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2 September, 04:47

An airplane in australia is flying at a constant altitude of 2 miles and a constant speed of 600 miles per hour on a straight course that will take it directly over a kangaroo on the ground. how fast is the angle of elevation of the kangaroo's line of sight increasing when the distance from the kangaroo to the plane is 3 miles? give your answer in radians per minute.

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  1. 2 September, 08:41
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    We model the problem as a rectangle triangle.

    We have on the one hand:

    hypotenuse = z = 3

    On the other hand:

    vertical side = y = 2

    The speed of the plane is:

    dz / dt = 600

    The flight height is constant, then:

    dy / dt = 0

    The angle with the horizontal will be:

    sin (θ) = 2/3

    θ = arcsin (2/3) = 0.729728 radians

    On the other hand:

    sin (θ) = y / z

    z sin (θ) = y

    We derive both sides with respect to time:

    dz / dt sin (θ) + z cos (θ) dθ / dt = dy / dt

    Substituting values:

    (600) (2/3) + (3) cos (0.729728) dθ / dt = 0

    Rewriting:

    400 + [3 / cos (0.729728) ] dθ / dt = 0

    400 + 4.0249 dθ / dt = 0

    We cleared dθ / dt:

    dθ / dt = - 98.38 radians / hour

    = - 99.38 / 60 radians / min

    = - 1.66 radians / min = 2PI - 1.66 radians / min

    = 4.62 radians / min

    Answer:

    The angle of elevation of the Kangaroo's line of sight is increasing at:

    = 4.62 radians / min
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