VCE Specialist 3/4 Question Thread!

[lzxnl]

Is there a way to express 3^1/4 x 2^1/2 as 12^1/4?
Thanks.

3^1/4 x 2^1/2 = 3^1/4 x 4^1/4 = 12^1/4

[zvezda]

Change the graph options to parametric and where it says put what you have for the i component, for put the j component.

Not sure which calc you have, but on the Ti-nspire (the old version at least), you can change to this mode by having the cursor in the entry bar, then pressing [ctrl]+[menu], then [2: Graph Type] and finally [2: Parametric]. The way of changing the mode might have changed slightly with the new calc though.

Yeah found it. A bit different but pretty much the same. Thanks b^3

[e^1]

Sorry if I'm asking heaps, and I don't know if this will be likely in a VCAA exam but I don't know how to do an itute question from 2013. (question 3c specialist exam). The answer's there, but it seems vague.

Thanks!

edit: should mention its from exam 1.
thanks for the help again! I didn't do GMA last year, but it was in the essentials books anyways.

[~T]

Sorry if I'm asking heaps, and I don't know if this will be likely in a VCAA exam but I don't know how to do an itute question from 2013. (question 3c specialist exam). The answer's there, but it seems vague.

Thanks!

edit: should mention its from exam 1.

If you visualise a ray going from the origin that represents the argument of a complex number, for this argument to be a maximum the ray will have to be a tangent to the circle given. This is very important.

We can start drawing a triangle to help us out with the angles (yay trig). If we draw a line from the origin to the centre of the circle, and a line from this centre to the point where our previously found tangent touches the circle, and a line from this point back to the origin, we have a useful triangle.

Remembering back to units 1/2 circle theorems, what is significant about a tangent to a circle? It is perpendicular to the radius of the circle. With this in mind, we have a right angle triangle, and we just need some side lengths.

The first side length we can find is the length from the origin to the centre of the circle, using our distance formula to the point (4, -3). The second length we know is the radius of the circle. Thus, we can use our tangent identity to find the small angle in our triangle at the origin. This comes out as arctan(4/3).

However, we have the entire angle, and we only want the part above the x axis (because we are calculating the argument, measured anticlockwise from said axis). So, we must subtract the part of this angle that lies below the axis. Simply using the point of the centre (4, -3), the tangent is rise/run so the angle is arctan(3/4).

We are now left with the maximum argument as arctan(4/3) - arctan(3/4). Here, use the subtracted angle formula for tan, to find a nicer expression for the angle.



*sorry for the lack of latex, but I hope my explanations suffice...

[Brytz]

Another question I cannot get my head around. Question 6 MC 2012 VCAA. I don't know where to start on this on.   

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2012/2012specmath2-w.pdf

[e^1]

Another question I cannot get my head around. Question 6 MC 2012 VCAA. I don't know where to start on this on.   

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2012/2012specmath2-w.pdf

Let




Testing solutions:
A. Rotating anticlockwise by 270 degrees =

B. Reflecting in both x and y axis =

Note:

• Reflection of (x,y) in the x-axis = (x, -y).
• Reflection of (x,y) in the y-axis = (-x, y).

C. Reflecting in y-axis =
Rotating anti-clockwise by 90 degrees then results into

Hence answer is C.

[zvezda]

Hey,
With q3 in the vcaa 2009 spec exam 1, am i missing something because i dont think that the answer they have is actually technically parellel to the other vector?? :/
Cheers

[Conic]

is parallel to , and is perpendicular to .

[duhherro]

Hi guys, need a bit of help on VCAA exam 2 2006

For Q1 c) How do you know that the A=pi x y^2 (circle) ?

And Q3b) ii) Was it okay to treat the question set up as like an incline plane question? Thats how I did it

And how exactly do you do 4d) ?

Thanks in advance!

[zvezda]

is parallel to , and is perpendicular to .

Yeah my bad i was confusing direction with parallel

[Jaswinder]

Hi guys, need a bit of help on VCAA exam 2 2006

For Q1 c) How do you know that the A=pi x y^2 (circle) ?

And Q3b) ii) Was it okay to treat the question set up as like an incline plane question? Thats how I did it

And how exactly do you do 4d) ?

Thanks in advance!

1- volume formula
3b) anything should be fine as long as its mathematically correct and gives you the right answer
4d - follow the field line which crosses the 0.5 on the y-axis

[duhherro]

1- volume formula
3b) anything should be fine as long as its mathematically correct and gives you the right answer
4d - follow the field line which crosses the 0.5 on the y-axis

Mmm thanks, are we dealing with a volme of a cone where height = radius so we can get V=1/3 x pi x r^3 ? Then just diff that to get SA ? and where radius is the y equation

[bonappler]

I was doing a question that wanted me to find the volume of an ellipse. What I found was when you rotate it in the x-axis you get a different answer to when you rotate it in the y-axis. Is this right or am I doing something wrong? Shouldn't the values I get be the same?

[brightsky]

that can happen if your ellipse isn't centred at the origin.

[bonappler]

What if it is?