Question

1. a man walks along a straight path at a speed of 5 $\frac{ft}{s}$. a searchlight is located on the ground 20 feet from the path and is kept focused on the man. at what rate is the searchlight rotating when the man is 15 feet from the point on the path closest to the searchlight? (angle $\theta$ is measured in radians for the units). (description: a man is drawn on a straight path labeled $x$. a segment of length 20 is drawn at a right angle from the straight path. another segment connects the segment drawn at a right - angle to the straight path where the man is, the angle is labled $\theta$.)

Explanation:

Step1: Establish a trigonometric relationship

Let $x$ be the distance of the man from the point on the path closest to the search - light. We know that $\tan\theta=\frac{x}{20}$, where $\theta$ is the angle of the search - light.

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

Using the chain - rule, we have $\sec^{2}\theta\frac{d\theta}{dt}=\frac{1}{20}\frac{dx}{dt}$.

Step3: Find $\sec^{2}\theta$ when $x = 15$

First, when $x = 15$, by the Pythagorean theorem, the hypotenuse of the right - triangle formed is $r=\sqrt{15^{2}+20^{2}}=\sqrt{225 + 400}=\sqrt{625}=25$. Then $\cos\theta=\frac{20}{25}=\frac{4}{5}$, and $\sec\theta=\frac{5}{4}$, so $\sec^{2}\theta=\frac{25}{16}$.

Step4: Substitute known values to find $\frac{d\theta}{dt}$

We are given that $\frac{dx}{dt}=5$ ft/s. Substituting $\sec^{2}\theta=\frac{25}{16}$ and $\frac{dx}{dt}=5$ into $\sec^{2}\theta\frac{d\theta}{dt}=\frac{1}{20}\frac{dx}{dt}$, we get $\frac{25}{16}\frac{d\theta}{dt}=\frac{1}{20}\times5$.
Solve for $\frac{d\theta}{dt}$:
\[

$$\begin{align*} \frac{25}{16}\frac{d\theta}{dt}&=\frac{1}{4}\\ \frac{d\theta}{dt}&=\frac{1}{4}\times\frac{16}{25}\\ \frac{d\theta}{dt}&=\frac{4}{25}\text{ rad/s} \end{align*}$$

\]

Answer:

$\frac{4}{25}$ rad/s