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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 0.6 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? (That is, find the angle's rate of change when the bottom of the ladder is 6 ft from the wall.)

Answer :

crushton
1) Set up an equation to model the situation.
    cos(Ф) = x/10

2) Take the derivative/gradient in terms of time (t) to find the related rates.
    d[cos(Ф)]/dt = d[x/10]/dt
    -sin(Ф)*dФ/dt = (1/10)*dx/dt
    dФ/dt = -(dx/dt)/(10*sin(Ф))

3) Substitute in the constants for the moment in time
    Given: dx/dt = 0.6; x=6 -> Ф = cos-1(6/10) = ~0.93rad
    Therefore: dФ/dt = -3/40 rad/sec

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