VCE Specialist 3/4 Question Thread!

[b^3]

We are trying to prevent the box from moving down the slope (as gravity is trying to pull it down), that means friction has to be in the opposite direction to which the block is wanting to move (this is the static friction type of friction). So that would be up the slope, in this case we are also appying a force up the slope, which is the tension.

In some cases the friction will be opposed to tension (e.g. a block on a flat slope being pulled by a rope), others it won't be (as in this case), it just depends on what the situation is.

[|ll|lll|]

a) Given z + 1/z = k, where k us a real constant, show that z either lies on the Re(z) axis, or the unit circle centred at the origin.
b) if z lies on the Re(z) axis, show that |k| ≥ 2.

What will everyone's approach be on this? I proved it using polar form, but not in a very elegant way so could someone shed some light? Thanks

[martin1106]

We are trying to prevent the box from moving down the slope (as gravity is trying to pull it down), that means friction has to be in the opposite direction to which the block is wanting to move (this is the static friction type of friction). So that would be up the slope, in this case we are also appying a force up the slope, which is the tension.

In some cases the friction will be opposed to tension (e.g. a block on a flat slope being pulled by a rope), others it won't be (as in this case), it just depends on what the situation is.

thank you

[rife168]

a) Given z + 1/z = k, where k us a real constant, show that z either lies on the Re(z) axis, or the unit circle centred at the origin.
b) if z lies on the Re(z) axis, show that |k| ≥ 2.

What will everyone's approach be on this? I proved it using polar form, but not in a very elegant way so could someone shed some light? Thanks

let
so



but this must equal to a real constant, so the imaginary part must equal zero







so then we get y=0 which implies that z lies on the Re(z) axis

or we get which shows that z lies on the unit circle centred at the origin



just give me a sec to look at the second part

[polar]

If you consider the real component,


For z to lie on the Re(z) axis, y=0









x must be a real number otherwise it won't lie on the Re(z) axis, therefore, there must be at least 1 real solution

[rife168]

For part b) Polar's answer is a nice way to do it (although it should be instead of ), but I'm bored, so for something more visual:

z lies on the Re(z) axis implies that y=0

reduces to



now if we look at then we realise that the range of is all values that k can take. So now we need to find the range of

The graph of looks like this:



We can verify the values of those turning points by letting 
From that we get which gives us turning points at

So it follows that the range of is

which is equivalent to the statement:

For a proper exam response you might want a bit more explanation but whatever..

[Bhootnike]

Cnan somone please give me worked solutions to this pleaseeeeeeeee

http://imgur.com/mLyVd

[b^3]

I'm not sure if I've intepretted what you're asking right but anyways, hope it helps.

When we had a repeated root, we need to split the partial fractions to have a and a term.
Then you multiply by the factors on the bottom, expand out, collecting like terms and equate the coefficents so that you can find A, B and C.
For the example above

[StumbleBum]

I'm not sure if I've intepretted what you're asking right but anyways, hope it helps.

That's how i interpreted what he was asking.
Also im impressed, you did all the latex faster than i did it by hand and scanned it onto my computer. Was just about to upload it and then you posted.

[Lasercookie]

This is the method I use for partial fractions, I find it a lot quicker and easier: http://en.wikipedia.org/wiki/Heaviside_cover-up_method



Multiply all sides by (x-2) --> this will get A by itself, and let x = 2, this will get rid of the B and C terms (stuff multiplied by zero is zero).

This means we end up with

Multiply all sides by (x+1)^2 and let x = -1


Now for B:
Multiply all sides by (x+1)
We get:
If let x = -1, we'll have division by zero. This is a problem.

Let's go back to

We know the values for A and C now though. Let's say we let x = 0, and sub that into the original thingo here (as well as A = 3 and C = 4):

(I'm not showing all my working here, it's just a bit of subbing in and simplifying things down)


Easy enough to find B now. From there we can find

So

[martin1106]

 Question 3c) from vce 2008 exam2, why it is unnecessary to consider the y-component ? 

 my solving process:      

 and       
                   
 Therefore it takes secs for the train to complete one circuit of the track as its start point is (0,0)

_____________________________________________________________________________________________

Umm... I probably see my problem...
because the train passes (0,0) twice, so is the time taken for completing a half of the journey?
 
If it is the case, why simply just need to find the period of x component ?

[Jenny_2108]

If it is the case, why simply just need to find the period of x component ?

Its about how you understand the question and show your working, not about how you get the answer

If you find the period of x component, then you can get the same answer (you should include the explanation as well, just to be clear in this case, the train starts from O) but in other cases, I don't think you can use this method

[martin1106]

If it is the case, why simply just need to find the period of x component ?

Its about how you understand the question and show your working, not about how you get the answer

If you find the period of x component, then you can get the same answer (you should include the explanation as well, just to be clear in this case, the train starts from O) but in other cases, I don't think you can use this method

what's the correct method to solve this question?

[Jenny_2108]

If it is the case, why simply just need to find the period of x component ?

Its about how you understand the question and show your working, not about how you get the answer

If you find the period of x component, then you can get the same answer (you should include the explanation as well, just to be clear in this case, the train starts from O) but in other cases, I don't think you can use this method

what's the correct method to solve this question?

I suggest you do the 1st method (x, y component) to be safe 

[martin1106]

 Thanks Ennjy