VCE Methods Question Thread!

[ahat]

ahat,
Instead of copying and pasting the answers you can also just press enter when it is highlighted. Be warned that this is effectively copy and paste, so anything you had already copied will be overwritten.

Sweet man, that's awesome. Cheers.

[Zealous]

Hey guys!
I need some help with a question from Neap 2012 (lost most of my marks on this part).

The only issue I have is with part g, it has to do with the relative cost. I'm not sure how they calculated it (after checking the solutions).

Image: http://i1282.photobucket.com/albums/a531/Ovazealous/HELP_zps620cf543.png
(linked it as it's quite a long question)

the answer is...

Spoiler

So, if anyway has the time to help me out (it could take a while to come up with a solution) it would be greatly appreciated!

[Damoz.G]

I've done that Exam as well, and came across that question.

My teacher said not to worry about it, because VCAA would word it in a much better format than NEAP has asked the question.

[Eugenet17]

During the Christmas holidays 42 students from a group of 85 VCE students found vacation employment while 73 went away on holidays. Assuming that every student had at least a Job or went on a holiday, what is the probability that a randomly selected student worked throughout the holidays (that is, did not go away), given that he/she had a Job?

Help please!

[b^3]

Let be the event that a student found vacation employment.
Let be the event that the student went away on holiday.


EDIT: Thanks SocialRhubarb.

[SocialRhubarb]

Okay, so firstly we know that everyone either had a job, went on holiday, or did both.

Logically, since we only have 85 students, 42 working and 73 on holiday, we will have some number of students who did both.

The number of students who both worked and went on holiday is given by 42+73-85=30.

We can see this by recognising that students who do both are counted twice if we simply add the the number of students working and the number on holiday, or you can use the probability formula:

but that seems a bit unnecessarily complicated to me.

So the number of people who both went on holiday and worked is 30. The number of people who worked is 42. This means that the number of people who worked but didn't go on holiday must be 12.

The question asks, given that a student has a job, what is the chance they don't go on holidays, which is given by

The probability of students working is 42/85, not 45/85. : )

[b^3]

The probability of students working is 42/85, not 45/85. : )

Thanks man, ah silly errors

(It actually felt quite odd doing that question, something felt wrong and I read through it 4 times before posting to be sure. Surely enough it was staring me right in the face).

[rhinwarr]

What's the difference between an increasing function and a strictly increasing function? From the answers it looks like increasing is when f'(x)>0 and strictly increasing is when f'(x)>=0?

[Damoz.G]

What's the difference between an increasing function and a strictly increasing function? From the answers it looks like increasing is when f'(x)>0 and strictly increasing is when f'(x)>=0?

Strictly Increasing INCLUDES the Turning Point, so you use Square/Closed Brackets. "[]"

For example, assume there was a TP was at (4,7), and it was a increasing from negative infinity.
Strictly Increasing: (- infinity, 4]
Increasing: (- infinity, 4)

[b^3]

VCAA put out a bulletin about this back in my year. This should also be helpful:
http://www.vcaa.vic.edu.au/Documents/bulletin/2011AprilSup2.pdf

[Eugenet17]

thanks for the help guys

[rhinwarr]

Thanks.
Can someone explain why sqrt(x^2) = |x| and (sqrt(x))^2 = x? Or is it the other way round.

[lzxnl]

sqrt(1^2)=sqrt((-1)^2)=|1|=|-1|
Here, by convention, the square root takes the positive root, so no matter what sign x was, square rooting x^2 makes x positive.

(sqrt(1))^2 = 1
and (sqrt(-1))^2 = -1 (if you're willing to accept that the notion of sqrt -1 exists)

Squaring is many to one, so it is well defined, but its inverse is not a function, so we arbitrarily take a particular branch of the square root, in this case the positive branch. Thus when you square a number, there is no ambiguity, but when you square root a number, ambiguity exists in the form of which sign to take, and convention has it to remove the sign completely. Thus explaining the difference in the order of operations.

[SocialRhubarb]

Hey guys!
I need some help with a question from Neap 2012 (lost most of my marks on this part).

Part g:

Spoiler

Okay, so first, let's establish an equation for the cost of the pipeline - we'll call it c(x):

x is the distance along the shore, away from S, that the pipeline meets land.

The amount of pipeline on land is quite simple to calculate, it's just 8-x, and the distance it travels underwater can be worked out using Pythagoras' theorem, giving us the expression:



We want to find the minimum of this function, so we'll differentiate it:



Now, it makes sense that x can take values from 0 to 8. What we want is for the minimum to occur at x=8, which will occur if the function is either at a turning point, or still decreasing when x=8. If it were increasing, the minimum of the graph would occur to the left of x=8, and hence the most cost efficient method of drilling is no longer direct from P to R.



Rearranging for k:



The stem of the question also tells us that k is greater than 1, so the range of values k can take is: .

Alternatively, from a less mathematical and more practical standpoint, you could realise that as k gets larger, it becomes less and less efficient to lay the pipeline underwater and it would be cheaper to shorten the underwater distance if k gets really big. Recognising this, we could find the value of k when c'(8)=0 and state that k would always be less than that.

[Zealous]

I've done that Exam as well, and came across that question.

My teacher said not to worry about it, because VCAA would word it in a much better format than NEAP has asked the question.

Haha, I know the Neap's have a lot of "out there" questions, but I still really wonder how they come to the solution =p

Part g:

Spoiler

Okay, so first, let's establish an equation for the cost of the pipeline - we'll call it c(x):

x is the distance along the shore, away from S, that the pipeline meets land.

The amount of pipeline on land is quite simple to calculate, it's just 8-x, and the distance it travels underwater can be worked out using Pythagoras' theorem, giving us the expression:



We want to find the minimum of this function, so we'll differentiate it:



Now, it makes sense that x can take values from 0 to 8. What we want is for the minimum to occur at x=8, which will occur if the function is either at a turning point, or still decreasing when x=8. If it were increasing, the minimum of the graph would occur to the left of x=8, and hence the most cost efficient method of drilling is no longer direct from P to R.



Rearranging for k:



The stem of the question also tells us that k is greater than 1, so the range of values k can take is: .

Alternatively, from a less mathematical and more practical standpoint, you could realise that as k gets larger, it becomes less and less efficient to lay the pipeline underwater and it would be cheaper to shorten the underwater distance if k gets really big. Recognising this, we could find the value of k when c'(=0 and state that k would always be less than that.

Thanks so much! The biggest realisation for me was that you had to let . I differentiated then I was just like "wuht..". Makes a lot of sense and you've explained it better than the solutions.