Calculus some1 plz!!!!

A plane flies horizontally at an altitude of 5 km and passes directly over a tracking telescope on the ground. When the angle of elevation is π/3, this angle is decreasing at a rate of π/4 rad/min. How fast is the plane traveling at that time? (Round your answer to two decimal places.)

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  1. x = 12
    x' = -13/10
    y' = ? when x = 4
    m = 2
    y = ?

    Use proportiions.

    12 / 2 = 4 / y
    6 = 4 / y
    2/3 = y

    12 / x = 4 / y

    Differentiating both sides with respect to time, we get:

    -12 / x^2 * x' = -4 / y^2 * y'
    -12 / 4^2 * x' = -4 / (2/3)^2 * y'
    -12 / 16 * x' = -9 y'
    -3/4 * -13/10 = -9 y'
    39 / 40 = -9 y'
    -39 / 360 = y'
    -0.11 m/s = y'

    --------------------------------------…

    y = 5
    a = pi/4
    a' = -pi/3
    x' = ? when a = pi/4
    x = ?

    Let's first find x.

    tan(a) = opposite / adjacent
    tan(a) = y / x
    tan(pi/4) = 5 / x
    1 = 5/x
    5 = x

    Now we can rewrite as:

    tan(a) = 5 / x

    Now let's differentiatite both sides with respect to time. Doing so gives:

    a' * sec^2(a) = -5 / x^2 * x'
    -pi/3 * (1+tan^2a) = -5 / 25 * x'
    -pi/3 * (1 + tan^2(pi/4)) = -1/5 * x'
    -pi/3 * 2 = -1/5 * x'
    -2pi/3 * -5 = x'
    10pi/3 = x'
    10.47 km/min = x'

    Hope this helped.

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  2. sadly, it didn't work :( thanks for trying

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  3. x=h*cot(θ)
    dx/dθ=-h*csc²(θ)
    Given
    h=5 km
    dθ/dt = -π/4
    θ=π/3
    dx/dt=(dx/dθ)*d(θ/dt)
    =-h*csc²(θ)*dθ/dt
    =-5csc²(π/3)*(-π/4)
    =-5*(√3)/2*(-π/4)
    =1.083π/min.
    =3.40 km/min (not a fast plane!)

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