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Applying inverse trig function

  1. Sep 16, 2009 #1
    Hey people, first time posting. This isn't really homework, but I want to know how to apply inverse trig functions. So, while looking over some problems in my textbook, I came across the following, which I really have no idea how to do.

    1. The problem statement, all variables and given/known data
    A ladder 10ft long leans against a wall. The bottom of the ladder (on the floor) slides away from the wall at 2ft/sec. How fast is the angle between the ladder & the wall changing when the bottom of the ladder is 6ft from the base of the wall?

    2. Relevant equations

    3. The attempt at a solution
    Since the ladder is 10ft (hypothenuse), the distance from the ladder to wall is 6ft (opposite), then I can figure the wall is effectively 8ft (adjacent).

    I assume the angle [tex]\Theta[/tex] can be figured with any inverse trig function.

    I know the answer (1/4 rad/s), but that's not really important. I want to be able to understand how to approach this type of problem.
    Last edited: Sep 16, 2009
  2. jcsd
  3. Sep 16, 2009 #2


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    Since the triangle stays in contact with the wall, you can say that [tex]x^{2}+y^{2}=10^{2}[/tex] at all times, and [tex]x=10\cos\theta, y=10\sin\theta[/tex].
  4. Sep 16, 2009 #3


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    This is a "related rates" problem. The idea is to use the geometry to get an equation in terms of the relevant variables then differentiate with respect to time. That gives the related rates equation in which you substitute the instantaneous values (the 6 and 8).

    You have x^2 + y2 = 100. You are given information about x and want information about theta so it would be good to get the y out and the theta in to your equation. In your setup

    [tex] y = 10 \cos(\theta)[/tex]

    So substitute that in to get an equation with x and theta. Differentiate it implicitly remembering that everything is a function of t. Then substitute what you know and solve for what you want to know. :-)
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