed to PR.MQ?

[SocialRhubarb]

Btw Nilu, why PR.QM opposed to PR.MQ?

Same thing.

[lzxnl]

QM=-MQ
PR.QM=-PR.MQ
but they're both zero anyway

[ahat]

I get Q10.c. now, thanks to your detailed solutions . But can I use De moivre theorem to solve this?
As for Q9a. is there a way to solve the area using calculus?What about 0 to pi. Is it possible in that case? Im a bit confused

There is, but doing it by hand is excessively complicated (I tried it once), and one mark should instantly tell you this is not the case. Doing it from 0 to pi would still introduce the same problem of finding the red area (because of symmetry, this is the same as finding twice the area of 0 to pi/2). Remember, your suggestion was to take the strips from the y axis. This means that the strips will be oriented horizontally and not vertically, therefore, you're going to be finding the area enclosed by the y-axis and the curve, not the x-axis and the curve such as the question requires. Hope that makes sense.

As for using De Moivre's theorem, I do not believe so. We can't express az³ + bz²... etc easily in the form rcis(a), so we can't apply the theorem. The information in the question would require us to find the values of a,b,c before this was possible anyway.

^and thanks Nilu and SocialRhubarb.

[sin0001]

Hey could someone please explain why the answer to the attached question is not D? The solutions say A
Thanks!

[abcdqd]

The distance between them is the difference of their displacements at t=20. Displacement at t=20 is calculated by the SIGNED area of the graph from t=0 to t=20. Option D gives you the UNSIGNED area (distance travelled) of f(t), since it places a negative sign in front of the second integral which is already negative, thus making it positive and adding on to the first integral. Therefore answer must be A.

[sin0001]

The distance between them is the difference of their displacements at t=20. Displacement at t=20 is calculated by the SIGNED area of the graph from t=0 to t=20. Option D gives you the UNSIGNED area (distance travelled) of f(t), since it places a negative sign in front of the second integral which is already negative, thus making it positive and adding on to the first integral. Therefore answer must be A.

I misinterpreted it as the difference in their distances, when it says 'they are travelling in a straight line'
Right, thanks heaps! That made sense

[LOLs99]

Hey could someone please tell me why I cant get a solution from CAS for Q4 e. http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2008specmaths2-w.pdf

[Phy124]

Are you solving or something else?

[SocialRhubarb]

Try hitting approximate instead of the regular enter key.

[ahat]

nSolve may work

[LOLs99]

Yup I did it. just left out the 10 when doing the question. Sorry guys

[simba]

Super confused as to how to work this out D:
The answer is B

[ahat]

Super confused as to how to work this out D:
The answer is B

This is the progress I made
(gosh this latex was a nightmare)

1. Draw a right angled triangle so you can express arcos as arctan.
2. So, arctan = < screw it, that 2 should be under the sqareroot.
and so, arctan =

3. Our expression now looks like:

4. Apply double angle tan formula, we'll let

5.

6.

7. Which gets me to  

(by removing the tan and multiplying top and bottom by 4, so I guess it's the answer - if we let |m| = m)
You don't know how painful that was ^ (latex)

[brightsky]

so we have the expression tan(2arccos(2/sqrt(m^2+4))
let a = arcos(2/sqrt(m^2+4))
cos(a) = 2/sqrt(m^2+4)
now draw a right angled triangle. label one of the non-right angles a. cos(a) = adj/hyp = 2/sqrt(m^2+4) so make the hypotenuse sqrt(m^2+4) and the adjacent to the angle a, 2. you will see that tan(a) = opp/adj = m/2.
now
tan(2arccos(2/sqrt(m^2+4))
= tan(2a)
= 2tan(a)/(1-tan^2(a))
= 2(m/2)/(1-(m/2)^2)
= m/(1-m^2/4)
= 4m/(4-m^2)
which is B.