VCE Specialist 3/4 Question Thread!

[b^3]

If you leave the one under the as (well really but you know what I mean) and the one under the as then it will still follow the basic form , it's just that the hyperbola is now on the other side of the asymptotes.
Have a look at the links below, and use the sliders to move it around e.t.c.
Positive , negative : https://www.desmos.com/calculator/dcr4o00o9b
Positive negative : https://www.desmos.com/calculator/cmbdhw0ail
Both on the same axes: https://www.desmos.com/calculator/7tlpu69l2v
E.g. for the question above we would have so and so option C.
I.e. https://www.desmos.com/calculator/h85ui63pba

[satya]

If where asked to graph a relation in the complex plane and it has a restriction Arg(z) <  π/2, do we graph it between - π < Arg(z) <  π/2?

Considering this graph: y = x/2 + 1 where Arg(z) <  π/2. This line in graphed in the first and third quadrant.
When it's restricted to Arg(z) <  π/4, it's only graphed in quadrant one. Could someone please explain why?

if arg(z)  <pi/2 then you should shade between 0 to pi/2 becasue arg(z) is less than pi/2
and for the line without any restriction it is in first and third quadrant, but when Arg(z) <pi/4 then it's only between 0 to pi/4 hence the line only exists in the frist quad.

[papertowns]

If you leave the one under the as (well really but you know what I mean) and the one under the as then it will still follow the basic form , it's just that the hyperbola is now on the other side of the asymptotes.
Have a look at the links below, and use the sliders to move it around e.t.c.
Positive , negative : https://www.desmos.com/calculator/dcr4o00o9b
Positive negative : https://www.desmos.com/calculator/cmbdhw0ail
Both on the same axes: https://www.desmos.com/calculator/7tlpu69l2v
E.g. for the question above we would have so and so option C.
I.e. https://www.desmos.com/calculator/h85ui63pba

Wow those links are pretty awesome, thanks for that!
And yeah okay, leaving it the same gives the correct answer thanks again!

[papertowns]

Help with this one please :')

[b^3]

Lets say our point is ( is positive real number and is negative) real number, then we have

Now since we know that , then so dividing our components by something that is greater than will give something smaller than our original components. Now we also note that initially is positive and is negative, and as we have and   in our result we need something that is in the fourth quadrant but has components that are smaller than our original components, which is closest to option C since it's in the right quadrant and the other options that have the right quadrant have larger components.
You could also look at it as taking the conjugate, which will flip it across the axis and then taking 1 over a complex number will scale down the components when the magnitude is greater than one, and change the size of the imaginary component. In this case that is 4th quad --> flip across when taking conjugate --> 1st Quad ---> Flip across and scale components according to ---> 4th quad with smaller components.

The flipping with the conjugate you should know but it's probably easier to re-derive what happens with the 1 on .

[brightsky]

can also let z=rcist. 1/conj(z) = 1/rcis(-t) = 1/r*cis(t). so angle doesn't change. the modulus just gets smaller.

[duhherro]

Hey guys,

for 2009 exam 2, Q5g) How come you can't do 1200 = integral from 0 to T v(equation) dt. To work out the time to fall to 1200m ?

[papertowns]

Need help on spesh exam 2 q1e part ii the area of the shaded region :/

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2011specmath2-w.pdf

[barydos]

Hey guys,

for 2009 exam 2, Q5g) How come you can't do 1200 = integral from 0 to T v(equation) dt. To work out the time to fall to 1200m ?

From what I recall when I did this question, it didn't really work on my CAS calc either, but then i attempted to do integral from 180 (the time to reach roughly 880m) to T and let that integral = 1200 minus 880 (the exact value tho), and then i was able to solve for T, not sure why though.

Need help on spesh exam 2 q1e part ii the area of the shaded region :/

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2011specmath2-w.pdf

Have you tried to find the integral of the circle (make the y the subject, and choose the positive alternative)  (the circle hat was shifted to the right) from x=0 to x=1, then subtract the triangular area (the triangle with verticies (0,0) (1,1) and (1,0)), then multiply this region by 2 (because the area shaded is symmetrical about the y=x line.

[brightsky]

Need help on spesh exam 2 q1e part ii the area of the shaded region :/

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2011specmath2-w.pdf

let the area of a quarter of the circle with radius 1 be A, and the area of the shaded region be R.
A + A - R = 1 (forms a square with dimensions 1*1)
2A - R = 1
R = 2A - 1
now what is A?
easily found. pi*(1)^2/4 = pi/4
so R = pi/2 - 1

alternatively, you can use integration. this second technique will be a bit difficult if you did not have a calculator, but it's exam 2 so should be fine.

[duhherro]

could anyone help me on Q3 of 2009 exam 2, uhhh for part Q3)  what is the difference between time at 12 and time at 14, 14 hits the ground but 12 becomes a loop ?


And still need help on my previous question about the integrals of Q5)g) of 2009

Thanks!

[sin0001]

could anyone help me on Q3 of 2009 exam 2, uhhh for part Q3)  what is the difference between time at 12 and time at 14, 14 hits the ground but 12 becomes a loop ?

Yea you've got the main idea there. As the child spirals downwards, it takes 12 seconds to complete a single turn; which can be found from the period of the 'i' and 'j' components. And the 'k' component represents the vertical height relative to the ground level- in other words, when the 'k' component equals 0 (at t=14s), the child is at the same height as the origin, so at the ground level.

[Professor Polonsky]

(VCAA 2011 MC Q2)

[nnl]

get the asymptote equation for each, and if the gradient of each asymptote is not -1 and 1 then they're not perpendicular

[SocialRhubarb]

At a glance, I'd say it's E, as it's the only one where the coefficients of x^2 and y^2 are not 1.