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Beginner Triangle Problem

  1. Jun 4, 2009 #1
    Beginner "Triangle" Problem

    1. The problem statement, all variables and given/known data

    A ladder 4m long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a speed of 30cm/s, how quickly is the top of the ladder sliding down the wall when the bottom of the ladder is 2m away from the wall?



    2. Relevant equations

    r^2 = x^2 + y^2

    dx/dt = 0.3 m/s

    dr/dt = ?

    dy/dt = 0

    x = 2

    r = 4

    y ~ 3.46

    3. The attempt at a solution

    I made two attempt at the question.

    #1:

    r^2 = x^2 + y^2

    2*r*dr/dt = 2*x*dx/dt + 2*y*dy/dt

    2*4*dr/dt = 2*2*dx/dt + 2*3.46*0

    8*dr/dt = 4*0.3

    > dr/dt = 0.15 m/s <

    #2:

    r^2 = x^2 + (r^2 - x^2)^2

    r^2 = x^2 + (r^4 - 2*r^2*x^2 + x^4)

    I'm not going to type the intermidiate steps, if you want to know how I arrived at an equation please ask.

    dr/dt * [4r^3 - 2r - 4rx^2] = [ 4*x*r^2*dx/dt - 4*x^3*dx/dt - 2*x*dx/dt ]

    dr/dt * [184] = [27.6]

    > dr/dt = 0.15 m/s <

    >.< The answer in the textbook is: sqrt(3)/10 ~ 0.17 m/s

    Any suggestions as to where I went wrong?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 4, 2009 #2

    zcd

    User Avatar

    Re: Beginner "Triangle" Problem

    Given that the formula given is r^2 = x^2 + y^2, you can assume that r is the hypotenuse. The question is asking for rate at which the ladder falls down (dy/dt). Even if a ladder slides down a wall, it is unlikely that the length of the ladder changes at all (dr/dt = 0).
     
  4. Jun 4, 2009 #3
    Re: Beginner "Triangle" Problem

    Ahhh, yes that makes sense. I was confusing dr/dt with dy/dt. Thanks a bunch.
     
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