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Jane is two miles offshore in a boat and wishes to reach a coastal village 6 mi down a straight shoreline from the point nearest the boat She can row 2 mph and can walk 5 mph where should she land?
Wiki User
∙ 14y ago
She should aim for a point that is sqrt(16/21) downshore.
Let t represent the total time it will take her to get to the village and let y represent the distance downshore that she lands. The total time that it takes her to get to the village will be the sum of her time rowing in the water and the time she walks. Distance=rate*time, so time=distance/rate.Using the Pythagorean theorem, we can see her distance in the water will be sqrt(y2+4). Her distance walking on land is (6-y). Her total time will be the time she is rowing plus walking on land. This gives us:
t=sqrt(y2+4)/2+(6-y)/5
Plugging in the two extremes for y, you will see that t=2.2 for y=0 and t=3 for y=6. Graphing this function will also show you that it has a minimum value somewhere just to the right of y=0. To find out where the minimum is, you need to differentiate the equation and set it equal to zero.
dt/dy=(1/2)(1/2)(y2+4)(-1/2) (2y)-1/5
Setting this equal to zero,
(1/2)(1/2)(y2+4)(-1/2)(2y)-1/5=0
(1/2)(1/2)(y2+4)(-1/2)(2y)=1/5
5y=2sqrt(y2+4)
25y2=4(y2+4)
y2=16/21
y=sqrt(16/21), or approximately 0.873, and t is approximately 2.12. (So for all her work figuring this out, Jane saves about four minutes off the time it would take to row straight to shore and walk the six miles :-)
Wiki User
∙ 14y ago
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