VCE Specialist 3/4 Question Thread!

[barydos]

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2008specmaths2-w.pdf
For question 3b) it's only worth 2 marks, but where do the 2 marks go to?
1 for shape? 1 for...something?

In the assessment report, they do not label the intercepts nor the max/min values, so I'm wondering if I'm "wasting" my time labelling all these features.
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/specialist_maths_2_assessrep_08.pdf

[Jeggz]

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2008specmaths2-w.pdf
For question 3b) it's only worth 2 marks, but where do the 2 marks go to?
1 for shape? 1 for...something?

In the assessment report, they do not label the intercepts nor the max/min values, so I'm wondering if I'm "wasting" my time labelling all these features.
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/specialist_maths_2_assessrep_08.pdf

Hmm good question...I would probably say one mark for shape (given that many didn't sketch the whole graph) and I have no idea where the other mark goes But I would still label maybe just the intercepts, if they are fairly straightforward and you don't need to do much working out for them, just to be on the safe side

[barydos]

Hmm good question...I would probably say one mark for shape (given that many didn't sketch the whole graph) and I have no idea where the other mark goes But I would still label maybe just the intercepts, if they are fairly straightforward and you don't need to do much working out for them, just to be on the safe side

Ahh okay, thank you

[ahat]

how do we find asyptotes of graphs like on the casio classpad? also is there a way of getting something like this in cartesian form through classpad? thanks

For asymptotes, you have many choices. You could let let the function equal 1/0 and then solve for your variable. For the complex question, substitute z for x + yi and then solve for y. Actually, after doing that, simplify it to |x + (y - 2)| first, then solve

[Alwin]

how do we find asyptotes of graphs like on the casio classpad? also is there a way of getting something like this in cartesian form through classpad? thanks

For the latter, you can actually type in:
Then: interactive > complex > cExpand which gives:

[stolenclay]

Old (from 2 days ago), but interesting.

I feel like there's a more elegant solution, but I haven't found it, please post one if you guys find something.

Diagram again

Solution

(Area of a triangle)

Let . Then .

[duhherro]

Could someone explain to me Q3)a of 2010 exam 2? Thanks!

[eddybaha]

basically you have to diff the P they give you and then show that left hand side equals right hand side when k=800. then just sub in the initial conditions into p and verify it satisfies the conditions.

[Lejn]

Could someone explain to me Q3)a of 2010 exam 2? Thanks!

To verify the initial condition, you simply have to show that P(0)=what it should.

To verify that P is a correct solution, we sub P into the dp/dt equation given (this is the 'by substitution' part), and we also derive the P(t) function given. If both derivatives equal each other, then we have verified that the given P(t) is a correct solution, as we have shown that its derivative is equivalent to the differential equation itself.

[Daenerys Targaryen]

A cricket ball is hit from ground level at an angle of 45* to the horizontal with an initial velocity of 20ms. The time taken, in secs, for it to return to the ground is?

[Stevensmay]

A cricket ball is hit from ground level at an angle of 45* to the horizontal with an initial velocity of 20ms. The time taken, in secs, for it to return to the ground is?

Resolve your initial velocity into a vertical component.
Use v = u+at to find the time when v = 0, this will be at the top of the balls flight.

Double this time to account for the trip back down.

Finished.

Spoiler

Initial vertical velocity is











But this is the time to apogee, so we need to double it for total flight time.

Total time is seconds.

[ahat]

A cricket ball is hit from ground level at an angle of 45* to the horizontal with an initial velocity of 20ms. The time taken, in secs, for it to return to the ground is?

Use the formula

Could someone explain to me Q3)a of 2010 exam 2? Thanks!

I've posted this before, but have a look at the solution attached. May help.

[Daenerys Targaryen]

Resolve your initial velocity into a vertical component.
Use v = u+at to find the time when v = 0, this will be at the top of the balls flight.

Double this time to account for the trip back down.

Finished.

Spoiler

Initial vertical velocity is











But this is the time to apogee, so we need to double it for total flight time.

Total time is seconds.

what the heck is an apogee.. haha thanks mate
that makes sense

Use the formula



I've posted this before, but have a look at the solution attached. May help.

what the hell is that formula!??!

[Daenerys Targaryen]

Also..

The vectors u=mi + j + k and v=i + mj + k and w=i + j + mk, where m is a real constant, are linearly dependent for what value of m?

[ahat]

Also..

The vectors u=mi + j + k and v=i + mj + k and w=i + j + mk, where m is a real constant, are linearly dependent for what value of m?

Hint:





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


You should be able to solve from there.

what the hell is that formula!??!

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