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October 03, 2025, 10:12:44 am

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2613835 times)  Share 

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Daenerys Targaryen

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Re: Specialist 3/4 Question Thread!
« Reply #2700 on: November 09, 2013, 09:07:28 pm »
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Hint:





Therefore, to be a linearly dependant set of vectors, then:


You should be able to solve from there.

This is from a set of projectile motion equations. They are super useful, I suggest you learn them!

Would you perhaps be able to list them for me? <3

And gracias amigo :)
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Stevensmay

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Re: Specialist 3/4 Question Thread!
« Reply #2701 on: November 09, 2013, 09:08:23 pm »
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Apogee is just the top of an arc.

And I'm getting m = 1 or -2, damn.

duhherro

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Re: Specialist 3/4 Question Thread!
« Reply #2702 on: November 09, 2013, 09:20:10 pm »
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Hey guys , is there an other way of working out the area for the 2 circles in Q1e)ii of 2011? Besides the suggested solution way?

Stevensmay

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Re: Specialist 3/4 Question Thread!
« Reply #2703 on: November 09, 2013, 09:24:49 pm »
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Hey guys , is there an other way of working out the area for the 2 circles in Q1e)ii of 2011? Besides the suggested solution way?

You could do it as the area of a segment.







But we have two segments put together so total area is




This method isn't that different to VCAA's but not many other options.

duhherro

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Re: Specialist 3/4 Question Thread!
« Reply #2704 on: November 09, 2013, 09:31:12 pm »
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You could do it as the area of a segment.







But we have two segments put together so total area is




This method isn't that different to VCAA's but not many other options.


Yeah isn't that what VCAA wanted? Could you do it in the traditional way of top function - bottom? I tried to do that but seemed much complicated as you are dealin with circle equations

Stevensmay

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Re: Specialist 3/4 Question Thread!
« Reply #2705 on: November 09, 2013, 09:32:55 pm »
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VCAA seemed to be going on about triangles and something didn't it?

You would have to do a bit of rearrangement to solve it via integration.

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Re: Specialist 3/4 Question Thread!
« Reply #2706 on: November 09, 2013, 09:35:29 pm »
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Hint:





Therefore, to be a linearly dependant set of vectors, then:



That condition isn't necessary for linear dependence; for example, if the set is linearly dependent (because it contains the zero vector), but the above equation doesn't hold if even though the set is dependent. The proper condition for dependence is that at least one of the following is true:





This is somewhat cumbersome since you generally would need to check each and every vector in the set, so we have the more symmetrical equivalent condition:

The vectors are linearly dependent if and only if for some scalars which are not all zero. A similar definition applies for an arbitrary number of vectors.

This is more useful when we actually know the vectors, since by equating components we get three simultaneous equations which we can attempt to solve for . It doesn't particularly help for this question, I just thought I'd point out that ahat's test won't always work unless you test every vector.

For this question, all you really need is that if a set of vectors contains two of the same vector, then it is linearly dependent. This is because, if we have the vectors , we can simply write , which satisfies the definition of linear dependence since not all of the coefficients are zero.

From there, notice that all three vectors are the same if , so they will be linearly dependent for this value. It's not necessarily the only value, but I'm guessing that was a multiple-choice question.
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Lejn

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Re: Specialist 3/4 Question Thread!
« Reply #2707 on: November 09, 2013, 10:04:13 pm »
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Yeah isn't that what VCAA wanted? Could you do it in the traditional way of top function - bottom? I tried to do that but seemed much complicated as you are dealin with circle equations

I was baffled by finding the area with a formula, so I took that approach. I'm unsure if we would've been given full marks since we could get the two circles for their first quadrant values and integrate, since integration of circles isn't a method we need to know (how do you do that again? Let u=sinx?)

Aelru

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Re: Specialist 3/4 Question Thread!
« Reply #2708 on: November 09, 2013, 11:16:30 pm »
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If anyone does have a copy of heinemann 2008 exam 2, I was wondering if someone could explain the reasoning behind that answer.
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ahat

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Re: Specialist 3/4 Question Thread!
« Reply #2709 on: November 09, 2013, 11:56:17 pm »
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DO NOT USE THIS EQUATION IN A SPESH EXAM.

Firstly, you will not be awarded method marks and secondly, this formula only applies to certain situations, it's usually useless, do it as Stevensmay said.

What, seriously?!? I've been using these formulae for the whole year - have a CAS program and everything D: How do you know?
I am a mathhole

Aelru

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Re: Specialist 3/4 Question Thread!
« Reply #2710 on: November 10, 2013, 01:01:06 am »
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Another question~ :

Regarding collision questions in vector calculus, what's the appropriate amount of working?
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Stevensmay

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Re: Specialist 3/4 Question Thread!
« Reply #2711 on: November 10, 2013, 01:03:48 am »
+1
Another question~ :

Regarding collision questions in vector calculus, what's the appropriate amount of working?
Equate i components, solve for t.
Equate j components, solve for t.

Write out final t values and explain why you discarded some.

That should be enough for 3 marks.

Aelru

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Re: Specialist 3/4 Question Thread!
« Reply #2712 on: November 10, 2013, 02:27:45 am »
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Equate i components, solve for t.
Equate j components, solve for t.

Write out final t values and explain why you discarded some.

That should be enough for 3 marks.

haha , to add on, what do you think is a suitable explanation?  I find that in some exams, I write out.. half of what is required.
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~T

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Re: Specialist 3/4 Question Thread!
« Reply #2713 on: November 10, 2013, 10:29:07 am »
+1
DO NOT USE THIS EQUATION IN A SPESH EXAM.

Firstly, you will not be awarded method marks and secondly, this formula only applies to certain situations, it's usually useless, do it as Stevensmay said.
Firstly, I must point out that I agree with the overwhelming sentiment here; you really don't need to use it, and there are few cases in which it will actually apply. I am, however, going to tentatively disagree with the rest.

You would still be awarded method marks. My tutor  - who has been an examiner for a million years - stressed to me that if your method is correct and you show all steps in your method, you cannot be marked down. So in this case, if the formulae apply, then your method is correct. If you state the formula and substitute your values in, then you have shown all steps of working. Thus, you cannot be marked down. There have been a few cases this year where I've used non-standard methods that I prefer or believe are quicker, and that has been his message.

Personally, I wouldn't use them. I just felt I should point out that I highly doubt you would be marked down. The only issue is that it's extremely unlikely that VCAA would put a simple enough question on the exam, especially exam 2, and you're better off just doing it the standard way so that it doesn't matter if it comes up or not.
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Re: Specialist 3/4 Question Thread!
« Reply #2714 on: November 10, 2013, 12:03:10 pm »
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What, seriously?!? I've been using these formulae for the whole year - have a CAS program and everything D: How do you know?

Besides, isn't it v^2sin(2 theta)/g ??

My question: http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2007specmaths2.pdf
For question 1. f)
Would I still be awarded the mark if I subbed z1 into LHS, and simplified
and then subbed to RHS and simplified, so LHS=RHS, or is it required that I sub in the x-value of z1 into y = sqr(3)x+2 (cartesian equivalent of the relation) and show that it equals the y-value of z1?
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