VCE Methods Question Thread!

[appleandbee]

Can someone please help me solve steady-state stuff WITHOUT a calculator. I've seen these type of questions pop up in two Exam 1s. Sorry about the confusing mathematical notation because I don’t know how to compute mathematical stuff. 

A Markov chain has a transition matrix T= [First Row, First Column]-0.2, [Second Row, First Column]-0.8, [First Row, Second Column]-0.9, [Second Row, Second Column]-0.1. In the long term, the steady state matrix will be [First Row] x, [Second Row] 1-x when 0 is less or equal to x which is less or equal to 1. Find the value of x.

[Zues]

just a question with transformations
if i have the point (x,y) and dilate by a factor of k from y axis why is the point (kx,y). similarily with a dilation from the x axis (x,ky) ?

i know dilations just stretch/narrow it down, and do not shift it up or down.

[brightsky]

Can someone please help me solve steady-state stuff WITHOUT a calculator. I've seen these type of questions pop up in two Exam 1s. Sorry about the confusing mathematical notation because I don’t know how to compute mathematical stuff. 

A Markov chain has a transition matrix T= [First Row, First Column]-0.2, [Second Row, First Column]-0.8, [First Row, Second Column]-0.9, [Second Row, Second Column]-0.1. In the long term, the steady state matrix will be [First Row] x, [Second Row] 1-x when 0 is less or equal to x which is less or equal to 1. Find the value of x.

Use the definition of the steady state. If a system is already in the steady state, then pre-multiplying the steady state by the transition matrix is going to do nothing at all. Hence, all you need to do to find x is solve the matrix equation below:

[0.2, 0.9; 0.8; 0.1] [x;1-x] = [x;1-x]

Just a bit of trivia: separate entries in the same row with a comma "," and entries in different rows with a semi-colon ";". This also works on the calculator.

just a question with transformations
if i have the point (x,y) and dilate by a factor of k from y axis why is the point (kx,y). similarily with a dilation from the x axis (x,ky) ?

i know dilations just stretch/narrow it down, and do not shift it up or down.

Think about what happens graphically. If you dilate a point (x,y) by a factor of k from the y-axis, you literally stretch the point by a factor of k away from the vertical axis. The y-coordinate of the point stays put, but the x-coordinate is changed by a factor of k. Hence, the image is (kx,y). Similarly, if you dilate a point (x,y) by a factor of k from the x-axis, you literally stretch the point by a factor of k away from the horizontal axis. The x-coordinate of the point stays put, but the y-coordinate is changed by a factor of k. Hence, the image of (x,ky).

EDIT: Typo rectified.

[appleandbee]

Use the definition of the steady state. If a system is already in the steady state, then pre-multiplying the steady state by the transition matrix is going to do nothing at all. Hence, all you need to do to find x is solve the matrix equation below:

[0.2, 0.9; 0.8; 0.1] [x;1-x] = [x;1-x]

Just a bit of trivia; separate entries in the same row with a comma "," and entries in different rows with a semi-colon ";". This also works on the calculator.

Thanks!

A binomial random variable, X has a mean of 3 and a variance of ¾. Find Pr (X=2).

[brightsky]

Thanks!

A binomial random variable, X has a mean of 3 and a variance of ¾. Find Pr (X=2).

E(X) = np = 3...(1)
Var(X) = np(1-p) = 3/4...(2)

Solve (1) and (2) simultaneously to get p = 3/4 and n = 4.

Hence, Pr(X=2) = "4 choose 2" * (3/4)^2 (1/4)^2. Alternatively, use the binompdf function on the calculator; input the values n = 4, p = 3/4 and x = 2 and press enter.

[keltingmeith]

EDIT: Beaten by brightsky, but see below for a more detailed summary.

Thanks!

A binomial random variable, X has a mean of 3 and a variance of ¾. Find Pr (X=2).

Bit of a tricky one - most straightforward to do this is to get an equation in terms of n and p and solve simultaneously so you can do the last bit. So, we know:

and . We also know that for , then and . So:

and

Substituting appropriately:



Putting this into equation 1:



So, , which means:

[Zues]

Think about what happens graphically. If you dilate a point (x,y) by a factor of k from the y-axis, you literally stretch the point by a factor of k away from the vertical axis. The y-coordinate of the point stays put, but the x-coordinate is changed by a factor of k. Hence, the image is (x,ky). Similarly, if you dilate a point (x,y) by a factor of k from the x-axis, you literally stretch the point by a factor of k away from the horizontal axis. The x-coordinate of the point stays put, but the y-coordinate is changed by a factor of k. Hence, the image of (kx,y).

i cant get my head around . The y-coordinate of the point stays put, but the x-coordinate is changed by a factor of k. Hence, the image is (x,ky). and the other part as well. if the x coordinate changes, why isnt it kx,y?

also

find the image of the graph y = 1 / x^2 under the transofmration
[1/2  0]
[0     4]

[brightsky]

i cant get my head around . The y-coordinate of the point stays put, but the x-coordinate is changed by a factor of k. Hence, the image is (x,ky). and the other part as well. if the x coordinate changes, why isnt it kx,y?

also

find the image of the graph y = 1 / x^2 under the transofmration
[1/2  0]
[0     4]

Sorry, typo. First one should be (kx, y) and second one should be (x, ky).

With the second question, the transformation matrix encapsulates two dilations: a dilation by a factor of 1/2 from y-axis and a dilation by a factor of 4 from the x-axis. Hence, equation of image is y=4/(2x)^2 = 1/x^2.

[Zues]

ok that makes a bit of sense now.

why did you have to square the 2 in the denominator?

[brightsky]

ok that makes a bit of sense now.

why did you have to square the 2 in the denominator?

A few ways to think about this.

First way:

1. Dilate by a factor of 1/2 from the y-axis. Replace x with x/(1/2) = 2x. y = 1/x^2 becomes y = 1/(2x)^2 = 1/(4x^2).
2. Dilate by a factor of 4 from the x-axis. Replace y with y/4. y = 1/(4x^2) becomes y/4 = 1/(4x^2), which is equivalent to y = 1/x^2.

Second way:

[x';y'] = [1/2,0;0,4] [x;y]
[x';y'] = [1/2x;4y]
x' = 1/2x, which implies x = 2x'
y' = 4y, which implies y = y'/4

Substituting these into the original equation yields:
y'/4 = 1/(2x')^2
y' = 1/x'^2

Guys, why is it that any is equal to 0?
Is it because is equal to  ,

hence = 0?? or did i just make that stupid theory up

Your reasoning is perfectly legit. Another way to think of it is this. Can you think of a number such that e to the power of that number is 1? Of course, the only number, such that e to the power of that number is 1, is 0. I find that this is the best way to think of logarithms. Below are a few other examples.

i) What is the value of log_2(4)? Ask: can I think of a number such that 2 to the power of that number is 4? Of course, the answer is 2, so log_2(4) = 2.

ii) What is the value of log_10(1000)? Ask: can I think of a number such that 10 to the power of that number is 1000? Of course, the answer is 3, so log_10(1000) = 3.

[Zues]

is addition of ordinates/product functions/subtraction of ordinates in the methods study design?

[Zues]

e.g. they have asked to sketch g(x) + h(x) , in spesh they said that each one acts as an aymptote, its been a while since ive done this chapter, but in the answers it has the resultant graph coinciding with each of these "asymptotes"

[keltingmeith]

e.g. they have asked to sketch g(x) + h(x) , in spesh they said that each one acts as an aymptote, its been a while since ive done this chapter, but in the answers it has the resultant graph coinciding with each of these "asymptotes"

Yes - not only are all three of those in the study design, they're explictly stated for methods, but NOT explictly stated for specialist. (note: the questions you've asked for spec and addition of ordinates are still examinable, though. They're just mentioned differently in the spec study design)

Honestly, this sentence just makes me think you don't understand addition of ordinates. The idea behind addition of ordinates -NEVER- states that adding functions together makes one an asymptote. In specific examples (eg, ), you may see one of the functions form an asymptote - but that's not the rule.

The idea behind addition of ordinates is this:

If f(x)=g(x) + h(x), and g(2)=3 while h(2)=6, then f(2)=3+6=9.

[Zues]

sorry, i was meant to say the asymptotes are x=0 and y = x

[keltingmeith]

sorry, i was meant to say the asymptotes are x=0 and y = x

There are no asymptotes - as I said, addition of ordinates DOES NOT imply that every addition of functions has asymptotes.

Consider f(x)=x and g(x)=x. Then, h(x)=f(x)+g(x) is just h(x)=2x - using your thought process that there must be an asymptote somewhere in this function - but we both know that y=2x is a straight line with gradient 2. Of course it doesn't have an asymptote!