VCE Specialist 3/4 Question Thread!

[nhmn0301]

Can anyone help with this MC


Question 6

If z is any complex number, then -i^3z   is found by
a.   rotating z by 3pi/2  in an anticlockwise direction about the origin
B.   rotating z by pi/2  in an anticlockwise direction about the origin
C.   rotating z by 3pi/2  in an anticlockwise direction about the origin and reflecting it in the   axis
D.   rotating z by  pi/2 in an anticlockwise direction about the origin and reflecting it in the Im(Z)   axis
E.   reflecting z in the line y=x

-i^3 = - (-i) = i
Hence, iz is equivalent to rotating z by pi/2 anticlockwise.

[M_BONG]

-i^3 = - (-i) = i
Hence, iz is equivalent to rotating z by pi/2 anticlockwise.

How is rotating something by pi/2 equivalent to multiplying it by i?

[psyxwar]

How is rotating something by pi/2 equivalent to multiplying it by i?

i=cis(pi/2)

Multiplying by cis (pi/2) means the argument increases by pi/2, ie. rotation pi/2 anticlockwise.

[nhmn0301]

This seems like an easy question but I still don't get why the answer is C?

[psyxwar]

v=x+2
a=dv/dt = dv/dx * dx/dt = dv/dx * v
dv/dx = 1
therefore a=(x+2)(1)=x+2

[nhmn0301]

v=x+2
a=dv/dt = dv/dx * dx/dt = dv/dx * v
dv/dx = 1
therefore a=(x+2)(1)=x+2

Ah yeah true, silly me. Thanks !

[nhmn0301]

Could someone be kind enough to point me in the right direction for question 2biii in the attachment? Please don't give me the answer, just a small clue would be good.

Is it a hard question, or am I overthinking it? Also, is there some form of trig identity involved?

Thanks!

Draw a force diagram first. You see that there is no applied force in this question, but there is parallel force going downward, hence, friction will go in a direction that opposes motion. Since the question states the system is in equilibrium, you know the net force is equal to 0, aka F=ma = 0. Oh, and rmb N = mgcos(theta).There no trig identity involved. Hope this hint helps! All the best!

[M_BONG]

Draw a force diagram first. You see that there is no applied force in this question, but there is parallel force going downward, hence, friction will go in a direction that opposes motion. Since the question states the system is in equilibrium, you know the net force is equal to 0, aka F=ma = 0. Oh, and rmb N = mgcos(theta).There no trig identity involved. Hope this hint helps! All the best!

Sorry I wanted help with part b iii.

Looks like you're describing bii
(Correct me if I'm wrong

[psyxwar]

Sorry I wanted help with part b iii.

Looks like you're describing bii
(Correct me if I'm wrong

yeah you need double angle.

[nhmn0301]

Sorry I wanted help with part b iii.

Looks like you're describing bii
(Correct me if I'm wrong

Yes sorry about that. I couldn't think anything properly today O_O.. Just a bit more hint: after cancelling out the "m" term, bring everything under one denominator and you should see the double angle formula !

[lzxnl]

Generally, you'll find that questions involving only gravity and friction have accelerations independent of the mass. Why? Well, the weight force is mg and any frictional force is some constant times mg. In either case, once you divide through by m, the masses will all cancel.

Trig identities are in your formula sheet, but you really should know them

[M_BONG]

How far are you expected to simplify for derivative problems?

I have a 2 marker, which is involves deriving tan inverse. And since it's only worth 2 marks, I did all the steps required (4 steps) and didn't bother simplifying. The marking scheme for Heffernan simplifies it. Do I not get an answer mark if I don't simplify?

[keltingmeith]

How far are you expected to simplify for derivative problems?

I have a 2 marker, which is involves deriving tan inverse. And since it's only worth 2 marks, I did all the steps required (4 steps) and didn't bother simplifying. The marking scheme for Heffernan simplifies it. Do I not get an answer mark if I don't simplify?

To memory, Heffernan have a marking guide with their solutions (they'll say at what points they'd distribute the marks), so I suggest checking that. However, you shouldn't need to simplify tricky things unless you're told that you should.

[Zlatan]

Could someone help me with this question ??

[keltingmeith]

Just a really annoying application of multiplying/dividing by complex numbers/De Moivre's theorem:



At this point, we see that both our angle is in the wrong domain, and that our radial value is negative. Both of these we don't like - but we'll fix the first one before the second. A negative sign is equivalent to multiplying by , which is a rotation of 180 degrees. We'll go backwards, and hope this fixes our arguement:

And now we have a number that agrees with our rules.