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Last question on integration (and i am stuck)

  1. Jul 24, 2009 #1
    in plan view, the shoreline of a bay may be approximated by a curve f(x) = -x2 + 2x between one headland at x = 0 and another at x = 1km, where (f) is also measured in km. at low tide, the edge of the sea just reaches each headland, following a straight line between them. calculate in km2 the area of the bay uncovered by water at low tide?

    does anyone get this ??

    because i do not, whatsoever, i would try, but dont know where to start either let alone answer it :(
     
  2. jcsd
  3. Jul 24, 2009 #2

    Mark44

    Staff: Mentor

    Have you drawn a graph of the shore? Do you know what a headland is?
     
  4. Jul 24, 2009 #3
    bare with me, i am a diagnosed dyslexic sorry, i know what a head land is, but i can not seem to go anywhere, im not comprehending this info lol
     
  5. Jul 24, 2009 #4

    Mark44

    Staff: Mentor

    OK, then have you drawn a graph of the function f(x) = -x2 + 2x?

    Also, are you sure about the information you've given in this problem? The graph doesn't show any headland at x = 1 km.
     
  6. Jul 24, 2009 #5
    that is what the question says on my sheet mate, im sure i aint dyslexic and it is the bloody lectures that cant write down questions properly haha
     
  7. Jul 24, 2009 #6

    HallsofIvy

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    I would assume that the bay curves from (0,0) to (1,1) so the area sought is between the graphs y= x and y= -x^2+ 2x.

    [tex]\int_0^2 x-(-x^2+ 2x) dx[/tex]

    JakePearson, what is your purpose in telling us in one post that "i am a diagnosed dyslexic" and in the very next "im sure i aint dyslexic". If you are not dyslexic (and if you can spell "dyslexic" you probably aren't!) why tell us that you were diagnosed that way?

    I see nothing wrong with the way the problem is given: there is one "headland" that extends to (0,0) and another that only extends to (1,1).
     
    Last edited: Jul 24, 2009
  8. Jul 24, 2009 #7

    Mark44

    Staff: Mentor

    OK, draw a graph, and assume that the lecturer meant to say that there was a headland at x = 2 km, (and write this down on your work that you hand in), and then work the problem. Other than the misinformation given, it's pretty straightforward.
     
  9. Jul 24, 2009 #8

    HallsofIvy

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    Why assume that? Is there any reason to think that the line between headlands is the x-axis?
     
  10. Jul 25, 2009 #9

    Mark44

    Staff: Mentor

    Yes. The graph of y = -x2 + 2x naturally defines a bay with headlands at x = 0 and x = 2. the point at (1, 1) doesn't look much like a headland to me. In fact, it looks like the the point in the bay that is most inland, if that makes any sense.

    I believe that there is a good chance that OP's lecturer erred in saying that there is a headland at x = 1.
     
  11. Jul 25, 2009 #10
    Here's the graph...

    WHy can't we just integrate the formula -x^2 + 2x between 0 and 1?
     

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