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Back again, need help one more time.

  1. Sep 26, 2004 #1
    After working hours on my first assignment LOL (i really want to do good in school) i hav ebecame stumpted.
    The question is:
    A ladder, 2 meters long, is standing flat against a vertical wall. The foot of the ladder is pulled away from the wall. As this happens, the top of the ladder stays in contact with the wall. The process stops when all of the ladder is on the ground. What is the locus of the mid-point of the ladder? (I.e., give a function which describes what happens to the mid-point of the ladder.) im not sure how to approach this problem so im posting here, im going to work on it for a while then go to bed. If someone could help me out that would be great. thx u
  2. jcsd
  3. Sep 26, 2004 #2
    damn im rusty, i thnk i got it... y= sqr(2^2 -x^2)/2 correct me if im wrong
  4. Sep 26, 2004 #3
    nm im wrong
  5. Sep 26, 2004 #4
    Hang on for 10 minutes. Hints are coming soon.
  6. Sep 26, 2004 #5
    cool ill stay up for a bit, im still working on it !
  7. Sep 26, 2004 #6
    Upper triangle :
    [tex] x= cos \theta[/tex]
    Lower triangle:
    [tex]y = sin \theta[/tex]
    Use trigo identity.

    Attached Files:

  8. Sep 26, 2004 #7
    hmm ill think about this for a minute, thx for help. hmm what do you mean by use trig identity
    Last edited: Sep 26, 2004
  9. Sep 26, 2004 #8
    This will be the answer actually.
  10. Sep 26, 2004 #9
    wouldn't the answer be y = sqr(1-cos^2(x))

    im really really sick right now... so im going to bed, ill check this out in the morning, i hope i am well for school :| thanks
  11. Sep 26, 2004 #10
    No. the answer is [tex]x^2+y^2 =1[/tex]
  12. Jan 29, 2005 #11
    yes dats ryt its a cercle (r= 1m)
  13. Jan 29, 2005 #12


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    Set up your coordinate system so that the wall of the house is the positive y axis and the ground is the positive x axis. The coordinates of the upper end of the ladder are (0, Y) and the coordinates of the lower end are (X,0) where X2+ Y2= 2. The midpoint of the ladder is at (X/2, Y/2). Letting x= X/2, X= 2x so
    X2= 4x2 and let y= Y/2 so Y= 2x and Y2= 4y2. Then X2+ Y2= 4x2+ 4y2= 2 so the locus of the midpoint is given by x2+ y2= (1/2) (x and y both non-negative). That is,a the midpoint moves on a circle (actually the portion in the first quadrant) with center (0,0) and radius [itex]\frac{\sqrt{2}}{2}[/itex].
  14. Jan 29, 2005 #13


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    i guess halls of ivy meant X^2 + Y^2 = 4?