Derivatives, rates of change (triangle)

[physics604]

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 $\frac{\pi}{3}$, this angle is decreasing at a rate of -$\frac{\pi}{3}$ 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.

 

[Simon Bridge]

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

Check the time-derivative of the equation: $x=y\tan\theta$

 

[physics604]

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.

 

[Simon Bridge]

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