vcaa 2011 exam 2 question 4F help!!

[dfgjgddjidfg]

have no idea how to do this.

[Moko]

Climbtoohigh:"Don't know if I can but I'll give it a shot

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.

[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