# Derivatives, rates of change (triangle)

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

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

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.

#### Attachments

• 234.png
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Simon Bridge
Homework Helper
That's good: $$\frac{dx}{dt}=y\sec^2\theta\frac{d\theta}{dt}$$