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A ladder 23 feet long leans up against a house.

The bottom of the ladder starts to slip away from the house at 0.16 feet per second.How fast is the tip of the ladder along the side of the house slipping when the ladder is 7.2 feet away from the house? (Round to 3 decimal places.)

ft/sec


Consider the angle the bottom of the ladder makes with the ground.
How fast is the angle changing (in radians) when the ladder is 7.2 feet away from the house?

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  1. distance of foot of ladder from wall ---- x
    distance of top of ladder from ground --- y
    given: dx/dt = .16 ft/sec
    find: dy/dt when x = 7.2 ft

    7.2^2 + y^2 = 23^2
    y = 21.844

    x^2 + y^2 = 23^2
    2x dx/dt + 2y dy/dt = 0
    we know all the values except dy/dt
    2(7.2) + 2(21.844) dy/dt = 0
    dy/dt = -7.2/21.844 ft/sec
    = appr -0.33 ft/sec

    the negative sign tells us the top of the ladder is moving downwards

    second part of problem:
    cos θ = x/23
    x = 23 cosθ
    dx/dt = -23sinθ dθ/dt

    when x = 7.2
    cosθ = 7.2/23
    θ = 71.7572..°
    sinθ = .94973...

    dx/dt = -23sinθ dθ/dt
    solve for dθ/dt

    explain the negative value of dθ/dt

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