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Optimization and related rates trig.

  1. Nov 19, 2009 #1
    1. The problem statement, all variables and given/known data

    A woman at point A on the shore of a circular lake with radius 2 miles wants to arrive at point C diametrically opposite A on the shore of the lake in the shortest time possible. She can walk at 4 mph and row a boat at 2 mph. To what point on the shore of the lake should she row before walking o minimize time? Verify your answer.

    2. Relevant equations

    Distance = velocity x time
    t=d/v
    total time = d in water/2mph + d on land/4mph

    3. The attempt at a solution

    I know that I must use angles to find the point B but I have no idea exactly what to do. If i could find the relation I could easily optimize it, but I really have no idea how to do it.
     
  2. jcsd
  3. Nov 19, 2009 #2

    LCKurtz

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    Your equation: total time = d in water/2mph + d on land/4mph is a good start. Draw a picture showing the person traveling a little way around the circumference to a point B. Call the center of the lake O. Call angle ACB [itex]\theta[/itex] and notice that angle AOB is [itex]2\theta[/itex]. You should be able to get the walking distance arc AB and the rowing distance BC in terms of [itex]\theta[/itex] to put in your equation for time. Then you are on your way.
     
  4. Nov 19, 2009 #3
    so if arc AB = r*2theta (i think) would BC = 2/sin(pi-2theta)?
     
  5. Nov 19, 2009 #4
    never mind, that was dumb, its not a right triangle, BC would be r*(pi-2theta)
     
  6. Nov 19, 2009 #5
    as of now i have gotten:

    time=((r(pi-2theta)/2)+r*2theta/4
    r=2 so i have:
    time = pi-(pi*theta)+2pi
    I have no idea how to optimize... any help?
     
  7. Nov 20, 2009 #6

    LCKurtz

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    You don't have the equation for BC correct. You are going to need some trig functions. In your figure, drop a line from B perpendicular to the diameter AC and call the intersection P. Now CBP is a right triangle and the rowing part BC is its hypotenuse. You can get OP from the little triangle OPB with a trig function and CP = OP + r. Now you can get the rowing distance CB from triangle CPB with another trig function. You have the walking arc length correct.
     
  8. Nov 20, 2009 #7
    I figured it out, thanks for the help
     
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