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Posted by on Friday, June 22, 2012 at 8:12am.

Suppose a ladder of length 36 feet rests against a wall, and we pull the base of the ladder horizontally away from the wall. For a given distance x that we pull the ladder horizontally, let g(x) be the height of the ladder on the wall, under the assumption that the other end of the ladder always contacts the wall. We know from Pythagorean theorem that for every x:
x2 + [g(x)]2 = 36^2

At what rate is the height changing when
x = 8?
When the ladder is 6 feet from the ground (g(x) = 6)?

  • Calculus - , Friday, June 22, 2012 at 8:23am

    for easier typing, I will let y = g(x)

    then x^2 + y^2 = 1296
    2x dx/dt + 2y dy/dt = 0
    dy/dx = -x dx/dt / y

    when x=8, y = √1232 or appr 35.1
    dy/dx = (-8/√1232)(dx/dt)

    for the second part, let y = 6, then x = √1260 or appr 35.5
    etc

    are you sure that you were not given the rate at which the ladder was pulled away ?

  • Calculus - , Friday, June 22, 2012 at 8:46am

    The first part of the problem was find a formula for g'(x) in terms of
    x and g(x)?

    I got -x/g(x)

  • Calculus - , Friday, June 22, 2012 at 9:00am

    At what point are you in the study of related rates?
    Are you not differentiating with respect to "time" ?

    If not, then you would simply have
    x^2 + y^2 = 1296
    2x + 2y dy/dx = 0
    y dy/dx = -x
    dy/dx = -x/y or -x/g(x), so that agrees with your answer.

  • Calculus - , Friday, June 22, 2012 at 9:07am

    I figured it out, for each part I just plug in the given x and the g(x) answer you gave me to the formula -x/g(x) and simplify.


    Thanks for your help!!

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