Derivatives, rates of change (triangle)

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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?



Homework Equations


$$tanθ=\frac{opp}{hyp}$$

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.
 

Answers and Replies

  • #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##
 
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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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  • #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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