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Derivatives, rates of change (triangle)

  1. Dec 8, 2013 #1
    1. 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 [itex]\frac{\pi}{3}[/itex], this angle is decreasing at a rate of -[itex]\frac{\pi}{3}[/itex] rad/min. How fast is the plane traveling at that time?



    2. Relevant equations
    $$tanθ=\frac{opp}{hyp}$$
    3. The attempt at a solution

    First I find the length of the top of the triangle, x. $$tan\frac{\pi}{3}=\frac{x}{5}$$ $$x=5\sqrt{3}$$
    Then I take the derivative.
    $$tanθ=\frac{x}{y}$$ $$sec^2x\frac{dθ}{dt}=\frac{y\frac{dx}{dt}-x\frac{dx}{dt}}{y^2}$$ The y, or altitude, is always constant so $$sec^2x\frac{dθ}{dt}=\frac{y\frac{dx}{dt}}{y^2}$$ $$sec^2\frac{\pi}{3}×-\frac{\pi}{6}×5=\frac{dx}{dt}$$ $$\frac{dx}{dt}=-\frac{10}{3}\pi$$

    The answer from my textbook is $$\frac{10}{9}\pi$$ What did I do wrong? Any help is much appreciated.
     
  2. jcsd
  3. Dec 8, 2013 #2

    Simon Bridge

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    Is the angle of elevation measured from the vertical or the horizontal?

    Check the time-derivative of the equation: ##x=y\tan\theta##
     
  4. Dec 8, 2013 #3
    I checked the time derivative of x=tany but I still get $$\frac{dx}{dt}=sec^2θ×\frac{dθ}{dt}×y$$

    I've also attached a drawing.
     

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  5. Dec 9, 2013 #4

    Simon Bridge

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    That's good: $$\frac{dx}{dt}=y\sec^2\theta\frac{d\theta}{dt}$$

    Now you have your answer.
    Hint:
    ##dx/dt## is the linear speed.
    ##\sec(\pi/3)=2##

    That leaves only the interpretation ... is the angle ##\theta## in that formula the angle of elevation?
    See: http://www.mathwords.com/a/angle_elevation.htm
     
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