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What route will take least amount of time

  1. Nov 12, 2008 #1
    1. The problem statement, all variables and given/known data
    You are standing at the edge of a slow-moving river which is one mile wide and wish to return to your campground on the opposite side of the river. you can swim at 2mph and walk at 3mph. You must first swim across the river to any point on the opposite bank. From there walk to the campground, which is one mile from the point directly across the river from where yous tart your swim. What route will take the least amount of time?

    2. Relevant equations
    I don't know....


    3. The attempt at a solution
    the easiest way is to swim straight across then walk... taking
    30 min (swimming at 2 mph 1 mile distance) + 20 (walking at 3mph 1 mile distance) = 50 minutes
     
  2. jcsd
  3. Nov 12, 2008 #2
    Re: Extrema

    If you swim it all the way, it takes you [tex]\frac{\sqrt{2}}{2}[/tex] hours which is about 42 minutes. Now can we do better than that by placing ourselves at an optimal point on the shoreline?

    So we set up our scenario. You can swim and you will reach a point on the opposite shore, call this point x. Then the distance you have to walk is 1-x. Draw a picture, and note that the distance you swam is given by pythagoras as sqrt(1+x^2). To find the amount of time you spend walking and swimming for some arbitrary point on the shore x is going to be the distance you swam divided by 2mph + the distance you walk divided by three mph. Now you have a formula telling you the total time it will take you for some arbitrary place you decide to swim to on the other shore.

    You want to minimize the time you spend total. Can you use some calculus to do this from here? Think about how you find maxima and minima on a curve, and then use that trick to find what x needs to be in order for you to have a minima in the function expressing time.
     
  4. Nov 12, 2008 #3
    Re: Extrema

    I'm sorry but i feel completely lost. Why are u using pathagoras? if your campground is a straight line from which you started.... and i know where to find extrema just can't figure out the algebra behind this problem....
     
  5. Nov 12, 2008 #4
    Re: Extrema

    You're okay. Draw a picture of a river with a stick man on one side, and a campground (down the river) on the other side. Now draw a line straight across the river. Label this line 1 as it denotes a distance of 1 has been travelled if you cross the river directly. Now from that lightly pencilled dot move along the edge of the river towards the campground and make a big Xo. This Xo is special and you'll use it a lot. Look at your drawing, and you should see that the distance from the lightly pencilled to the campground is 1 mile, the distance from the lightly pencilled dot to the Xo is well let's just call it X. Now the distance from Xo to the campground is 1-X. Check it out, if you add 1-X to X you get 1, which just says that you can go from the lightly pencilled dot to the campground and travel one mile, or you can go from the lightly pencilled dot to the Xo and from the Xo to the campground and it is also a distance of 1 mile.

    Now, you swam to the point Xo right? Well, how far did you swim? To find this out, you need pythagoras theorem, aka the distance formula. Which says that the distance from stick man to Xo is the square root of (1^2 (the distance directly across the river) + x^2 (the length you travelled down the shoreline by swimming)). Hang in, you're almost done.

    You now know the distance you have to swim. It is [tex]{\sqrt{x^2 + 1}[/tex]. You also know that once you hit the shore on the other side you will have to walk the distance 1-x in other words the distance along the shore to the campground that you did not cover by swimming.

    Now, you want to minimize the total trip time. So you need to figure out how long you are going to spend swimming and how long you are going to spend walking. To do this add the time you spend swimming which is equal to the distance you swim divided by how fast you swim to the time you spend walking which is equal to the distance you walk divided by how fast you walk. Now you have a formula that tells you how long your trip will take for however far down the shore you wish to swim. T(x)

    Can you see what to do from here.

    Hint: take the derivative and find the minima of T. Solve for x.
     
  6. Nov 12, 2008 #5
    Re: Extrema

    I'm not sure if we have the same picture of the problem (mine is probably wrong)
    well look at my attachment thats what i see...
    [​IMG]
     

    Attached Files:

  7. Nov 12, 2008 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Re: Extrema

    You seem to be under the impression that the camp ground is one mile along the line, extended, directly across the river. If that were the case, the only reasonable thing to do would be to swim all the way across the river and then walk the rest of the way! Are you sure you gave the exact statement of the problem? Does it not say that the campground is on the river one mile from the point directly across?
     
  8. Nov 12, 2008 #7
    Re: Extrema

    can someone draw how the picture is suppose to be?... or at least explain it better than the problem did
     
  9. Nov 12, 2008 #8

    Mark44

    Staff: Mentor

    Re: Extrema

    You can draw the picture. Draw a horizontal line with endpoints A and B. This segment represents the width of the river, 1 mile.

    Now draw a vertical line from B to another point C. This segment represents the distance from B to your campsite.
    Code (Text):

    A---------------B
                    |
                    |
                    |
                    |
                    |
                    C
     
    You start swimming at point A. Point B is directly across the river and 1 mile away from A. Point C is 1 mile along the river (preferably downstream) from point B.
     
  10. Nov 12, 2008 #9
    Re: Extrema

    so could you directly swim from point a to point c? thus using pathagoras' theorem which would be the fastest way... please can someone confirm this.. this problem was driving me nuts
     

    Attached Files:

    • 1.JPG
      1.JPG
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      6.2 KB
      Views:
      55
  11. Nov 12, 2008 #10
    Re: Extrema

    To do this add the time you spend swimming which is equal to the distance you swim divided by how fast you swim to the time you spend walking which is equal to the distance you walk divided by how fast you walk. Now you have a formula that tells you how long your trip will take for however far down the shore you wish to swim. T(x)

    could someone explain this in terms of an equation? i got something= sqrt(x^2+1)/2mph
     
  12. Nov 12, 2008 #11

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Re: Extrema

    That would be the shortest way- but you can walk faster than you can swim so it is not necessarily the fastest way.

    In your picture, mark a point between the perpendicular across the river and the camp and call the distance from the perpendicular to that point "x". Suppose you swim directly to that point and then walk to camp. How far would you swim (as a function of x)? How much time would that require? How far would you walk (as a function of x)? How much time would that require? Add those times to find the total time required, again as a function of x. What value of x makes that minimum?
     
  13. Nov 12, 2008 #12

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Re: Extrema

    "Something"? What is that "something"? What does "x" mean?
     
  14. Nov 12, 2008 #13
    Re: Extrema

    can you just write out the equations I am not getting them :( grrr i've been working on this problem for the longest i understand y its like that just don't know how to write the 2 equations...
     
  15. Nov 12, 2008 #14
    Re: Extrema

    T= sqrt(x^2+1)/2mph
     
  16. Nov 12, 2008 #15
    Re: Extrema

    (1-x)/3 = sqrt(x^2+1)/2mph
     
  17. Nov 12, 2008 #16
    Re: Extrema

    i then took the derivative of that and got that he swims for 24 minutes and walks for 4 making it the fastest time available
    i hope thats right ...:(
     
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