VCE Methods Question Thread!

[TrueTears]

The essentials textbook suggests otherwise??

Both are correct.

Just depends on how you define your i and j entries: http://en.wikipedia.org/wiki/Markov_chain#Steady-state_analysis_and_the_time-inhomogeneous_Markov_chain

[~T]

Is there any way to stop the TI-nspire from giving solutions with hyperbolic functions? It's fairly straight-forward to convert them but it can be frustrating sometimes, depending on the question.


EDIT: Figured out how to deal with it. For anyone else who is wondering, you can convert to exponential form using:

Menu - Algebra - Convert Expression - Convert to exp
(Menu - 3 - A - 3)


EDIT2.0: b^3's method works as well, but it will do the typical expand things like make negative exponents as denominators and such.

[b^3]

Not sure if it will always work, but try Expand on the result you get, should put it into and terms for you. It just requires another step, I don't think there is a way to set it to just go straight to that though.

EDIT: 3100^th post.

[Damoz.G]

From memory, Essentials defines the matrix as to give the steady state probabilities that you mentioned.

If is your success and is your failure then the long run/steady state probabilities are

So you can think of it as taking the entry from the opposite diagonal that corresponds to the row for the event you want, and dividing it by the sum of the opposite diagonal.
So if you want the steady state probability of the event that you defined to be the top row of the transition matrix then you want to find .

EDIT: When I say opposite diagonal I mean the bottom left to top right diagonal.
This will probably still work, it's just a different method of looking at it and defining it.

Yeah, that's what it has in the Essentials Textbook.

Congrats on 3,100 posts!

[oceanblues]

Hey guys, a little help with this question (attached)? Vcaa exam 1 2007 q7
I always seem to struggle with these types of questions

[b^3]

We have an integration by recognition question here. So we need to firstly find a derivative, then we use will that to find an antiderivative of something contained inside that original derivative.
So firstly finding the derivative.

Now notice that the term that we need to integrate is contained in the right hand side of the equation. Now what happens if we integrate both sides? The will just give us back again, and we can split the integral up on the right hand side into two integrals, one of which is what we're looking for.

(And no I didn't forget the , since it asks for "an antiderivative" you don't have to have it in this case).
Hope that helps.

[oceanblues]

Thanks it did

[achre]

CAS Active VCAA 2008, SA question 3f.
I don't understand how they found the coordinates for C or the equation for CD.

[b^3]

You know that when , at point the coordinate is , that is he has 8 units of antidode in his body at point . Now he takes a capsule which adds 16 units to the amount in his body at that point in time. So the coordinate of point is just the coordinate of point , plus the 16 units. . Since this is at , we have the coordinate .

[darklight]

Hey guys,

What's the mathematical reasoning behind this:
If g'(x) and h'(x) exist, g'(x) > h'(x), then the graph of y=g(x) and the graph of y=h(x):
A. Do not Intersect.
B. Intersect Exactly Once.
C. could intersect more than once
D. Intersect no more than once
E. Have a common tangent at each point of intersection

Answer = D. Does the sign of the inequality matter? As in g'(x) > h'(x) produce the same answer as g'(x) < h'(x) ?

[b^3]

If the gradient of one is always greater than the gradient of the other, then we can have two situations.
- The curve with the larger gradient is below the curve with the smaller gradient starting from . As we increase , as the curve with the larger gradient will 'climb' more than the curve with the lower gradient (including negative, think of it as negative climbing), to the point where the two curves meet and cross over. Now we have the curve with the larger gradient on top, and as the gradient is larger, it cannot 'curve back towards' the other curve. Even if we have a negative gradient for the larger curve, the other curve will have a gradient that is more negative, and the two will get further and futher away from each other, and so won't cross again.
-The curve with the larger gradient is above the curve with the smaller gradient starting from . As we increase , the curve with the larger gradient will still 'climb' more than the other curve for each increase in . So it won't get closer to the other curve and won't cross at all.

So that means they can either intersect times or , that is no more than once.

It's really best to draw a few sketches out for this situation, which explains it better than words. But if you were to look at , it's a similar situation, same reasoning as above except that the first part of the example will have the reasoning from the second part and vice versa. Just think of it as swapping the functions give for and around, one will still be moving away from the other, and depending on whether the curve with the larger gradient (where a small positive is larger than a large negative), will change whether they cross once or twice.

I hope that makes sense, it's a little convoluted, really need to draw things out to explain it.

EDIT: Tried to draw out an example, didn't come out clear though. Also yeah I know the tangents are really rough (a few of them are horrible )

Image

[lzxnl]

Hey guys,

What's the mathematical reasoning behind this:
If g'(x) and h'(x) exist, g'(x) > h'(x), then the graph of y=g(x) and the graph of y=h(x):
A. Do not Intersect.
B. Intersect Exactly Once.
C. could intersect more than once
D. Intersect no more than once
E. Have a common tangent at each point of intersection

Answer = D. Does the sign of the inequality matter? As in g'(x) > h'(x) produce the same answer as g'(x) < h'(x) ?

So...the point is that g(x) increases faster than h(x)
The significance of this? Think of it in terms of the function i(x)=g(x)-h(x). Intersection points become x intercepts.
From the question, i'(x)>0 always
Now we have a few scenarios. If i(x)<0 at some point, then it will increase. The most that can happen is that i grows to intersect with the axes. As i'(x)>0, it can't turn around and intersect the axis again, so there is at most one intercept. Similarly if g'(x)<h'(x); define i(x)=h(x)-g(x) and we have the same reasoning.

[Damoz.G]

I can't find the mistake, but its whenever you divide or multiply by a negative.

EDIT: Beaten

[abcdqd]

BasicAcid the mistake is on your first line. When you reciprocate both sides, you have to flip the inequality

[abcdqd]

Cheers mate

Could you give me a list of all the other rules that cause the inequality to flip?

To my knowledge, taking anything to the negative power or multiplying/dividing be a negative number requires flipping. You should also watch out when dividing/multiplying by logs, because if the number inside the log is less than 1, it makes it negative (e.g. )