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Math Help - Related Rates

  1. #1
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    Related Rates

    A girl flies a kite at a height of 300 ft, the wind carrying the kite horizontally away from her at a rate of 25 ft/sec. How fast must she let out the string when the kite is 500 ft away from her?

    i set up a triangle

    y(t) = 300, y'(t)= 0

    x(t) = 400 (since its a 3-4-5 triangle) x'(t) = 25

    s(t) = 500 s'(t) = ? <--- hypotenuse

    so the cos(theta) = 300/500 = .927295 radians

    now, how can i get s'?
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  2. #2
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    Angry

    That's not the way to do it. As it is, you are analysing it at the condition when the lenght of the string is already 500 ft.
    You must analyze it in the conditions before the length of the string is 500 ft.

    Here it is assumed that the string is in a straight line, like the hypotenuse of a right triangle. [In reality, it is not--it sags a lot.]

    Your right triangle should be
    ----x at the top, horizontal leg.
    ----y=300 as the vertical leg
    ----z as the hypotenuse

    So,
    z = sqrt(x^2 +300^2)
    z = sqrt(x^2 +90,000)

    You are to find dz/dt when z = 500 ft.

    z = sqrt(x^2 +90,000)
    Differentiate both sides with respect to time t,
    dz/dt = (1/2)(x^2 +90,000)^(-1/2) *2x *dx/dt
    dz/dt = x (dx/dt) / sqrt(x^2 +90,000) --------------(i)

    When z = 500ft, what is dz/dt?
    You are solving for that.

    When z = 500ft, what is dx/dt?
    dx/dt is always 25 ft/sec anytime.

    When z = 500ft, what is x?
    z = sqrt(x^2 +90,000)
    500 = sqrt(x^2 +90,000)
    Square both sides,
    250,000 = x^2 +90,000
    x^2 = 250,000 -90,000
    x^2 = 160,000
    x = 400 ft
    [Your 3-4-5 triangle.]

    Hence, at z=500ft,
    dz/dt = x (dx/dt) / sqrt(x^2 +90,000) --------------(i)
    dz/dt = 400(25) / sqrt(400^2 +90,000)
    dz/dt = 400(25) / sqrt(250,000)
    dz/dt = 400(25) /500
    dz/dt = 20 ft/sec

    Therefore, she must let the string out at the rate of 20 feet per second at that time. -----------------answer.
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  3. #3
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    Hello, mistykz!

    A girl flies a kite at a height of 300 ft,
    the wind carrying the kite horizontally away from her at a rate of 25 ft/sec.
    How fast must she let out the string when the kite is 500 ft away from her?
    I wouldn't introduce trigonometry . . .
    Code:
          B        x        C
          * - - - - - - - - * →
          |              *
          |           *
      300 |        * S
          |     *
          |  *
        A * - - - - - - - - - - - - -

    The girl is at A.
    The kite is at height of 300 feet: AB = 300
    . . and is moving from point C at the rate of \frac{dx}{dt} = 25 ft/sec.
    The string has length: S \,=\,AC

    Pythagorus says: . S^2\:=\:x^2+300^2 .[1]

    Differentiate with respect to time: . 2S\left(\frac{dS}{dt}\right) \:=\:2x\left(\frac{dx}{dt}\right)

    . . Hence, we have: . \frac{dS}{dt} \:=\:\frac{x}{S}\cdot\frac{dx}{dt} .[2]


    When S = 500, [1] becomes: . 500^2 \:=\:x^2+300^2\quad\Rightarrow\quad x \:=\:400

    Substitute into [2]: . \frac{dS}{dt} \:=\:\frac{400}{500}(25) \:=\:20 ft/sec

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