a woman standing on a cliff is watching a motorboat through a telescope as the boat approaches the shoreline…

Explanation:

Step1: Define variables

Let $x$ be the horizontal distance of the boat from the shore and $\theta$ be the angle between the telescope - line - of - sight and the horizontal. We know that $\tan\theta=\frac{100}{x}$.

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

Using the chain - rule, $\sec^{2}\theta\frac{d\theta}{dt}=-\frac{100}{x^{2}}\frac{dx}{dt}$.

Step3: Find $\sec^{2}\theta$ when $x = 200$

When $x = 200$, $\tan\theta=\frac{100}{200}=\frac{1}{2}$. Then, using the identity $\sec^{2}\theta=1 + \tan^{2}\theta$, we have $\sec^{2}\theta=1+\left(\frac{1}{2}\right)^{2}=1+\frac{1}{4}=\frac{5}{4}$.

Step4: Substitute known values into the differentiated equation

We are given that $\frac{dx}{dt}=- 30$ (negative because $x$ is decreasing). Substitute $\sec^{2}\theta=\frac{5}{4}$, $x = 200$, and $\frac{dx}{dt}=-30$ into $\sec^{2}\theta\frac{d\theta}{dt}=-\frac{100}{x^{2}}\frac{dx}{dt}$. $\frac{5}{4}\frac{d\theta}{dt}=-\frac{100}{200^{2}}\times(-30)$.

Step5: Solve for $\frac{d\theta}{dt}$

First, simplify the right - hand side: $-\frac{100}{200^{2}}\times(-30)=\frac{100\times30}{200^{2}}=\frac{3000}{40000}=\frac{3}{40}$. Then, solve for $\frac{d\theta}{dt}$: $\frac{d\theta}{dt}=\frac{3}{40}\times\frac{4}{5}=\frac{3}{50}=0.06$ radians/sec.

Answer:

$0.06$ radians/sec