question you are stationary on the ground and are watching a bird fly horizontally towards you at a rate of…

Explanation:

Step1: Define variables

Let $x$ be the horizontal distance between you and the bird, and $\theta$ be the angle of elevation. We know $\tan\theta=\frac{40}{x}$.

Step2: Differentiate both sides with respect to time $t$

Using the chain - rule, $\sec^{2}\theta\frac{d\theta}{dt}=-\frac{40}{x^{2}}\frac{dx}{dt}$.

Step3: Find $\sec^{2}\theta$ when $x = 170$

First, when $x = 170$, $\tan\theta=\frac{40}{170}=\frac{4}{17}$. Then, $\sec^{2}\theta=1 + \tan^{2}\theta=1+\left(\frac{4}{17}\right)^{2}=1+\frac{16}{289}=\frac{289 + 16}{289}=\frac{305}{289}$.

Step4: Substitute known values

We know that $\frac{dx}{dt}=- 5$ m/s (negative because $x$ is decreasing as the bird moves towards you). Substitute $\sec^{2}\theta=\frac{305}{289}$, $x = 170$, and $\frac{dx}{dt}=-5$ into $\sec^{2}\theta\frac{d\theta}{dt}=-\frac{40}{x^{2}}\frac{dx}{dt}$. $\frac{305}{289}\frac{d\theta}{dt}=-\frac{40}{170^{2}}\times(-5)$. $\frac{305}{289}\frac{d\theta}{dt}=\frac{40\times5}{28900}$. $\frac{d\theta}{dt}=\frac{40\times5}{28900}\times\frac{289}{305}$. $\frac{d\theta}{dt}=\frac{200}{30500}=\frac{2}{305}$ rad/s.

Answer:

$\frac{2}{305}$