A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per se

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A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second. (a) What is the velocity of the top of the ladder when the base is given below

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Thu Thảo 4 years 2021-08-10T13:40:09+00:00 1 Answers 45 views 0

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  1. Cherry
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    2021-08-10T13:41:19+00:00
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    This question is incomplete, the complete question is;

    A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second. (a) What is the velocity of the top of the ladder when the base is given below

    7 feet away from the wall

    15 feet away from the wall

    20 feet away from the wall

    Answer:

    – When x is 7 feet away from the wall, \frac{dr}{dt} = -7/12 ft/s

    – When x is 15 feet away from the wall, \frac{dr}{dt} = -3/2 ft/s

    – When x is 20 feet away from the wall, \frac{dr}{dt} = -8/3 ft/s

    Explanation:

    Given the data in the question;

    Let rft and xft represent the distance of the top of ladder from the ground and the distance of the bottom of ladder from the wall respectively.

    Now using the Pythagoras Theorem ( right angled triangle ).

    (hypotenuse)² = ( adjacent )² + ( opposite )²

    so

    r² = ( 25 )² – x²

    r² = 625 – x² ——- let this be equation 1

    now, we differentiate with respect to t

    2r \frac{dr}{dt} = 0 – 2x \frac{dx}{dt}

    r \frac{dr}{dt} = -x \frac{dx}{dt}

    \frac{dr}{dt} = -x \frac{dx}{dt} × 1/r

    \frac{dr}{dt} = -x/r \frac{dx}{dt}

    Now, from equation, r² = 625 – x²

    r = √( 625 – x² )

    Then

    \frac{dr}{dt} = -x / √( 625 – x² ) \frac{dx}{dt}

    given that, The base of the ladder is pulled away from the wall at a rate of 2 feet per second i.e \frac{dx}{dt} = 2 ft/s

    so;

    \frac{dr}{dt} = ( -x / √( 625 – x² ) ) × 2

    \frac{dr}{dt} = -2x / √( 625 – x² )

    ∴

    When x is 7 feet away from the wall;

    \frac{dr}{dt} = -2(7) / √( 625 – (7)² )

    \frac{dr}{dt} = -14 / √( 625 – 49 )

    \frac{dr}{dt} = -14 / √576

    \frac{dr}{dt} = -14 / 24

    we simplify

    \frac{dr}{dt} = -7/12 ft/s

    Therefore, When x is 7 feet away from the wall, \frac{dr}{dt} = -7/12 ft/s

    When x is 15 feet away from the wall

    \frac{dr}{dt} = -2(15) / √( 625 – (15)² )

    \frac{dr}{dt} = -30 / √( 625 – 225 )

    \frac{dr}{dt} = -30 / √400

    \frac{dr}{dt} = -30 / 20

    we simplify

    \frac{dr}{dt} = -3/2 ft/s

    Therefore, When x is 15 feet away from the wall, \frac{dr}{dt} = -3/2 ft/s

    When x is 20 feet away from the wall

    \frac{dr}{dt} = -2(20) / √( 625 – (20)² )

    \frac{dr}{dt} = -40 / √( 625 – 400 )

    \frac{dr}{dt} = -40 / √225

    \frac{dr}{dt} = -40 / 15

    we simplify

    \frac{dr}{dt} = -8/3 ft/s

    Therefore, When x is 20 feet away from the wall, \frac{dr}{dt} = -8/3 ft/s

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