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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Moko on October 28, 2012, 10:41:59 pm

Title: 2011 VCAA exam 2 last question
Post by: Moko on October 28, 2012, 10:41:59 pm: 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!
Title: Re: 2011 VCAA exam 2 last question
Post by: FlorianK on October 28, 2012, 10:53:28 pm: Quote from: Moko on October 28, 2012, 10:41:59 pm
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.
Title: Re: 2011 VCAA exam 2 last question
Post by: Moko on November 02, 2012, 10:03:05 am: .
Title: Re: 2011 VCAA exam 2 last question
Post by: polar on November 02, 2012, 07:31:57 pm: if he goes directly from his camp at $(0,0)$ to the plant then $x = \frac{\sqrt{7}}{2}$ since the coordinates of the plant is $(\frac{\sqrt{7}}{2}, \frac{3}{4})$

the function that describes how long he takes to go from his camp to the plant is $T= \frac{1}{2}\sqrt{x^4-x^2+1} + \frac{1}{4}k(7-4x^2)$

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 $x= \frac{\sqrt{7}}{2}$ gives $\frac{dT}{dx} = -\sqrt{7}k + \frac{5\sqrt{259}}{74}$ thus, solving $\frac{dT}{dx}\le 0$ for k gives $k \ge \frac{5\sqrt{37}}{34}$
Title: Re: 2011 VCAA exam 2 last question
Post by: D.H on November 02, 2012, 11:42:56 pm: Quote from: polar on November 02, 2012, 07:31:57 pm
if he goes directly from his camp at $(0,0)$ to the plant then $x = \frac{\sqrt{7}}{2}$ since the coordinates of the plant is $(\frac{\sqrt{7}}{2}, \frac{3}{4})$

the function that describes how long he takes to go from his camp to the plant is $T= \frac{1}{2}\sqrt{x^4-x^2+1} + \frac{1}{4}k(7-4x^2)$

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 $x= \frac{\sqrt{7}}{2}$ gives $\frac{dT}{dx} = -\sqrt{7}k + \frac{5\sqrt{259}}{74}$ thus, solving $\frac{dT}{dx}\le 0$ for k gives $k \ge \frac{5\sqrt{37}}{34}$

I had trouble with this question as well.
But why do you solve dy/dx <= 0 rather than dy/dx = 0?
Title: Re: 2011 VCAA exam 2 last question
Post by: Jenny_2108 on November 02, 2012, 11:45:26 pm: Quote from: D.H on November 02, 2012, 11:42:56 pm
I had trouble with this question as well.
But why do you solve dy/dx <= 0 rather than dy/dx = 0?

Quote from: polar on November 02, 2012, 07:31:57 pm
the question requires that T is as small as possible, and since x is kept as a constant, only k can be varied
Title: Re: 2011 VCAA exam 2 last question
Post by: abeybaby on November 03, 2012, 12:08:41 am: 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