a lighthouse stands 475 m off a straight shore and the focused beam of its light revolves (at a constant…

Explanation:

Step1: Define variables

Let $x$ be the distance along the shore from $P$ to the point where the beam hits the shore. Let $\theta$ be the angle between the line from the lighthouse to $P$ and the line from the lighthouse to the point where the beam hits the shore. We know that $\tan\theta=\frac{x}{475}$, and $\frac{d\theta}{dt} = 4\times2\pi$ radians per minute (since it makes 4 revolutions per minute).

Step2: Differentiate the tangent - relation

Differentiate $\tan\theta=\frac{x}{475}$ with respect to time $t$. Using the chain - rule, we have $\sec^{2}\theta\frac{d\theta}{dt}=\frac{1}{475}\frac{dx}{dt}$.

Step3: Find $\sec^{2}\theta$ at the given point

When the beam is at point $Q$ which is $x = 175$ m from $P$, $\tan\theta=\frac{175}{475}=\frac{7}{19}$. Then, using the identity $\sec^{2}\theta=1 + \tan^{2}\theta$, we get $\sec^{2}\theta=1+\left(\frac{7}{19}\right)^{2}=1+\frac{49}{361}=\frac{361 + 49}{361}=\frac{410}{361}$.

Step4: Solve for $\frac{dx}{dt}$

We know that $\frac{d\theta}{dt}=8\pi$ radians per minute. Substituting $\sec^{2}\theta=\frac{410}{361}$, $\frac{d\theta}{dt}=8\pi$ into $\sec^{2}\theta\frac{d\theta}{dt}=\frac{1}{475}\frac{dx}{dt}$, we have $\frac{410}{361}\times8\pi=\frac{1}{475}\frac{dx}{dt}$. Then $\frac{dx}{dt}=475\times\frac{410}{361}\times8\pi$. First, $475\times\frac{410}{361}=\frac{475\times410}{361}=\frac{194750}{361}\approx539.47$. Then $\frac{dx}{dt}=539.47\times8\pi\approx539.47\times25.133\approx13557.7$. Rounding to the nearest integer, $\frac{dx}{dt}\approx13558$ m/min.

Answer:

$13558$ m/min