VCE Methods Question Thread!

[zvezda]

Hey,
With the vcaa exam 2 for 2011, im not 100% sure about the very last question.
Why must dT/dx be smaller than or equal to zero??
Thanks in advance

[Phy124]

if he goes directly from his camp at to the plant then since the coordinates of the plant is

the function that describes how long he takes to go from his camp to the plant is

the question requires that T is as small as possible, and since x is kept as a constant, only k can be varied. so, differentiating T and substituting gives thus, solving for k gives

k determines how fast he can swim. Big k-values means the time he spends swimming is really big, ie, swims slowly.
small k values means he doesnt spend much time swimming and so, he swims quickly.

recapping: big k values, slow swimmer, LESS SWIMMING MORE RUNNING
small k values, fast swimmier, MORE SWIMMING LESS RUNNING

we just found out, that if k=5root(37)/74, he should do ZERO swimming, so if k is even bigger, how much swimming should he do? even less than 0 kms of swimming, which he cant, so we just say he runs directly there for k=>5root(37)/74

[Rectophobia]

I'm not sure about their reasoning either, but I reasoned the answer with the following:
When you make dT/dx = 0, you're finding the minimum k value such that it is quickest for Tasmania to run straight to the desalination plant from his camp. Looking at the information given in part c, you are told that the time taken in the river is proportional to the value of k. From this you can reason that a larger k value yields a longer time spent in the water. Given that the k value found when dT/dx was a minimum, any k value larger than this will cause river travel to be even more inefficient and hence values larger than the one obtained must also be included as it would still be quicker to run straight to the desalination plant.

[lzxnl]

When I did this question last year, my reasoning was something like this.

At k=5sqrt(37)/34, dT/dx = 0 only at the right endpoint (and x=0, but we don't care about that). Also, dT/dx is negative for all other values of x in the domain of T. Therefore, T is always decreasing to get to x=sqrt(7)/2. Increase k, and dT/dx never even gets to zero. Therefore, it is always decreasing and the right endpoint is thus the minimum value of T in the domain.

dT/dx=(2x^3-x)/(2sqrt(x^4-x^2+1)) - 2xk

Notice how if you increase k, dT/dx always decreases.

[clıppy]

VCAA 2011 Tech MC Q21

For two events, P and Q, Pr(P n Q) = Pr(P' n Q). P and Q will be independent events exactly when
A - Pr(P') = Pr(Q)
B - Pr(P n Q') = Pr(P' n Q)
C - Pr(P n Q) = Pr(P) + Pr(Q)
D - Pr(P n Q') = Pr(P n Q)
E - Pr(P) = 1/2

How does does this happen?
More generally, are there some basic rules to remember that you can apply when working with probability. For example, they ask you to find the Pr(A' u B) or Pr(A' n B). In TF's i can really only do these questions with a karnaugh map.

[b^3]

This is still VCAA's way of doing it but hope it explains it a bit more, I added in what each step is doing or where it's getting each step from.



Hence Option E.

EDIT:
For the part, we can look at the venn diagram below. We want “Whats in but not ” (which is just the blue shaded region), which is just what is in (both the blue shaded and dual shaded reigons) minus what is in the intersection (the dual shaded region).

OR

For independent events, Pr(A n B) = Pr(A) x Pr(B)

question says Pr( P n Q) = Pr( P' n Q)

so for them to be independent events , Pr(P n Q) will be Pr(P) x Pr(Q), Pr(P' n Q) will be Pr(P') x Pr(Q)

Pr(P) x Pr(Q) = Pr(P') x Pr(Q)

Pr(P) = Pr(P')  (1)
Pr(P) + Pr(P') =1  (2)
Sub 1 into 2
2Pr(P) =1
Pr(P)=1/2

is this way ok?

The way I normally approach them is to write down what you know, and see if there are any equivalent expressions (are they independent events, mutually exclusive, 1 take the opposite of what you're looking for or just something that gives the same thing via a venn diagram) , then try to convert all the equivalent expressions into one kind of expression, then see if you can factorise and use the null factor law. Sometimes you might need to do a little sketch of a venn diagram or a karnaugh map to get one expression in terms of other expressions.

[zvezda]

When I did this question last year, my reasoning was something like this.

At k=5sqrt(37)/34, dT/dx = 0 only at the right endpoint (and x=0, but we don't care about that). Also, dT/dx is negative for all other values of x in the domain of T. Therefore, T is always decreasing to get to x=sqrt(7)/2. Increase k, and dT/dx never even gets to zero. Therefore, it is always decreasing and the right endpoint is thus the minimum value of T in the domain.

dT/dx=(2x^3-x)/(2sqrt(x^4-x^2+1)) - 2xk

Notice how if you increase k, dT/dx always decreases.

The thing is, i sort of understand the logic, but whats the point of finding k anyway when he has already run to the desalination plant? K should be completely redundant shouldnt it?

[lzxnl]

This is to show that for x=sqrt(7)/2, T is at its minimum value in the domain.

[clıppy]

VCAA TF 2012 Q4c

I always get stumped on these kinds of questions. Are they conditional, or are they able to be worked out through a tree diagram?
Also, how come the answers say Monday 1, Tuesday 3 and Monday 3, Tuesday 1 are different outcomes, but Monday 2, Tuesday 2 is the same?

[Lejn]

VCAA TF 2012 Q4c

I always get stumped on these kinds of questions. Are they conditional, or are they able to be worked out through a tree diagram?
Also, how come the answers say Monday 1, Tuesday 3 and Monday 3, Tuesday 1 are different outcomes, but Monday 2, Tuesday 2 is the same?

It's conditional probability, 'over these two days' links the second statement to first, and since they have the ability to receive no calls except we are told they DO, then it's a given condition. You can use a tree diagram to find the events where they receive two calls and the events where they receive any calls to finally find your answer with the given probability formula.

[lzxnl]

THIS QUESTION.


The notorious 2012 VCAA exam 1 question that 1% of the state got right.
Yeah, note the wording: he receives "calls" on each day. Plural.

I think they mean that you can have Monday 1, Tuesday 3 and vice versa, but only one combination of Monday 2 and Tuesday 2

[Jaswinder]

question 3a, 2008 VCAA exam 2, when it asks for the time, why do they round up? Shouldn't they round the time down? Cuz if its more than the exact time, he would die? thanks

[clıppy]

THIS QUESTION.


The notorious 2012 VCAA exam 1 question that 1% of the state got right.
Yeah, note the wording: he receives "calls" on each day. Plural.

I think they mean that you can have Monday 1, Tuesday 3 and vice versa, but only one combination of Monday 2 and Tuesday 2

He receives "calls" on each day. Is this in relation to it being conditional, or how we interpret how many calls a day he can receive?

It's conditional probability, 'over these two days' links the second statement to first, and since they have the ability to receive no calls except we are told they DO, then it's a given condition. You can use a tree diagram to find the events where they receive two calls and the events where they receive any calls to finally find your answer with the given probability formula.

So whenever a question usually has a statement that is relating to the first part of the question, we should try and see if its a conditional question or not?

[Lejn]

Well given how cautiously you should be reading all the questions, it should be fairly apparent. Keeping this in mind, carefully inspect statements that are a kind of precursor to a question. A real thing that should keep you on your toes is the amount of marks - simply drawing a tree diagram and giving some Probabilities is definitely not worth 3 marks.

[Alwin]

question 3a, 2008 VCAA exam 2, when it asks for the time, why do they round up? Shouldn't they round the time down? Cuz if its more than the exact time, he would die? thanks

yes, they should. I think I was debating this with someone last time, one of the inconsistencies of VCAA...
usually they round correctly...

Oh, and not related to anything:

THIS QUESTION.

The notorious 2012 VCAA exam 1 question that 1% of the state got right.

Hi Mr top 1% of the state! *waves*

I strongly dislike this question

quote author=Alwin link=topic=128232.msg691534#msg691534 date=1382569219]
I hate this question so much now. Only mark I lost on Exam 1 coz of the wording AND when I was LaTeXing my solution up freaking internet died and lost everything I was typing -.-
So sorry, no pretty looking LaTeX:

> Pr(total 4 calls | calls on Mon and Tue) = Pr(total 4 calls n calls on Mon and Tue) / Pr(calls on Mon and Tue)
   ^ By applying the conditional probability formula

> Pr(calls on Mon and Tue) = Pr(X > 0 on Mon and Tue) = Pr(X>0) x Pr(X>0) = 0.8 x 0.8 = 0.64
   ^ Finding the denominator

> Pr(total 4 calls n calls on Mon and Tue) == probability of ways for calls to be on both days that sum to 4
>> Pr(1 on Mon and 3 on Tue) x Pr(2 on Mon and 2 on Tue) x Pr(3 on Mon and 1 on Tue) = 0.1x0.2 + 0.5x0.5 + 0.1x0.2 = 0.29
      ^Finding the numerator

> Pr(total 4 calls | calls on Mon and Tue) = 0.29/0.64 = 29/64
   ^Subbing the numerator and denominator into the formula

Hope it makes sense!
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