VCE Specialist 3/4 Question Thread!

[ahat]

The gradients at -2 and 2 will be infinity I believe, and the x-intercepts are still continuous anyway.

Don't know where the other message went, but like he said, the hyperbola is undefined in the region (-2,2), but as the relation is still continuous at these points, you can find the gradient.

[Stevensmay]

ahat, I removed mine as I had answered a different question to the one posed.

[lzxnl]

Differentiate the equation with respect to x.
You have y^2 = 4x^2-4
y=+-sqrt(4x^2-4)
dy/dx = -+4x/sqrt(4x^2-4)
So really, you're looking at the range of values this thing can take. It's looking ugly at the moment, and as we know it's positive and negative, let's square it for simplicity.
(dy/dx)^2 = 16x^2/(4x^2-4) = 4 + 16/(4x^2-4)
I trust you can find the range of this function. Firstly, note that dy/dx has a restricted domain such that the denominator of the fraction is always positive (4x^2-4>0 from the square root expression). Thus, (dy/dx)^2 must be strictly greater than 4. This means that |dy/dx|>2. AKA dy/dx>2 or dy/dx<-2
Which is dy/dx E R\[-2,2]

Or if you don't like algebra, visualise the graph in your head. As x^2>y^2/4, the graph will be under the asymptotes. You can probably work out really quickly that the asymptotes are of the form y=+-2x. So there's a 2 somewhere.
Now, as the function is always under the asymptotes, it always approaches the asymptotes from below. That means its slope is always larger than 2 (in the first quadrant). Likewise, in the fourth quadrant, its slope is always smaller than -2 (draw it out; it'll make sense). At the vertices, the slope is +-infinity, and the asymptotes have slope +-2. You can now see that the accepted values of m are (-inf, -2) U (2, inf).

[abcdqd]

Where's this question from, if I may ask?

vcaa 2008 exam 1

[Alwin]

Anyone have any ideas on some kind of way to think about the asymptote equations for hyperbolae in the form and that you sometimes get as multiple choice questions? When it just gives you a hyperbola equation and asks for the two asymptote equations in the aforementioned form...

I expanded the usual form:



and got:



Which maybe sort of helps a bit. Perhaps if I practise using this... there're a couple of things that you can quickly gather from it to rule out some solutions I guess. The plus/minus bh bit is nice, because it usual rules out most; the difference between the two constants must be 2bh. Stuff like that. Thoughts? Not worth it, just expand every time?

For conics in the form:

1. usually on the MC part of the exam, they will rearrange the equations of the asymptotes in the form: _x + _y  = _
I find usually the quickest way is to substitute (h, k) into each of the options will give you the correct answer, or at least it will eliminate most of the options and only leave you with 1 or 2. Then, you rearrange for y and check the gradient.
There will be only 1 option in which the equation of the asymptotes pass through (h, k) and have the right gradients.

2. if you prefer formulas tho, the two asymptotes can be represented by the equations:



In other words, which is what you already have My only concern is if the options in the MC section aren't in that form, you'll have to fiddle around with numbers again, which is why personally I prefer the first method.

But either way is fine, just make sure that if it's a SA or ER question so appropriate working

[zvezda]

Hey,
Theres a question on the AN guide; tech free question 1.
In the denominator of my log expression, i have x+3/2 instead of 2x+3 and i cant see anything wrong with what i did. I initially had 1/(x-1) - 1/(x+3/2) as the expression to antiderive (with 2/5 outside the integrand).
Help is much appreciated

[brightsky]

yeah don't know what the question is but there is a likelihood that the answer you obtained is equivalent to that given in the study guide. ln(2x+3) = ln(2(x+3/2) = ln(2) + ln(x+3/2). if there is a constant of integration, then that ln(2) goes.

[jono88]

How would you expand (cos(theta)+isin(theta))^3 in the form of a+ib or (cos(theta)+isin(theta))^5 in the form a+bi?

[jono88]

casio classpad wont expand to the required answer :/

[ahat]

How would you expand (cos(theta)+isin(theta))^3 in the form of a+ib or (cos(theta)+isin(theta))^5 in the form a+bi?

I don't know about Casio steps, but I'm sure it'd be similar.

First, convert the cartesian form to polar form, i.e. cis(Ө) and use demoivre's theorem, i.e. cis(5Ө). Expand, so cos(5Ө) + isin(5Ө) and then simplify.

[Stevensmay]

I'm not sure if my method is correct. What I would do.

Correct the given expression into where z is something in polar form.
So now we have and we want

Using De Moivre's theorem we get  
Expanding this gives us

I know with the TI Nspire range it is possible to define cis as cos(x) + isin(x), not sure if this will help with the classpads.

[jono88]

question says to write the expansion and the solutions has cos(theta) (4cos^2(theta) -3) + isin(4cos^2(theta-1)
something ridiculous like that which you have to get right to follow on to the next part of b) hence use de moirvres theorem (\cos(3 \theta) + i \sin(3 \theta)
) as you have done to show that sin(3theta)=3sin(theta)-4sin^3(theta)

[Stevensmay]

Doesn't sound like that great of a question honestly.
Maybe binomial expansion for the higher powers.

One thing to watch out for would be the i terms when expanding, could cause havoc if you leave one out or simplify it incorrectly.

[jono88]

yeh its from a 2013 tssm paper...most ridiculous complex numbers question ive ever done.........

[lzxnl]

Those questions are quite common.
The main point is to use cos(nt) = Re(cis(nt))=Re((cis(t))^n) and likewise for sines
Again, make sure the expansion is correct...don't forget that i^3 = -i, for instance.