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Related Rates: The ladder problem

  1. Sep 29, 2011 #1
    1. The problem statement, all variables and given/known data

    An 18 meter ladder is sliding down a vertical wall at a rate of 2.5 m/s. Find the speed of the lower end of the ladder when it is 12 meters from the wall.

    2. Relevant equations

    Pythagorean Theorem

    3. The attempt at a solution

    Let h = height of the wall
    L = length of the ladder
    b = distance from the wall to the bottom of the ladder

    dh/dt = 2.5 m/s
    db/dt = ? when b = 12

    18^2 = 12^2 + h^2
    h = rad(180)

    L^2 = h^2 + b^2
    18^2 = 2h dh/dt + 2b db/dt
    0 = 2h (2.5) + 2(12) db/dt
    0 = 2[rad(180)] (2.5) + 2(12)(db/dt)

    -2.79 m/s = db/dt

    Could anyone verify that this is right? Please correct me if I'm wrong.
     
    Last edited: Sep 29, 2011
  2. jcsd
  3. Sep 29, 2011 #2
    You need to solve for h at the given instant in time (you can do this easily given the Pythagorean relationship; this should involve no differentials). Also, note that the length of the ladder does not change at all as time progresses. Therefore, [itex]\frac{dL}{dt}[/itex] will be zero.

    Also, there is an error in your setup. Is [itex]\frac{dh}{dt}[/itex] increasing or decreasing?
     
  4. Sep 29, 2011 #3
    I suppose dh/dt would be decreasing, making my final answer positive?
     
  5. Sep 29, 2011 #4
    Yup! Draw a picture if it helps, but this should be evident from daily experiences. What the positive answer is saying, of course, is that the length of the bottom leg of the triangle is increasing.
     
  6. Sep 29, 2011 #5
    Right, thank you for your help! I need to be more aware of positives and negatives.
     
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