Question

a plane flies horizontally at an altitude of 7 km and passes directly over a tracking telescope on the ground. when the angle of elevation is $\frac{pi}{4}$, this angle is decreasing at a rate of $\frac{pi}{4}$ rad/min. how fast is the plane traveling (in km/min) at that time? (round your answer to two decimal places.)

Explanation:

Step1: Establish a trigonometric relation

Let $x$ be the horizontal distance of the plane from the point directly above the telescope and $h = 7$ km be the altitude of the plane. We know that $\tan\theta=\frac{h}{x}$, since $h = 7$, we have $\tan\theta=\frac{7}{x}$, which can be rewritten as $x = \frac{7}{\tan\theta}=7\cot\theta$.

Step2: Differentiate with respect to time $t$

Using the chain - rule, $\frac{dx}{dt}=- 7\csc^{2}\theta\frac{d\theta}{dt}$.

Step3: Substitute the given values

We are given that $\theta=\frac{\pi}{4}$ and $\frac{d\theta}{dt}=-\frac{\pi}{4}$ rad/min. When $\theta = \frac{\pi}{4}$, $\csc\theta=\sqrt{2}$. Substitute $\theta=\frac{\pi}{4}$ and $\frac{d\theta}{dt}=-\frac{\pi}{4}$ into the equation $\frac{dx}{dt}=-7\csc^{2}\theta\frac{d\theta}{dt}$.
$\csc^{2}\theta = (\sqrt{2})^{2}=2$. Then $\frac{dx}{dt}=-7\times2\times(-\frac{\pi}{4})=\frac{7\pi}{2}\approx10.99$ km/min.

Answer:

$10.99$