VCE Specialist 3/4 Question Thread!

[lzxnl]

A complex number z satisfies the inequality

Find the least possible value of |z|.

Find the greatest possible value of Arg z.

For any complex numbers a and b, |a-b|>=|a| - |b|, assuming of course that |a|>|b| otherwise the result is trivial
You're not MEANT to use this in spesh, but it's an easy check

[psyxwar]

For any complex numbers a and b, |a-b|>=|a| - |b|, assuming of course that |a|>|b| otherwise the result is trivial
You're not MEANT to use this in spesh, but it's an easy check

sorry could you elaborate a bit more on this (in relation to the q)?

[kinslayer]

sorry could you elaborate a bit more on this (in relation to the q)?

It's using the triangle inequality. It goes something like this:

Let  and assume



It follows that . If you want the specific point where z = 2 then you can flip the two inequalities in the chain to equalities and they will give the same conditions as the ones laid out in Zealous' graphical argument (z on the boundary of the circle, z the same distance away from w as the origin).

[lzxnl]

Not sure if I am confusing myself again.. but isn't the magnitude of the vector resolute of vector v parallel to u the same as the scalar resolute of vector parallel to u?

Not quite. If the dot product of your two vectors is negative, then the scalar resolute will be negative. Draw up a diagram and you'll see. The scalar resolute is the (signed) length of the vector resolute.

[keltingmeith]

Ok so when we get applications questions like "using a vector method to find the minimum distance" why do we use vector resolute THEN using that, finding the magnitude of that vector resolute?

Why wouldn't we just use scalar resolute?

Because the scalar resolute can be negative (as |zxn| said). We don't like to have negative distance.

[keltingmeith]

So just always mod the scalar resolute when it's negative?

AFAIK, you'll be fine to do that. |zxn| may know some instance where it won't work, but I certainly can't think of one.

[lzxnl]

AFAIK, you'll be fine to do that. |zxn| may know some instance where it won't work, but I certainly can't think of one.

No. The scalar resolute has the same sign as the dot product. You can't just mod it.
Here is a very silly example. Take the vector (1,0) and the vector (-1,0)
The scalar resolute of the first vector in the direction of the second is -1, which is negative.

[keltingmeith]

No. The scalar resolute has the same sign as the dot product. You can't just mod it.
Here is a very silly example. Take the vector (1,0) and the vector (-1,0)
The scalar resolute of the first vector in the direction of the second is -1, which is negative.

Yes, so you take the modulus of that, which is 1. This is also the magnitude of the vector resolute, which was what I thought Zezima meant by "mod the scalar resolute".

[lzxnl]

I don't actually know anymore what he meant
I thought he meant "when finding the scalar resolute just mod it when it's negative"

That's what I was rebutting

If you want the length of the vector resolute, sure, just find the absolute value of the scalar resolute and that'd be fine. But it's NOT the scalar resolute.

[bts]

The motion of a particle is described by the vector equation r(t) = 3 cos ti + 3 sin tj + k,
t ≥ 0.
1) find the Cartesian equation?


help please!

thx

[lzxnl]

The motion of a particle is described by the vector equation r(t) = 3 cos ti + 3 sin tj + k,
t ≥ 0.
1) find the Cartesian equation?


help please!

thx

Let x = i component
y = j component
z = k component, which is a constant 3
If you just consider x and y, what do you have?

[bts]

thx lzxnl - apparently there is no cartsian version for this 3D equation

find r.(t) and r..(t)
for r(t) = tan(t)i + cos2(t)j

i've attached the solutions, but have no idea how they got that

thank you

[nhmn0301]

thx lzxnl - apparently there is no cartsian version for this 3D equation

find r.(t) and r..(t)
for r(t) = tan(t)i + cos2(t)j

i've attached the solutions, but have no idea how they got that

thank you

There is a cartesian form for that, but VCE won't bother you too much about it as far as I'm aware. You would have a a circle (x^2 + y^2 =1 ), centre on the z-axis (the k direction) and 1 unit away from the xy plane.

They are just differentiating sec^2(x). you can write it as 1/cos^2(x) and diff to get :  -2/cos^3(x)  x (-sinx)   =   2tan(x)/cos(x) = 2tan(x)x sec(x). the j component is just normal differentiation.

[lzxnl]

thx lzxnl - apparently there is no cartsian version for this 3D equation

find r.(t) and r..(t)
for r(t) = tan(t)i + cos2(t)j

i've attached the solutions, but have no idea how they got that

thank you

Cartesian? It's just x^2 + y^2 = 9 (not 1, nhmn0301) centred on (0,0,1). AKA take your circle radius 3 centred on the origin and move it up by 2.

[zsteve]

Hi guys, I'm going to do Spec next year, but this year I'm doing some preparation with a Singaporean math text... and its questions are hard.
Could anyone give me some help with this one?
Find, in terms of n, the value of the sum:
, Hint

Although it's not a question from spesh, it definitely deserves to be!!

Edit: Entering it on my CAS gives the algebraic result: