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Math Calculus

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The foot of a 30-foot ladder is being pulled away from a vertical wall at a rate of 1 foot per minute. When the top of the ladder is 6 feet from the ground, at what rate is the top of the ladder moving down the wall?

  • Math Calculus - ,

    Let the foot of the ladder be x ft from the wall, and let the top of the ladder by y ft above the ground

    x^2 + y^2 = 30^2
    2x dx/dt + 2y dy/dt = 0
    x dx/dt + y dy/dt = 0

    when y = 6
    x^2 + 36 = 900
    x = √864

    given: dx/dt = 1 , x = √864 , y = 6

    √864(1) + 6dy/dt = 0
    dy/dt = -√864/6 ft/min or appr. -4.9 ft/min

    (the negative sign shows that y is decreasing or the ladder is moving down the wall)

  • Math Calculus - ,

    Reini - How did you ended up getting x dx/dt + y dy/dt=0?

    my professor said something with
    a^2 + b^2 = c^2
    then differentiate it so 2a(da/dt) + 2b(db/dt) = 2c(dc/dt)

    so a^2+b^2=c^2
    a^2 + 6^2 = 30^2
    so a=sqrt(864)

    then I differentiate
    let da/dt = 1
    b = 6
    c=30

    2a(da/dt)+2b(db/dt)=2c(dc/dt)
    2sqrt(864)(1)+2(6)(db/dt)=2(30)(0)
    =24sqrt(6)+12db/dt = 0

    but how do i sold for db/dt?

    is this way correct?

    thanks

  • Math Calculus - ,

    that is exactly what I did, except I used x and y instead of a and b
    remember that your c is a constant, so its derivative is zero (I used the actual 30^2)

    I had defined x as the distance along the ground, and we were told the base of the ladder moved out"at a rate of 1 foot per minute"
    this is the same as saying dx/dt = 1

    from your
    24sqrt(6)+12db/dt = 0
    12db/dt = -24√6
    db/dt = -24√6/12 = -4.9 , the same answer as I had

    so yes, you are correct

  • Math Calculus - ,

    Thank you very much for your help :)

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