Calculus

posted by Hailey

An airplane flys 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.

  1. Reiny

    I made a sketch showing the distance covered by the plane after it passed over the kangaroo as x miles
    let the angle of elevation be Ø

    I have :
    tanØ = 2/x
    xtanØ = 2
    x sec^2 Ø dØ/dt + tanØ dx/dt = 0

    when the distance between the kangaroo and the plance is 3
    x^2 + 2^2 = 3^2
    x = √5

    when x = √5 , dx/dt = 600, tanØ = 2/√5, and sec^2 Ø = 49/5
    √5(49/5) dØ/dt + (2/√5)(600) = 0
    dØ/dt = (-1200/√5)/(49√5/5)
    = -1200/49 radians/hr
    = -20/49 radians/min

    notice the angle is decreasing
    check my arithmetic, I should write it out on paper first.

  2. David Kelman

    tan(Θ) = 2/x
    dx/dt = -600
    y = 2

    Start with tan(Θ) = 2/x.Taking the derivative leaves you with:

    sec^2(Θ)*dΘ/dt = -2/x^2*dx/dt

    Solve for dΘ/dt.

    dΘ/dt = -2/x^2 * dx/dt * 1/sec^2(Θ)

    when the distance from the plane to the kangaroo is 3, x = sqrt(5); so Θ = arcsin(2/3).

    Use the equation above and plug in dx/dt, x, and theta, then divide by 60 to get units to rads/min.

    the answer should be about 2.22222 rad/min.

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