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One end of a long 26 foot ladder is on the floor and the other end rests on a verical wall the bottom end of the ladder is pulled away from the wall qt a rate of 4 feet per second. how fast is the angle formed by the ladder and the floor changing when the ladder is 10 feet away from the wall

  • calculus - related rates? - ,

    "bepper" is not a subject I am familiar with.

    Let x be the distance of the bottom from the wall.

    x = (26 ft)*sin(theta)

    dx/dt = 26 cos(theta)* d(theta)/dt = 4 ft/s
    Solve for d(theta)/dt. It will be in radians per second.

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