VCE Specialist 3/4 Question Thread!

[Stevensmay]

I do remember doing a proof for that as part of year 11, but do not believe it is on the course. Would be classed as circle theorems.

[BubbleWrapMan]

It's listed as assumed knowledge on the study design (pages 160-161):

http://www.vcaa.vic.edu.au/Documents/vce/mathematics/mathsstd.pdf

[ahat]

Hey guys, could someone kindly explain the options for this question (why they're wrong).
Thanks.

Answer: (btw, how do you change spoiler titles?)

Spoiler

answer A

[lzxnl]

Hey guys, could someone kindly explain the options for this question (why they're wrong).
Thanks.

Answer: (btw, how do you change spoiler titles?)

Spoiler

answer A

Well...tan 35>1/sqrt3>1/1.8>0.5>0.4, so sin 35>ucos 35, particle accelerates down the slope.

There is a net force. A cannot be correct.

[Alwin]

Hey guys, could someone kindly explain the options for this question (why they're wrong).
Thanks.

Answer: (btw, how do you change spoiler titles?)

Spoiler

answer A

Since nliu has answered your q with such efficacy,

Code: [Select]

[spoiler=Ahat's custom spoiler name]
<Stuff inside the spoiler, Ahat's deepest darkest secrets (aka the answer A)>
[/spoiler]

[ahat]

Hahaha, right, cheers Nilu. I forgot that fundamental fact.

Thanks Alwin.

[ahat]

Apologies for posting so frequently, but help on this question would be really appreciated. This is probably my weakest aspect of Spec: (question attached)

a) Unsure, but I'm guessing linear independence.

b) From my knowledge, I knew "three 3d vectors are linearly dependent if they are coplanar" so therefore, I tried:
but just kept getting circular results.

c) Unsure again.

Whilst this was a non-CAS question, if you have the time, what would be the CAS approach,or alternative methods? E.g. I believe if you put all components into separate rows into a matrix and find the determinant, a determinant of 0 means dependance, non 0 is independance. Also, a function called ''rref" which I don't know how to use. Anyways, is the method I outlined, correct?
i.e. For:
Linear Dependance:
Linear Independence:

My deepest gratitude.

[lzxnl]

Apologies for posting so frequently, but help on this question would be really appreciated. This is probably my weakest aspect of Spec: (question attached)

a) Unsure, but I'm guessing linear independence.

b) From my knowledge, I knew "three 3d vectors are linearly dependent if they are coplanar" so therefore, I tried:
but just kept getting circular results.

c) Unsure again.

Whilst this was a non-CAS question, if you have the time, what would be the CAS approach,or alternative methods? E.g. I believe if you put all components into separate rows into a matrix and find the determinant, a determinant of 0 means dependance, non 0 is independance. Also, a function called ''rref" which I don't know how to use. Anyways, is the method I outlined, correct?
i.e. For:
Linear Dependance:
Linear Independence:

My deepest gratitude.

a. Collinear = on a straight line i.e. AB=k*AC

b. Coplanar when linearly dependent, yes. What you do is solve xa+yb=c. Just 1c, not zc. Try and solve that.

c. I don't have a quick way of doing this.
You have OB and OC, and you want to bisect BOC. What I would do is normalise the vectors such that k*|OB|=|OC|
Let k*OB = OD, |OD|=|OC|
So ODC is an isosceles triangle. Find the midpoint of DC; call it M. Then, OM will bisect angle DOC and hence BOC.

[ahat]

a. Collinear = on a straight line i.e. AB=k*AC

b. Coplanar when linearly dependent, yes. What you do is solve xa+yb=c. Just 1c, not zc. Try and solve that.

c. I don't have a quick way of doing this.
You have OB and OC, and you want to bisect BOC. What I would do is normalise the vectors such that k*|OB|=|OC|
Let k*OB = OD, |OD|=|OC|
So ODC is an isosceles triangle. Find the midpoint of DC; call it M. Then, OM will bisect angle DOC and hence BOC.

Absolutely perfect, thanks.
One more question though For the relation, I shaded the little intersection in the first quadrant as well. I can't figure out why it's not shaded.

[lzxnl]

z+zbar = 2x
1<=2x<=2
x E [1/2,1]


Erm...I would shade that too...what paper is this from? Answers can be incorrect for company exams.

[ahat]

itute 2011 Exam 1

I gave up with exam 2 because it was excessively difficult (in my opinion at least)
I noticed a mistake (I think) on the exam 2 solutions as well. It was asking for values of c for which a parabola would 'touch' an ellipse and included a value which touched and intersected the ellipse :/

Thanks for the help though.
And just to confirm, was this correct?:

Linear Dependance:
Linear Independence:

[09Ti08]

For your second rule. I would replace the = sign with sign. Your rule is only true if x=y=z=0. Logically, if you move zc to the right hand side and let z=-1, you'll basically end up with the first rule for linear dependence, which doesn't look right.

[ahat]

Can someone please explain the last equivalent? How did they remove the modulus on the bottom? Thanks.

[Stevensmay]

Now we want to get rid of our denominators, done by multiplying them out.



Now we have and still need to get rid of the so we multiply all by x.





Now to isolate each variable we want to make the other expressions evaluate to 0 by choosing appropriate x values to substitute in.





Substitute this value back in and repeat with a different x value.





Substitute this value back in and repeat with a different x value.






Substitute all our values back into the equation we found earlier to get


And done!

[Jaswinder]

the question asks to label all the forces acting on the two masses. How do we know which direction the friction is?
The answers suggest that it should be down the plane.

thanks guys