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Author Topic: VCAA 2011 Exam 2 Question 4f  (Read 1197 times)  Share 

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diligent18

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VCAA 2011 Exam 2 Question 4f
« on: November 06, 2012, 01:30:39 pm »
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I see a lot of threads regarding the VCAA 2011 papers so I hope this question hasn't been answered already.

Could someone please explain 4f (the very last question of the exam 2 paper for 2011) to me?
I've read the report but I cannot make head nor tail out of the VCAA's solution.

Would be greatly appreciated!
UMAT 2012: 99%tile   ATAR 2012: 99.55
2013: Monash MBBS I

Phy124

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Re: VCAA 2011 Exam 2 Question 4f
« Reply #1 on: November 06, 2012, 01:35:04 pm »
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These posts may help :)

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

I don't get the last question of 2011 Exam 2 part f)... why is dT/dx ≤0?
I don't even know actually, but this is what I PM'd Moko (too lazy to type again haha)

"The previous parts of the question demonstrated that the larger k is, the closer to the plant he will have to run to achieve minimum time. This is because the time taken to swim increases as k increases, so running would be more beneficial than swimming if k is bigger.

From this I gathered that for every point on the lake near the plant, there was an associated value of k for which he should run there to achieve minimum time. So I figured that there was a value of k for which minimum time would be achieved if he ran straight to the plant, i.e. k was too large for him to bother swimming because it would take too long. So I solved dT/dx = 0 for k, when x was equal to the x-coordinate of the desalination plant.

If k was any larger, however, it still made sense that he should run straight to the plant, since the swimming process would take even longer. So any value of k bigger than this would require him to run straight there."
2011
Mathematical Methods | Physics | Chemistry | English | Business Management

2012-2017
Bachelor of Science in Applied Mathematics and Bachelor of Civil Engineering (Honours) @ Monash University

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Transport Modeller @ Arup

diligent18

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Re: VCAA 2011 Exam 2 Question 4f
« Reply #2 on: November 06, 2012, 01:36:47 pm »
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Ah.. so it has been answered after all.
Shall have a read through.
Thanks for compiling these posts, rangaa.
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diligent18

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Re: VCAA 2011 Exam 2 Question 4f
« Reply #3 on: November 06, 2012, 01:45:14 pm »
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Got it!
+100 respect points for the 1% of the state who got it last year.
UMAT 2012: 99%tile   ATAR 2012: 99.55
2013: Monash MBBS I