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Homework Help: Optimization problem with a round lake

  1. Nov 28, 2012 #1
    1. A person from point A wants to get to point C diammetrically across a round lake. This person is on the shore and can walk at a rate of 4 mi/hr and row at a rate of 2 mi/hr. Which method should she use?

    2. radius = 2 mi, triangle with angle θ has the points ABC

    3. I started out by calculating the area of the semi circle which is about 6.6 mi. I found that she can walk there in less time than it takes to row. How do I prove this using differentiation.
  2. jcsd
  3. Nov 28, 2012 #2

    Ray Vickson

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    Set up a formula for the total time if she rows from A to B and walks from B to C; this will be an expression involving some description of the point B, such as its (x,y) coordinates, or its angle from the line AC, or something similar. (Often in such problems, one way is easier than another; the way to figure that out is to try several approaches. And, yes, that might require some false starts.)
  4. Nov 28, 2012 #3
    Don't I need to know the length of the arc subtended by the angle?
  5. Nov 28, 2012 #4
    One equation for walking would be [itex]\frac{\sqrt{2+x^{2}}}{4}[/itex] and the equation for rowing would be [itex]\frac{2-x}{2}[/itex]
  6. Nov 28, 2012 #5

    Simon Bridge

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    It is written to sound like |AC|=2R - i.e. it's across the width of the lake: shore-to-shore.
    But the question makes more sense if point C or point A is on the lake itself ... I'm betting point A since the alternative would involve dragging a boat at 4mi/hr.

    And yes - you'll want to know the length of arc subtended by some angle.
    Do the algebra first and then put the numbers in ... let the walking speed be v and the rowing speed u, derive the formula.
    What does "x" stand for in what you have written?

    Note: if you walk some arc-length s and row some linear distance d, then the time to complete the journey is $$T=\frac{s}{v}+\frac{d}{u}$$
    Last edited: Nov 28, 2012
  7. Nov 28, 2012 #6
    Point A is on the shore of the lake as well as points B and C. C is on the opposite side of A.
  8. Nov 28, 2012 #7
    x is some distance.
  9. Nov 28, 2012 #8
    One could row from A to B and walk from B to C.
  10. Nov 28, 2012 #9
    I found x to be [itex]\sqrt{\frac{2}{3}}[/itex] and T to be 1.11 hours.
  11. Nov 28, 2012 #10

    Simon Bridge

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    Can x be any distance? A distance around the shore? Or the distance betwen point A and point B in some coordinate system? You calculations are meaningless without context

    Like when I described distances I said what they were for.
    So in my description you'd row from A to B and d=|AB|, then walk around to C - which is a distance s.

    ... distance should at least have units.

    Technically - you only save time if the arclength AB is more than twice the chord (since you row the chord at half speed). But you have to use calculus :)
  12. Nov 28, 2012 #11
    Here is how I solved it [itex]T = \frac{2-x}{4} + \frac{\sqrt{2+x^{2}}}{2} = \frac{1}{2} - \frac{1}{4}x + \frac{1}{2}(2+x^{2})^{\frac{1}{2}}; T' = \frac{1}{2}\frac{1}{2}(2+x^{2})^{-\frac{1}{2}}(2x) - \frac{1}{4} = \frac{x}{2\sqrt{2+x^{2}}}-\frac{1}{4} = 0, \frac{x}{2\sqrt{2+x^{2}}}=\frac{1}{4}, 2x = \sqrt{2+x^{2}}, 4x^{2} = 2+x^{2}, 3x^{2} = 2, x^{2}=\frac{2}{3}[/itex]
  13. Nov 28, 2012 #12
    The answer could be wrong, since I forgot to take in account the length of the arc. I did this as if it were a right triangle.
  14. Nov 28, 2012 #13
    It looks like in the picture that the arc makes up a third of the semicircle.
  15. Nov 28, 2012 #14
    She would be rowing for [itex]\frac{2}{3}[/itex] of the way or [itex] 2\frac{2}{3}[/itex] miles and would walk for [itex]\frac{1}{3}[/itex] of the way or [itex]1\frac{1}{3}[/itex] miles.
  16. Nov 28, 2012 #15
    It would take her [itex]1\frac{2}{3}[/itex] hours to make it to the other end of the lake.
  17. Nov 28, 2012 #16

    Simon Bridge

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    Hmm ... that means that she has travelled a total of ##2\frac{2}{3}+1\frac{1}{3}=4\text{miles}## ... which is the diameter of the circle. But that would only happen by rowing the whole way (since rowing is the only direct way across the lake)! Does this make sense? Can she walk on water?

    I cannot tell where you made the mistake because you have not answered my question: what distance does x represent?

    For example - if we make the center of the lake the origin O of a Cartesian coordinate system, so that A=(0,R), C=(0,-R), and B=(x,y) ... then
    the distance rowed is ##d=|AB| = \sqrt{x^2+(y-R)^2}## and the equation of the lake-shore is ##x^2+y^2=R^2##.

    If ##\angle AOB = \theta## then the distance walked is given by ##s=(\pi-\theta)R##

    I'm not sure how you can get the arclength s easily in terms of rectangular coordinates, but you can get the rowing distance d in terms of ##\theta## using the cosine rule.
    Lessee: looks like it is easier to use y than x - but that's just a question of labels right?
    $$d=\sqrt{2R(r-y)};\; s=\cdots$$... something ... to do dT/dy=0 I'd have to look up the derivative of an arctangent or an arccosine or something like that...

    None of these equations look anything like yours ... so x does not appear to be a rectangular coordinate. So what is it?
    Last edited: Nov 28, 2012
  18. Nov 28, 2012 #17
    I found the length of the chord to be [itex]2\sqrt{3}[/itex]
  19. Nov 28, 2012 #18
    I see your point. I was just trying to do it a different way.
  20. Nov 28, 2012 #19
    The rowing distance is about 3.46.
  21. Nov 28, 2012 #20
    The arc length is 2.09.
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