Sunday

August 31, 2014

August 31, 2014

Posted by **Alison** on Saturday, October 20, 2012 at 8:51pm.

- Calc -
**Reiny**, Sunday, October 21, 2012 at 12:09amGeesh, I wish authors would stick to the traditional wording of the "cable through a river" problem, instead of trying to be "current"

anyway...

Make a sketch labeling his position B and the point on the road 32 km away as A. let the point 16 km down from A be C

Suppose he aims for a point on the road between A and C, call that P, let AP=x

So his path is BP +PC

then BP^2 = x^2 + 32^2

BP = (x^2 + 1024)^(1/2)

time to go BP = (x^2 + 1024)^(1/2)/48

time to go PC = (16-x)/80

T = (x^2+1024)^(1/2) /48+ 16/80 - x/80

dT/dx = (1/96)(x^2 + 1024)^(-1/2) (2x) - 1/80

= 0 for a min of T

2x/(96√(x^2+1024)) = 1/80

96√(x^2+1024) = 160x

3√(x^2+1024) = 5x

square both sides

9(x^2 + 1024) = 25x^2

9216 = 16x^2

x^2 = 576

x = √576 = 24

But we clearly expected x to be between 0 and 16, so let's investigate the two direct routes.

1. directly to A, then AP

T = 32/48 + 16/80 = .86666 hrs

2. directly to C

distance = √(16^2 + 32^2) = √1280

T = √1280/48 = .7454

3. If we "blindly" sub in our x = 24 we get a total time

of T = √1600/48 + (-8/80) = 11/15 = .73333

3. consider a point between A and C, say x = 4

distance through sand = √(32^2+4^2) = √1040

time through sand = √1040/48 = .67185 hrs

time along road = 12/80 = .15

total time = .82185

I am puzzles by these results, and I can only guess that I made some arithmetic error somewhere. Just can't seem to find it, perhaps somebody else can find it.

anyway , he has 50 minutes to get there.

my 3 answers are :

.866666.. hrs = 52 minutes , go boom!

.7454 hrs = 44.7 minutes, that would do it

.7333333.. hrs = 44 minutes, so would that.

- Calc -
**Steve**, Sunday, October 21, 2012 at 6:41amlet me take a whack at it.

If the jeep arrives at the road a distance x from the bomb, then

the distance on road is x

distance on sand is √(32^2 + (16-x)^2)

travel time is thus

t = x/80 + √(32^2 + (16-x)^2)/48

dt/dx = 1/80 - (16-x)/48√(x^2-32x+1280)

dt/dx = 0 when x = -8

Also puzzling. And the graph of t(x) appears not to have a minimum for x>0.

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