Post a New Question

Calculus II

posted by on .

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 - ,

    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

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question