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After blast-off, a space shuttle climbs vertically and a radar-tracking dish, located 1000 meters from the launch pad, follows the shuttle. How fast is the radar dish revolving 10 sec after blast-off if at that time the velocity of the shuttle is 100 m/sec and the shuttle is 500 meters above the ground?

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    h = height = 500 + 100 t
    b = base = 1000
    A = elevation angle. we want dA/dt

    tan A = h/b = 100(5+t) / 1000 = .5 +.1 t

    (1 /sec^2 A) dA/dt = 0 + .1

    dA/dt = .1 sec^2 A

    at this time ratio of b to h is 2/1
    so hypotenuse is sqrt 5
    cos A = 2/sqrt 5
    sec A = sqrt 5/2
    sec^2 A = 5/4
    so
    dA/dt = .1 (5.4) = .54 radians/second
    or
    31 degrees per second
    sec A = 1/cos A = 1/(1000/(1000^2

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