VCE Specialist 3/4 Question Thread!

[keltingmeith]

For your first question, the second way of thinking is the right one - the first I think you're getting confused with inverse trig functions?

For your second question - I have no clue what you're doing to get so confused, so I'm going to rearrange the equation this way and see if it works for you:

[knightrider]

Hello i will be completing specialist maths in 2016.

what will be the changes in the study design and where can i find this. I want to know what things will be changing and what will stay the same .

Thanks very much

[keltingmeith]

Hello i will be completing specialist maths in 2016.

what will be the changes in the study design and where can i find this. I want to know what things will be changing and what will stay the same .

Thanks very much

AFAIK, VCAA are still sussing out the changes. Looking over the old draft, there's not much changes that will influence you. Conic sections will no longer be covered as an assessable topic - however, you will still be expected to draw them if given their parametric equations. (i.e., no easy multiple choice questions on conics) There will be some new things (if it all passes) in differential equations, if you really want to get a head-start just google "separable differential equations" and you'll get the stuff you need.

Otherwise, everything is about the same. Important to note is that modulus is no longer a part of methods, but it IS a part of specialist now. Also, there is now a statistical inference section - once everything has been confirmed, we should be able to tell you more about the stats section. For now, don't worry about the changes too much - you'll have plenty of time to prepare for them, and your teachers will be aware of the changes and what they should/shouldn't teach you.

[knightrider]

AFAIK, VCAA are still sussing out the changes. Looking over the old draft, there's not much changes that will influence you. Conic sections will no longer be covered as an assessable topic - however, you will still be expected to draw them if given their parametric equations. (i.e., no easy multiple choice questions on conics) There will be some new things (if it all passes) in differential equations, if you really want to get a head-start just google "separable differential equations" and you'll get the stuff you need.

Otherwise, everything is about the same. Important to note is that modulus is no longer a part of methods, but it IS a part of specialist now. Also, there is now a statistical inference section - once everything has been confirmed, we should be able to tell you more about the stats section. For now, don't worry about the changes too much - you'll have plenty of time to prepare for them, and your teachers will be aware of the changes and what they should/shouldn't teach you.

Thanks. could you link me to the old draft eulerfan101

[grannysmith]

How do I find the coordinates of the POI of the following?

[keltingmeith]

How do I find the coordinates of the POI of the following?

First thing I'm going to do is put these in a form I can work with - note that this is not at all necessary and you can do it without completing the square. For the first one,



So, we have a circle with radius . To find our intercepts, we let x=0 and y=0. So, for the y-intercept:

Since we can't square root the negative number, we say there is no y-intercept.

For the x-intercept:

We will quickly run into the same problem, and say that there is no x-intercept.

Something else you could do - since we have a circle, we can immediately note that the radius is less than both the translations, and so there should be no intersections.

I'll let you try the second now that you have a way of doing it.

EDIT: I just realised I read it wrong. Gimme a minute. :3
EDIT EDIT: Leaving the top up there for interest~

Okay, what we have are two non-linear equations. So, solving for are much more difficult - and VCAA won't expect you to do this, but for SACs it's fair-game. Normally, I'd suggest just using your calculator to solve, but by hand:

Using the divided by four from above, we get the equations:


I'm just going to take the first from the second, giving us:


This is the line along which we will find all intersections of this circle - so, if we sub it in for our specific cases, we'll get values we can use:



You can then solve this quadratic, and input it into to find the corresponding y-values, and you've got the POI.

[grannysmith]

Okay, what we have are two non-linear equations. So, solving for are much more difficult - and VCAA won't expect you to do this, but for SACs it's fair-game. Normally, I'd suggest just using your calculator to solve, but by hand:

Using the divided by four from above, we get the equations:


I'm just going to take the first from the second, giving us:


This is the line along which we will find all intersections of this circle - so, if we sub it in for our specific cases, we'll get values we can use:



You can then solve this quadratic, and input it into to find the corresponding y-values, and you've got the POI.

Ah, thanks for that! Didn't realise I could just sub the into one of the original equations

[keltingmeith]

The trick to non-linear equations:

Do shit 'till you have numbers. If you don't know if it's okay, do it anyway, because otherwise you have nothing to work with~

[Zues]

bit of help on this one

[keltingmeith]

bit of help on this one

You could quickly look at the equation, say that in front of the x-term you need a -6, so a must be -3. Then factor out the 2 from the y, and then say you need a 4 in front of the y, and so b must be 2 (C). Or, you could do this long method:

We take the equation, and slowly mold it into the equation of an ellipse:


At this point, we ignore the RHS since that only gives a dilation, and instead just look at the LHS to get translation information.

So, this ellipse has centre at (-a, -b), and we want the centre to be at (3, -2), so a=-3 and b=2 (C)

[Zues]

You could quickly look at the equation, say that in front of the x-term you need a -6, so a must be -3. Then factor out the 2 from the y, and then say you need a 4 in front of the y, and so b must be 2 (C). Or, you could do this long method:

We take the equation, and slowly mold it into the equation of an ellipse:


At this point, we ignore the RHS since that only gives a dilation, and instead just look at the LHS to get translation information.

So, this ellipse has centre at (-a, -b), and we want the centre to be at (3, -2), so a=-3 and b=2 (C)

second line, how did you get 4a^2, and also 4b^2 on the left hand side (i got a^2 and b^2 respectively.
third line,  i see how you got centre (3,-2) but how do we know a is -3 and b is 2 from this i know obviously (x-3) .. (x-2), but why are these your a and b values. I thought in the general form of elipse a and b are your denominator values.

additionally, how did they get 3200 in this question

[Zues]

second line, how did you get 4a^2, and also 4b^2 on the left hand side (i got a^2 and b^2 respectively.
third line,  i see how you got centre (3,-2) but how do we know a is -3 and b is 2 from this i know obviously (x-3) .. (x-2), but why are these your a and b values. I thought in the general form of elipse a and b are your denominator values.

additionally, how did they get 3200 in this question

crap.. dont worry about the third line thing, i got that. just dont get the attachment and also the second line stuff. Thanks !!

[keltingmeith]

second line, how did you get 4a^2, and also 4b^2 on the left hand side (i got a^2 and b^2 respectively.
third line,  i see how you got centre (3,-2) but how do we know a is -3 and b is 2 from this i know obviously (x-3) .. (x-2), but why are these your a and b values. I thought in the general form of elipse a and b are your denominator values.

additionally, how did they get 3200 in this question

Actually, looking at my working again, I have a mistake in the second line - shouldn't have or , and they should be just and . I forgot to divide by 2.

[Zues]

alright thanks Euler, last question I see where 40 square root 2 came from, but where did the 3 in the denominator come from?

[keltingmeith]

alright thanks Euler, last question I see where 40 square root 2 came from, but where did the 3 in the denominator come from?

You've gotta square root the whole fraction, can't just square root the top and not the bottom.