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Posted by on Sunday, March 18, 2012 at 4:56pm.

A 24 foot ladder is leaning against a building. Let x be the distance between the bottom of the ladder abd the building and let theta be the angle between the ladder and the ground. Suppose the bottom of the ladder is sliding away from the wall at a rate of 2 ft/sec. When x=6, how fast is theta (in Radians/sec) changing?

  • Calculus II - , Sunday, March 18, 2012 at 5:14pm

    cosØ = x/24
    -sinØ dØ/dt = (1/24)dx/dt

    when x = 6, opposite side is y
    y^2 + 6^2 = 24^2
    y = √540 and sinØ = √540/24

    dØ/dt = -1/(24sinØ)dx/dt
    = (-1/24)(24/√540)(2) = -2/√540 rad/sec

    simplify as needed

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