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Math Help - Possible Related Rates Problem w/ Ladder

  1. #1
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    Possible Related Rates Problem w/ Ladder

    Hello everyone,

    I did the following SAT problem correctly using Pythagoras' Theorem and algebra, but I would like to try to do it with calculus. Is it possible to do this problem as a related rates problem?

    Thank you.

    ---

    1. A 25-foot ladder is placed against a vertical wall of a building, with the bottom of the ladder standing on concrete 7 feet from the base of the building. If the top of the ladder slips down 4 feet, then the bottom of the ladder will slide out ___ feet.

    ---

    Let:  y = height of the ladder on wall

     x = distance of the base of the ladder to the wall

    Therefore:

     \frac{dy}{dt} = 4  \frac{dx}{dt} = ?

     x^2 + y^2 = 25

     \frac{d}{dx}(x^2 + y^2 = 25)

     2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0

     2x\frac{dx}{dt} + 8y = 0

     \frac{dx}{dt} = \frac{-4y}{x}

    I am not sure what to substitute as x and y here.
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  2. #2
    MHF Contributor kalagota's Avatar
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    Taguig City, Philippines
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    Quote Originally Posted by scherz0 View Post
    Hello everyone,

    I did the following SAT problem correctly using Pythagoras' Theorem and algebra, but I would like to try to do it with calculus. Is it possible to do this problem as a related rates problem?

    Thank you.

    ---

    1. A 25-foot ladder is placed against a vertical wall of a building, with the bottom of the ladder standing on concrete 7 feet from the base of the building. If the top of the ladder slips down 4 feet, then the bottom of the ladder will slide out ___ feet.

    ---

    Let:  y = height of the ladder on wall

     x = distance of the base of the ladder to the wall

    Therefore:

     \frac{dy}{dt} = 4  \frac{dx}{dt} = ?

     x^2 + y^2 = 25

     \frac{d}{dx}(x^2 + y^2 = 25)

     2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0

     2x\frac{dx}{dt} + 8y = 0

     \frac{dx}{dt} = \frac{-4y}{x}

    I am not sure what to substitute as x and y here.
    you don't have to use it..

    note that the length of the ladder is 25, and the foot of the ladder is 7 units away from the base of the building..

    thus, from the base of the building to the top of the ladder is given by
    y = \sqrt{25^2 - 7^2}

    so, if the ladder slips down 4 units, the distance from the base of the building from the bottom of the ladder is given by
    x =\sqrt{25^2 - (y-4)^2}

    thus, the bottom of the ladder moves away x - 7 units from the base of the building..
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  3. #3
    Senior Member
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    Quote Originally Posted by scherz0 View Post
    Hello everyone,

    I did the following SAT problem correctly using Pythagoras' Theorem and algebra, but I would like to try to do it with calculus. Is it possible to do this problem as a related rates problem?
    I'm pretty sure not. There are no rates here to relate.
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  4. #4
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    All right, thank you!
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