A kite 100 ft above the ground moves horizontally at a speed of 11 ft/s. At what rate is the angle (in radians) between the string and the h

Question

A kite 100 ft above the ground moves horizontally at a speed of 11 ft/s. At what rate is the angle (in radians) between the string and the horizontal decreasing when 200 ft of string have been let out?

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Bình An 4 weeks 2021-08-30T16:00:11+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-08-30T16:01:42+00:00

    Answer:

    -2.26×10^-4 radians

    Explanation:

    The solution involves a right angle triangle

    Length is z while the horizontal is the height x

    X^2+ 100^2=z^2

    Taking the derivatives

    2x(dx/dt)=Z^2(dz/dt)

    Specific moments = Z= 200 ,X= 100sqrt3 and dx/dt= 11

    dz/dt= 1100sqrt3/200 = 9.53

    Sin a= 100/a

    Taking derivatives in terms of t

    Cos a(da/dt)=100/z^2 dz/dt

    a= 30°

    Cos (30°)da/dt= (-100/40000×9.5)

    a= -2.26×10^-4radians

    0
    2021-08-30T16:01:54+00:00

    Answer:

    0.11 rad/s

    Explanation:

    Given  that,

    kite is 100 ft high

    and moves horizontally at 7 ft/s

    Total string let out =200 ft

    String length(l), vertical(y) & Horizontal(x) distance of kite will form a right angle triangle

    L^2 = y^2 + x^2

    differentiate both side

    = 2y\frac{dy}{dt}  + 2x\frac{dx}{dt} \\y\frac{dy}{dt} = -x\frac{dx}{dt} \\100\frac{dy}{dt}  = \sqrt{3} * 100 * \frac{dx}{dt} \\\frac{dy}{dt} = 11\sqrt{3}

    Now Lcosθ = x

    diferentiate

    Lsin\theta * \frac{d\theta }{dt} = \frac{dx}{dt} \\200 * \frac{100}{200} * \frac{d\theta}{dt}  = 11\\\frac{d\theta }{dt }  = 0.11 rad/s

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