Calculus

A 10 foot ladder leans against a 20 foot wall. Someone begins pushing the base of the ladder toward the wall at the rate of one foot per second. How quickly is the top of the ladder moving up the wall after 2 seconds?

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  1. Let y be the wall height
    let x be the horizontal distance

    x^2+y^2=100

    2x dx/dt+2y dy/dt=0
    dy/dt=-x dx/dt

    so calculate the x,y positions after 2 seconds, I have no idea the starting position.

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    bobpursley
  2. This problem is the most popular lead-in question to "rate of change" questions for almost any Calculus text I have seen

    Let the ladder be x ft from the wall, and y m up the wall
    we know x^2 + y^2 = 100
    2x(dx/dt) + 2y(dy/dt) = 0

    given: dx/dt = 1 ft/s
    find dy/dt when t = 2 or
    in other words, when x = 2 ft since we know in 2 s it moved 2 ft.
    also when x = 2
    2^2 + y^2 = 100
    y = √96

    2(2)(1) + 2√96(dy/dt) = 0
    dy/dt = -4/(2√96
    = -.204 ft/s

    at that moment the top of the ladder is dropping (note the - sign) at .204 ft/s

    Just realized I read the question as the foot of the ladder moving away from the wall, so just change
    dx/dt to -1,
    the result will change to dy/dt = +.204

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  3. how can a 10 foot ladder lean on a 20 foot wall?

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