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September 20, 2025, 01:23:10 pm

Author Topic: 2011 VCAA exam 2 last question  (Read 2821 times)  Share 

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Moko

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2011 VCAA exam 2 last question
« on: October 28, 2012, 10:41:59 pm »
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Yep, I'm asking about the question that only 1% got right- the very last one (qs 4f). If you can do it please give me a detailed explanation of how u did it...I'll be eternally grateful!

FlorianK

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Re: 2011 VCAA exam 2 last question
« Reply #1 on: October 28, 2012, 10:53:28 pm »
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Yep, I'm asking about the question that only 1% got right- the very last one (qs 4f). If you can do it please give me a detailed explanation of how u did it...I'll be eternally grateful!
I'll write something up shortly, but essentially the k value is kind of the opposite of speed. If k is high he is swimming very slowly so when k is = or higher than a specific value than he'll go directly.

Moko

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Re: 2011 VCAA exam 2 last question
« Reply #2 on: November 02, 2012, 10:03:05 am »
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.

polar

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Re: 2011 VCAA exam 2 last question
« Reply #3 on: November 02, 2012, 07:31:57 pm »
+1
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

D.H

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Re: 2011 VCAA exam 2 last question
« Reply #4 on: November 02, 2012, 11:42:56 pm »
+1
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

I had trouble with this question as well.
But why do you solve dy/dx <= 0 rather than dy/dx = 0?

Jenny_2108

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Re: 2011 VCAA exam 2 last question
« Reply #5 on: November 02, 2012, 11:45:26 pm »
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I had trouble with this question as well.
But why do you solve dy/dx <= 0 rather than dy/dx = 0?

the question requires that T is as small as possible, and since x is kept as a constant, only k can be varied
« Last Edit: November 02, 2012, 11:47:17 pm by Ennjy »
2012: Bio | Chem| Spesh | Methods | ESL | Vietnamese
2013-2016: BActuarial studies/BCommerce @ ANU

Thanks to gossamer, TT, pi, laserblued, Thus for helping and supporting me during VCE

abeybaby

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Re: 2011 VCAA exam 2 last question
« Reply #6 on: November 03, 2012, 12:08:41 am »
+2
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

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