Calculus problem
Okay, this is the problem my teacher thought we could finish in the last ten minutes on wednsday:
A man is in a boat 2 miles from the shore. He then has to go to point Q, which is three miles down the coast and one mile from the shore. The man can row at two miles per hour and he can walk at 4 miles per hour. At what point on the coast should he row towards(how far down the coast) in order to reach point Q in the least time?
Well the teacher didn't finish the problem in time but he sure as hell tried. I am going to try and redo the problem and post the answer for my next blog entry, but if I don't, don't be dissapointed. It just means I never finished the damn thing.
1 Comments:
Okay I don't think I am going to waste my time in making another blog entry on this problem. It is a tough one I will tell you that. I will tell some of the equations I did get.
Since I need to find the least amount of time. I am going to have to find the derrivitive of the the equation of the ine between where the man is now and point Q. So he is the original equation I got:
((x^2 + 4)^(1/2))/2) + (((3-x)^2) + 1)^(1/2)/4)
so the derrivitive is going to be:
((x^3 + 4x)/4) + ((x^2 + 6x - 5)(2x - 6)/8)
which factors as:
((x^3 + 4x)/4) + ((2x^3 - 18x^2 + 26x - 30)/8)
which factors as:
((4x^3 - 10x^2 + 26x - 30)/8)
set equation equal to 0 to find critical points:
4x^3 - 10x^2 + 26x - 30 = 0
(2)(2x^3 - 5x^2 + 13x - 15) = 0
This is pretty much where I am stuck, I can't find the critical points from here. I am assuming that I made some mistake in finding the derrivitive or I made some kind of simple mistake in the algebra. Either way I am tired of working on this problem. If anyone out there likes caclulus and is good at it sees anything I did wrong, let me know.
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