VCE Methods Question Thread!

[nerdgasm]

Because not all of the final branches are relevant to the question - the finals series stops whenever one team has won 2 matches (if you like, you may think of it as a game of tennis where the first person to win 2 sets wins the match). Therefore, if team R wins the first two games (RR), or if team S wins the first two games (SS), a third game isn't played. There is no way to get (RRR), (RRS), (SSR) or (SSS) because in all of these cases, the third game isn't played.
On the other hand, if the first two games are split (RS) or (SR), then we do have a third game, and so you can get the possibilities (RSR), (RSS), (SRR) and (SRS). That's why those branches are included in the final tree diagram.

[cosine]

How would you do this question attached?

We know that the tangents pass through (0, 4) so this point is NOT the point to find the gradient of the curve.

Say is any point on the curve.

y' = 8x
The gradient of the tangent at x = p is:

y' = 8p

y = mx+c
y=8px + c

We know (0, 4) exists on the tangent;

4 = c

y = 8px + 4

Now we just gotta solve for p, which will tell us where exactly the point intersects, we know that exists on the tangent line:









So we said that the point p is the point on the curve, and we know that when the tangent at this point will intersect the point (0, 4) too. So now x = . Just sub this into the to work out the corresponding y-values.

To prove that this works, graph the function of and draw the tangents of y = and you should see they both intersect (0, 4)

[HopefulLawStudent]

Can someone please help me with part d? (MAV, 2013: Exam 2)

NB: I haven't supplied any of the answers for a-c but only because I don't think you really need them for d and the answer that you need, from c, is given in the question for part d anyway. But I can totally supply them if you need them.

Anyone? Please?

[knightrider]

We know that the tangents pass through (0, 4) so this point is NOT the point to find the gradient of the curve.

Say is any point on the curve.

y' = 8x
The gradient of the tangent at x = p is:

y' = 8p

y = mx+c
y=8px + c

We know (0, 4) exists on the tangent;

4 = c

y = 8px + 4

Now we just gotta solve for p, which will tell us where exactly the point intersects, we know that exists on the tangent line:









So we said that the point p is the point on the curve, and we know that when the tangent at this point will intersect the point (0, 4) too. So now x = . Just sub this into the to work out the corresponding y-values.

To prove that this works, graph the function of and draw the tangents of y = and you should see they both intersect (0, 4)

Thanks cosine  Really made sense !

[knightrider]

Because not all of the final branches are relevant to the question - the finals series stops whenever one team has won 2 matches (if you like, you may think of it as a game of tennis where the first person to win 2 sets wins the match). Therefore, if team R wins the first two games (RR), or if team S wins the first two games (SS), a third game isn't played. There is no way to get (RRR), (RRS), (SSR) or (SSS) because in all of these cases, the third game isn't played.
On the other hand, if the first two games are split (RS) or (SR), then we do have a third game, and so you can get the possibilities (RSR), (RSS), (SRR) and (SRS). That's why those branches are included in the final tree diagram.

Thanks nerdgasm    once again awesome explanation!

[popsicles]

Getting a bit stuck on this question:

The letters of the word featuring are randomly rearranged. Find the probability that the letters of the word feat are together, though not necessarily in the order shown.

Where is this question from? From inspection it seems very tedious and not really part of the methods course

[knightrider]

For this question and answer attached.

using   and Area where

for this question shouldnt the answer only be  maximum area when

because when    the area is negative.

So are the answers attached right or is my answer right ?

[schooliskool]

For this question and answer attached.

using   and Area

for this question shouldnt the answer only be  maximum area when

because when    the area is negative.

So are the answers attached right or is my answer right ?

Considering P is on the positive side of the x axis it should only be positive. Might be missing something, but yeah I'd go with the positive answer.

[knightrider]

i Have just edited my post above with the domain for p in the question

So this should clarify which of the answers is correct.

[knightrider]

Considering P is on the positive side of the x axis it should only be positive. Might be missing something, but yeah I'd go with the positive answer.

Thanks schooliskool 

i have edited my question with the relevant domains could you have another look please. 

[schooliskool]

Thanks schooliskool 

i have edited my question with the relevant domains could you have another look please. 

Yeah haha the domain is definitely important here mate!
Saying that the area is negative at p=- 2root3/2 is not correct. Since a parabola is symmetrical, the points +/- p will give the same area.
An easier way of thinking it is that they are the same triangle, just flipped on one side. That doesn't change the area haha

[knightrider]

Yeah haha the domain is definitely important here mate!
Saying that the area is negative at p=- 2root3/2 is not correct. Since a parabola is symmetrical, the points +/- p will give the same area.
An easier way of thinking it is that they are the same triangle, just flipped on one side. That doesn't change the area haha

Thanks schooliskool 

[kiddoes]

Hi all 
A general question; are we required to be able to do markov chains by hand - as in, evaluating markov chains where the power is greater than 1?
Thanks!

[knightrider]

Are we expected to find the steady state of a matrix by hand and if so how do we do this?

[Mc47]

Are we expected to find the steady state of a matrix by hand and if so how do we do this?

Yes

First matrix is your standard 2x2 matrix that you create from the info in the question. Multiply this by a 2x1 matrix with two random letters in it, I usually put an 'a' on the top and a 'b' below it. Then let this equal to another 2x1 matrix, again with 'a' on the top and 'b' on the bottom. Then use this to get your first equation (i.e. multiply the matrices and let it equal to the other matrix). Your second equation will be a+b=1. Then use simultaneous equations to solve for a and b and hence find the steady state solution