VCE Specialist 3/4 Question Thread!

[Robert123]

Ok still confused about general solutions...

This is a complex number, hence Arg Z is between - pi and pi

When I am solving for:
cos (npi/12) = 0

This is what I do,

npi/12 = pi/2 + 2k pi     k element of Z

rearranging:

npi = 6pi + 24 k pi

n = 6 + 24k.

Answer is 6 +12 k?


Isn't general solutions just getting the two angles which equal something and adding two pi to it? Or is it different for complex numbers because of the restriction on the argument?

The best way I found when doing these is to use a mental unit circle. Now on a unit circle, the cos value provide the x coordinate. Now this equal 0 at the angles of Pi/2 and 3pi/2. In your answer, you have neglected the other answer, 3pi/2, which, when visualised mentally you can see cos(x)=0 every half of the period, ie Pi.

Majority of the time, when you do general solutions, there will be two answers within the first period, thus, two General solution equations. In your scenario, these two general solution equatikns can nearly be combined into one single equation with half the period.

[M_BONG]

The best way I found when doing these is to use a mental unit circle. Now on a unit circle, the cos value provide the x coordinate. Now this equal 0 at the angles of Pi/2 and 3pi/2. In your answer, you have neglected the other answer, 3pi/2, which, when visualised mentally you can see cos(x)=0 every half of the period, ie Pi.

Majority of the time, when you do general solutions, there will be two answers within the first period, thus, two General solution equations. In your scenario, these two general solution equatikns can nearly be combined into one single equation with half the period.

I took 3pi/2 out because it's not part of the domain of Arg (Z) isn't that what you're supposed to do?

[keltingmeith]

I took 3pi/2 out because it's not part of the domain of Arg (Z) isn't that what you're supposed to do?

This is wrong - because 3pi/2 is outside of the domain, you have to use something equivalent which IS inside the domain. In this case, -pi/2.

[M_BONG]

This is wrong - because 3pi/2 is outside of the domain, you have to use something equivalent which IS inside the domain. In this case, -pi/2.

Huh? But pi/2 is in the domain, which is what I did? And I still got 6 + 24k not 6+12k as required?

[lzxnl]

Ok still confused about general solutions...

This is a complex number, hence Arg Z is between - pi and pi

When I am solving for:
cos (npi/12) = 0

This is what I do,

npi/12 = pi/2 + 2k pi     k element of Z

rearranging:

npi = 6pi + 24 k pi

n = 6 + 24k.

Answer is 6 +12 k?


Isn't general solutions just getting the two angles which equal something and adding two pi to it? Or is it different for complex numbers because of the restriction on the argument?

General cos solution:

cos x = y
x = +- arccos(y) + 2n*pi

Remember how cos(-x) = cos x?
So for your question, you should have cos(npi/12) = 0
npi/12 = +- pi/2 + 2k pi
n = +-6 + 24k
which is 6, 18, 30, 42 etc, coinciding with 6 + 12k

[M_BONG]

General cos solution:

cos x = y
x = +- arccos(y) + 2n*pi

Remember how cos(-x) = cos x?
So for your question, you should have cos(npi/12) = 0
npi/12 = +- pi/2 + 2k pi
n = +-6 + 24k
which is 6, 18, 30, 42 etc, coinciding with 6 + 12k

Oh I see the mistake. I only wrote 6 + 24k, totally ignoring -6 +24k (because I just removed 3pi/2 without making it -pi/2) and VCAA condensed +-6k + 24k into 6+12k....thanks..

[BLACKCATT]

Any help would be nice. Questions attatched

[M_BONG]

Can someone help with this question, not sure what its actually asking..

NEAp 2014 (5 marks)

The decrease in a thermometer's temperature reading is modelled by the differential equation dT/dt = -k (T - To)  where To is the outdoor temperature.

A thermometer kept at a constant room temperature of 25 degrees is placed outside.
After five minutes outside, the temperature is 15 degrees.
Five minutes later, the temperature reads 10 degrees.

If To is the thermometer's temperature t minutes after being placed outside, find To.

** not only is the wording really ambiguous, they give you two definitions of To - ie, outdoor temperature and the thermometer temperature...

Someone tell me what's going on please

[lzxnl]

Can someone help with this question, not sure what its actually asking..

NEAp 2014 (5 marks)

The decrease in a thermometer's temperature reading is modelled by the differential equation dT/dt = -k (T - To)  where To is the outdoor temperature.

A thermometer kept at a constant room temperature of 25 degrees is placed outside.
After five minutes outside, the temperature is 15 degrees.
Five minutes later, the temperature reads 10 degrees.

If To is the thermometer's temperature t minutes after being placed outside, find To.

** not only is the wording really ambiguous, they give you two definitions of To - ie, outdoor temperature and the thermometer temperature...

Someone tell me what's going on please

Looks to me they mean let 'T' be the thermometer's temperature if you're trying to use Newton's law of cooling here

So at t=5, the temperature is 15 degrees
At t=10, the temperature is 10 degrees
At t=0, the temperature 25 degrees

Does that clarify anything?

[M_BONG]

Looks to me they mean let 'T' be the thermometer's temperature if you're trying to use Newton's law of cooling here

So at t=5, the temperature is 15 degrees
At t=10, the temperature is 10 degrees
At t=0, the temperature 25 degrees

Does that clarify anything?

hmm, not exactly. I did assume that and tried going with it but didn't get anywhere, unless I am doing it all wrong...

Can you please try doing it the normal way then NEAP's way? When I did it the "normal way" I had To in a bunch of random places and couldn't solve it tech free...


BUT, if I numerically solve on the CAS i get the right answer. If you solve the "normal way" without numerically solving, CAS doesn't give you a value. Is NEAP being dodgy here where you have to know their particular method tech free to be able to solve the question?

[M_BONG]

Someone check my working please. I seem to have switched the sign the wrong way....
Just to clear things up

IF, for example,

You have 2loge | x -2 | = t

You sub in x = 0 and t = 0

You get, 0 = 2 loge |-2)

Does this now become 0 = -2 loge (2)??

If so, why is my working wrong. Cheers!

[psyxwar]

Someone check my working please. I seem to have switched the sign the wrong way....
Just to clear things up

IF, for example,

You have 2loge | x -2 | = t

You sub in x = 0 and t = 0

You get, 0 = 2 loge |-2)

Does this now become 0 = -2 loge (2)??

If so, why is my working wrong. Cheers!

lolwut

ln(|-2|) = ln(2)
ln(1/2)=ln(2^-1)=-ln(2)

[M_BONG]

lolwut

ln(|-2|) = ln(2)
ln(1/2)=ln(2^-1)=-ln(2)

Wait I must have things really confused.

I thought when you have ln(|-2|), you mod the argument so it becomes In (2) but since you have modded the argument, you make it negative ( -In(2) )....  LOLOL if that is wrong, I have made that mistake all year LOL.

[psyxwar]

Wait I must have things really confused.

I thought when you have ln(|-2|), you mod the argument so it becomes In (2) but since you have modded the argument, you make it negative ( -In(2) )....  LOLOL if that is wrong, I have made that mistake all year LOL.

are you retarded LOL kys

nah when you take the modulus it just means it's always going to be positive. you're overthinking it lol, |ln(x)| would do what you described (negative branches would be reflected across the x-axis).

[eagles]

Any ideas how to do question 21?

Thanks.