a lighthouse stands 550 m off a straight shore and the focused beam of its light revolves (at a constant…

Explanation:

Step1: Establish the relationship

Let (x) be the distance from (P) to the point where the beam hits the shore, and (\theta) be the angle between the line from the lighthouse to (P) and the line from the lighthouse to the point on the shore. We know that (\tan\theta=\frac{x}{550}), so (x = 550\tan\theta). Differentiating both sides with respect to time (t), we get (\frac{dx}{dt}=550\sec^{2}\theta\frac{d\theta}{dt}).

The light revolves (4) times per minute. Since one full - revolution is (2\pi) radians, (\frac{d\theta}{dt}=4\times2\pi = 8\pi) rad/min.

When (x = 275), (\tan\theta=\frac{275}{550}=\frac{1}{2}). Then (\sec^{2}\theta=1 + \tan^{2}\theta=1+\frac{1}{4}=\frac{5}{4}).

Step2: Calculate the speed

Substitute (\sec^{2}\theta=\frac{5}{4}) and (\frac{d\theta}{dt}=8\pi) into (\frac{dx}{dt}=550\sec^{2}\theta\frac{d\theta}{dt}).

(\frac{dx}{dt}=550\times\frac{5}{4}\times8\pi)

[ \begin{align*} \frac{dx}{dt}&=550\times10\pi\ &=5500\pi\ &\approx5500\times 3.14159\ &=17278.745\approx17279 \end{align*} ]

For the second part, from (\frac{dx}{dt}=550\sec^{2}\theta\frac{d\theta}{dt}=550(1 + \tan^{2}\theta)\frac{d\theta}{dt}) and (x = 550\tan\theta). As (x) (the distance between (P) and (Q)) decreases (i.e., (Q) is closer to (P)), (\tan\theta=\frac{x}{550}) decreases.

Since (\frac{dx}{dt}=550(1+\tan^{2}\theta)\frac{d\theta}{dt}) and (\frac{d\theta}{dt}) is constant ((8\pi)), when (x) (distance between (P) and (Q)) decreases, (\frac{dx}{dt}) decreases.

Answer:

The speed of the beam along the shore when it strikes (Q) is about (17279) m/min. The correct option for the variation of speed with the distance between (P) and (Q) is C. If (Q) were closer to (P), then the speed of the beam along the shore when it strikes (Q) would be less.