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Rate of Change of theta in ladder to wall.

  1. Oct 13, 2008 #1
    1. The problem statement, all variables and given/known data
    A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.3 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 6 ft from the wall? Evaluate your answer numerically.

    2. Relevant equations

    a^2+b^2=c^2 ?
    Theta(dot) is change of time respect to theta.

    3. The attempt at a solution

    In class he went over finding the rate of theta, getting the equation Theta(dot)=X/(10cos(45) I tried plugging in 6 for X, I just really dont know what to do here.
  2. jcsd
  3. Oct 13, 2008 #2


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    I would suggest that instead of using a2+ b2= c2, which does not involve an angle, you use [itex]a= bcos(\theta)[/itex] where a is the "hypotenuse" (the length of the ladder) and a is the "near side" (the distance from the foot of the ladder to the wall).
  4. Jan 3, 2012 #3
    theta=tan(y/x) right
    then what the formula to find the change in theta at any given point?
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