VCE Specialist 3/4 Question Thread!

[sin0001]

Edit: I have a question, if two particles are going to collide, do we equate their position vectors or velocity vectors? I always thought it was velocity, but this question says position. In what scenarios (not really sure how to describe this question) do we use velocity vector opposed to position vector for something? Thanks.

You equate the position vectors, because when two particles collide, they'll be at the same place at the same time. The only question I can think of, which would require you to equate the velocity vectors could be: Find the time at which the direction of motion of the particles is equal.
Hope that helped!

[SocialRhubarb]

The only question I can think of, which would require you to equate the velocity vectors could be: Find the time at which the direction of motion of the particles is equal.
Hope that helped!

Actually, if you were trying to find a point where the direction of motion is equal, you wouldn't be equating the velocity vectors because they can have different magnitudes while having the same direction. You'd probably finding points where the derivative of y with respect to x (dy/dx) is equal and then checking that they are both moving in the same direction along that path, as when dy/dx is equal they could actually also be going in exactly opposite directions.

[sin0001]

Actually, if you were trying to find a point where the direction of motion is equal, you wouldn't be equating the velocity vectors because they can have different magnitudes while having the same direction. You'd probably finding points where the derivative of y with respect to x (dy/dx) is equal and then checking that they are both moving in the same direction along that path, as when dy/dx is equal they could actually also be going in exactly opposite directions.

Touché!
Need help with Heffernan Exam 1 (2010):
For 9b (attached), I used both the roots of 4cis(pi/3), but in the solutions they have just used one...
I know I'm wrong (coz I'm getting 4 solutions for a quadratic), but can someone clarify why?
Solutions are attached too.

[Stevensmay]

Touché!
Need help with Heffernan Exam 1 (2010):
For 9b (attached), I used both the roots of 4cis(pi/3), but in the solutions they have just used one...
I know I'm wrong (coz I'm getting 4 solutions for a quadratic), but can someone clarify why?
Solutions are attached too.

Can I ask how you got two roots from
Using De Moivre's theorem we get

EDIT: Apologies, I should have been clearer and mentioned the equivalence the two solutions would eventually present.

[sin0001]

Can I ask how you got two roots from
Using De Moivre's theorem we get

I got 2cis(pi/6) and 2cis(7pi/6) as the roots

[lzxnl]

I got 2cis(pi/6) and 2cis(-5pi/6) as the roots

domain of the argument?

[ahat]

Could you please scrutinise my method and point out my mistakes for this question:

Three points:
P(0,1,1)
Q(1,3,-1)
R(2,2,5)
Find the coordinates of a point M on that is closest to point Q.
(I fear someone may experience heart palpitations reading my method)



M is parallel to and a scalar multiple of and is closest to Q when the two vectors are parallel perpendicular. represents a scalar.















Which is incorrect anyway

Answaar

Thanks

Can I ask how you got two roots from
Using De Moivre's theorem we get

There'll be two solution 180 degrees apart.

[sin0001]

I got 2cis(pi/6) and 2cis(-5pi/6) as the roots

domain of the argument?

Regardless of whether the second root was in the domain or not, the solutions only list the root: 2cis(pi/6) :/

[SocialRhubarb]

I know I'm wrong (coz I'm getting 4 solutions for a quadratic), but can someone clarify why?

That's because cis(7pi/6)=-cis(pi/6), and you have a plus/minus sign out the front, so using the second value doesn't give you any more solutions.

It's likely that two of your solutions are the same as the other two.

[sin0001]

That's because cis(7pi/6)=-cis(pi/6), and you have a plus/minus sign out the front, so using the second value doesn't give you any more solutions.

It's likely that two of your solutions are the same as the other two.

Ohhh
Yup, that's exactly what I did.
Thanks heaps!

[lzxnl]

Could you please scrutinise my method and point out my mistakes for this question:

Three points:
P(0,1,1)
Q(1,3,-1)
R(2,2,5)
Find the coordinates of a point M on that is closest to point Q.
(I fear someone may experience heart palpitations reading my method)



M is parallel to and a scalar multiple of and is closest to Q when the two vectors are parallel perpendicular. represents a scalar.















Which is incorrect anyway

Answaar

Thanks
There'll be two solution 180 degrees apart.

The main problem I see is that PM is a factor of PR, but OM means little.

I would have went PR.QM=0           insert vector arrows over all capitals
QM=OM-OQ=OP+PM-OQ=OP+k*PR-OQ=OP+k*(OR-OP)-OQ=OP(1-k)+k*OR-OQ
so this dotted with PR=0
(OP(1-k)+k*OR-OQ).(OR-OP)=0
OP=(0,1,1)
OQ=(1,3,-1)
OR=(2,2,5)

so we have ((0,1-k,1-k)+(2k-1,2k-3,5k+1)).(2,1,4)=0
or (2k-1,k-2,4k+2).(2,1,4)=0
4k-2+k-2+16k+8=0
21k=-4
k=-4/21

PM=k*PR
so OM=OP+PM=(0,1,1)+-4/21*(OR-OP)
=(0,1,1)+-4/21*(2,1,4)
=(0,1,1)+(-8/21,-4/21,-16/21)
=(-8/21,17/21,5/21) as requested

[ahat]

Thanks man I'm glad I was on the right track at least. If you don't mind, I have another question though:

[brightsky]

find the area of the segment cross-section and then multiply by length of 2 m.

we want area of segment DC. we know how to find area of sector DOC: simply a/(2*pi) * pi*(1)^2 = a/2. now we want to subtract this area by the area of triangle DOC. how do we find the area of the triangle? 1/2 * 1*1*sin(a) = 1/2 sin(a).

so the area of segment cross-section is a/2 - 1/2 sin(a) = 0.5(a - sin(a)) as req.

[ahat]

find the area of the segment cross-section and then multiply by length of 2 m.

we want area of segment DC. we know how to find area of sector DOC: simply a/(2*pi) * pi*(1)^2 = a/2. now we want to subtract this area by the area of triangle DOC. how do we find the area of the triangle? 1/2 * 1*1*sin(a) = 1/2 sin(a).

so the area of segment cross-section is a/2 - 1/2 sin(a) = 0.5(a - sin(a)) as req.

Thanks, I tried doing this as well, but doesn't the question want volume? Or have we found volume? That's what confused me. Also, actually, how do we know the angle of the sectors is alpha?
Also, for a question, there was this statement (these are vectors): |b| = 2|a| and then in the solution working, they have subbed for b, ie. b = 2a. Is this strictly correct?

[brightsky]

picture the scenario. you have a container with a semi-circle cross section, which is 2 m long (like a gutter of sorts). you start pouring water into this container. the volume of water will be the segment cross section multiplied by the length. and alpha is given in the question.

technically b = 2a is fine. b and a without the tilde means magnitude. but to be safe i'd put b with  tilde and magnitude symbols (I personally don't like the magnitude symbols because it looks too much like absolute value...norm is better but absolute value symbol suffices for VCE).