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Algebra questions

  1. Aug 17, 2008 #1
    1. The problem statement, all variables and given/known data
    a ladder (L meters) makes a right angled triangle against the wall, while the ladder slips, a person, Bob, climbs n up the ladder. So, the distance bob is climbing is n, and his distance from the top of the ladder is (L-n). The height of the ladder can be expressed as b+y, where y is the height of Bob on the ladder. The distance of the base of the ladder from the wall is x+a, where a is bob's distance from the base of the ladder. I need to express the ladder slipping against the wall while bob climbs n up the ladder. Express it using x,y,n & L.


    2. Relevant equations
    c2 = a2 + b2


    3. The attempt at a solution


    (L-n)/n = b/y , b = √(L-n)² - x²)

    (L/n) - 1 = √((L-n)^2 - x^2) / y

    ((L/n) - 1)² = ((L-n)² - x²) / y²

    (L/n)² + 1 - (2L/n) = ((L² + n² - 2Ln) - x²) / y²


    can someone solve it or explain to me what the above means.
    please and thank you~~
     

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    Last edited: Aug 17, 2008
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  3. Aug 17, 2008 #2

    HallsofIvy

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    What do you mean by "express the ladder"? Do you mean give his height as a function of distance he has climbed along the ladder? You also say that the ladder is slipping but say nothing about how fast it is slipping. Please state exactly what your problem says.
     
  4. Aug 17, 2008 #3
    the general equation for Bob's position as a ladder slides down the wall, given that Bob climbs n up the ladder.

    the rate at which the ladder is stated isnt specified.

    just express the equation in terms on x, y, n & L

    what i have as working out is something i came up with which doesnt really make sense to me anymore...
    i worked from: as n increases, y should increase as well
    dont know if thats correct.
     
  5. Aug 17, 2008 #4

    HallsofIvy

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    It should be obvious that Bob's position (height above the ground and distance from the wall) will depend upon how fast the ladder is sliding on the wall. Can you just put in another variable for that?
     
  6. Aug 17, 2008 #5
    nope, thats all they gave to work with

    a previous question talked about his path as a circle, but thats when bob was at 2.5m on the ladder.
     
  7. Aug 17, 2008 #6

    HallsofIvy

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    If you want the position as a function of the distance from the wall to the foot of the ladder, that is doable:

    Since let A be the distance from the wall to the foot of the ladder. Then the height of the ladder, B, is given by the Pythagorean theorem: [itex]B= \sqrt{L^2- A^2}[/itex].
    Now, the triangle given by x and y, his distance from the wall and distance above the ground, is similar to the triangle formed by the ladder:
    x/n= A/L so x= An/L and y/n= B/L so y= Bn/L= [itex]n\sqrt{L^2- A^2}/L[/itex]
     
  8. Aug 17, 2008 #7
    the equation must b expressed using x,y,n,L
    and the ladder slipping while bob moves n up or down must be taken into account

    i was thinking if his path could be modelled as a circle, and any other distance on the ladder becomes an ellipse model (if bob is at 3m on the ladder)

    maybe, we could use the (x+h)2/a2 + (y+k)2/b2 = 1
    (x+a)2 + (b+y)2 = L2 (the ladder)

    a = √(n2 - y2)
    b = √((L-n)2 - x2)

    sub those in,
    (x + √(n² - y²) )2 + ( √((L-n)² - x²) + y )2 = L2


    (its yr 12 maths by the way, so it cant be overly complicated, no? lol i cant get it :S)

    wait...your working out might be right, if we rearrange to get A = (Lx/n)

    y = n√(L² - (Lx/n)²) / L

    (yL/n)² + (xL/n)² = L²
    but what would this mean?
    y and x is dependent on bob's position on the ladder,
    (the ladder is 5 m but were not allowed to use that)
    when bob is at the top of the ladder, y² + x² = 25 ?? (but this equation is for when bob is 2.5 m on the ladder)

    i just confused you even more havent i?

    wat do you think?
     
    Last edited: Aug 17, 2008
  9. Aug 18, 2008 #8
    thnx newayz i think ill just do it that way~~
    thank you!!~


    what about an expression in terms of t, for the speed of the foot of the ladder when the top of the ladder is slipping at a constant rate of 4m/s?

    the ladder is still 5m and its height against the wall is 4m.

    i got v(x) = -4t
    or
    (using the chain rule)
    dx/dt = - [4t √(25-(4-2t2)2)] / [4-2t2]

    i let x = √(25-y2)

    i found the speed equation for the top of the ladder to be v(y) = 4t
    then antidifferentiated to get the position which is : y = 4-2t2 , i used this to get x interms of t for the above dx/dt

    is this correct?
     
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