VCE Specialist 3/4 Question Thread!

[StumbleBum]

I men't like this:



It doesn't give the same answer? or am I doing something wrong...

EDIT: OH GAWDD, realised that they are equal... didn't see the answers negative at the front

[WhoTookMyUsername]

yep effectively
a constant can be anything that is constant
k^1000000
ln(k)
log_100000_(K)

(k-1)^e^2

[nina_rox]

Can someone help me to solve:

z^3 - (2+2i)z^2 + 3iz + 1 - i = 0 given (z-i) is factor?

I tried division (short method) but doesn't seem to be working for me...

Any help would be reaally appreciated. Thank you!

[Jenny_2108]

Can someone help me to solve:

z^3 - (2+2i)z^2 + 3iz + 1 - i = 0 given (z-i) is factor?

I tried division (short method) but doesn't seem to be working for me...

Any help would be reaally appreciated. Thank you!

You can do long division but normally I prefer other methods

Method1: You can try with other solutions like z=-i,1,-1 etc then you can get one more solution
In this case, z-1 is another solution
so you have (z-i)(z-1)(z-a) => from there you can easily find a

Method2: I like this one because it works in all cases without trying to substitute many values of z or if solutions are not easily found like 1,-1,i,-i




now you solve:

[nina_rox]

Can someone help me to solve:

z^3 - (2+2i)z^2 + 3iz + 1 - i = 0 given (z-i) is factor?

I tried division (short method) but doesn't seem to be working for me...

Any help would be reaally appreciated. Thank you!

You can do long division but normally I prefer other methods

Method1: You can try with other solutions like z=-i,1,-1 etc then you can get one more solution
In this case, z-1 is another solution
so you have (z-i)(z-1)(z-a) => from there you can easily find a

Method2: I like this one because it works in all cases without trying to substitute many values of z or if solutions are not easily found like 1,-1,i,-i




now you solve:

Thank you so much! Just a question about the method 2 though, how did you know (z^2 + az + 1 + i) is the other factor? As in where did the 3z in the equation go and how do you know it was 1+i ? Thank you!    

[Jenny_2108]

Just a question about the method 2 though, how did you know (z^2 + az + 1 + i) is the other factor? As in where did the 3z in the equation go and how do you know it was 1+i ? Thank you!

The constant part is (1-i) so (-i) x *a number*= 1-i => this number = 1+i (my explanation sucks  )
Hope you can get what I mean :S

[nina_rox]

Just a question about the method 2 though, how did you know (z^2 + az + 1 + i) is the other factor? As in where did the 3z in the equation go and how do you know it was 1+i ? Thank you!

The constant part is (1-i) so (-i) x *a number*= 1-i => this number = 1+i (my explanation sucks  )
Hope you can get what I mean :S

Sorry for being annoying... I'm still confused. So do you multiply the constant by the solution you know?

So are you saying -i x (1+i) = -i+1 but..

[Jenny_2108]

So are you saying -i x (1+i) = -i+1 but..

but?

[nina_rox]

So are you saying -i x (1+i) = -i+1 but..

but?

Sorry! as in it doesn't equal to 1+i like you said...

[Jenny_2108]

So are you saying -i x (1+i) = -i+1 but..

but?

Sorry! as in it doesn't equal to 1+i like you said...

-i x (1+i)= -i-i^2= -i+1

[nina_rox]

So are you saying -i x (1+i) = -i+1 but..

but?

Sorry! as in it doesn't equal to 1+i like you said...

-i x (1+i)= -i-i^2= -i+1

Nevermind, thanks a load anyway.

[aznxD]

2010 Exam 1 - Q2 b)

Why do you take the negative value of |(v-4)/4|?

When t=0, v=0, sub in these values and you'll see that only satisfies the initial conditions.

[Lasercookie]

2010 Exam 1 - Q2 b)

Why do you take the negative value of |(v-4)/4|?

"initially at rest", so at t = 0, v = 0.

We end up getting to a step with

Sub in t = 0, . We know that v must equal zero, it's the negative case that'll get us this solution.

edit: oh didn't see that there was another page with aznxD's post.

[Special At Specialist]

Please help me with this question! I have no idea what to do...

The path of a particle is given by r = sin(t)*i - cos(2t)*j, 0 ≤ t ≤ 5pi/4
Find a vector in the direction of motion of the particle at t = pi/4

[b^3]

The vector in the direction of motion, will be in the direction of the velocity vector.
So finding the velocity vector

At

We might have to find the unit vector, depending on the question, well the vector above gives the direction anyway, but I'll do it just in case.