VCAA 2011 Exam 2 HELP!!!!

[ArthurJ]

        Hi guys , i just have a quick question.
   Im just doing the 2011 VCAA Exam 2 at the moment, and on Question 4 C , i basically did everything right, except i don’t understand where the 1/2 comes from!
   Its probably the most simple thing that im over looking, so im sort of embarrassed asking!
   Is it just because he is going 2km/h that the time taken for that past is reduced by half? But i fell like that’s wrong because that means im making the assumption that he swims at 1km/h.
   
Thank you for your help


Heres a link to make things easier ! : http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2011mmcas2-w.pdf

[b^3]

, , so , hence you need to divide the distance for that leg by the speed that he's travelling at. Then you add the time from the other leg, which as they've said is proportional to the difference between coordinates from the desalination plant and the point where he enters the river.

[ArthurJ]

Thank you I knew it was somthing  simple i overlooked!

Much appreciated, if youre still there could you explain to me why the last question isnt k=..., but instead k>...

[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

[YouAreNowReadingMyName]

The last question's extremely tricky. In terms of mathematically instead of logically as above, you want the minimum point to be at the endpoint. (Fuck I need to learn to use that fancy stuff. Oh well.) For it to be a minimum, though, you want the local minimum to be to the right of or at the endpoint. Therefore, the gradient at the endpoint is negative, and that's how you find the set of solutions. This also makes sense because for it to be a negative it needs to be lower than the value to the left and you must have a negative gradient to achieve this.