A high-speed motor torpedo boat is moving parallel to a straight shore line at 40 miles per hour (2/3 miles per minute), 1 and 1/2 miles from the shore, and is followed by a search-light beam which is trained on the boat from a station half a mile back from the shore. At what rate in radians per minute must the beam turn in order to follow the boat just as the boat passes directly opposite the station?

So, $$\operatorname{tan} \theta = \frac{x}{2}$$


$$(\operatorname{sec} \theta)^2 \frac{d \theta}{dt}= \frac{1}{2}\frac{d x}{dt}$$


$$\frac{d \theta}{dt} = \frac{1}{2}(\operatorname{cos} \theta)^2\frac{d x}{dt}.$$

If the boat is directly opposite the station, $\theta = 0$ degrees, and $(\operatorname{cos} 0)^2 = 1$ and

$$\frac{d \theta}{dt} = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}.$$

Note that I converted 40 miles per hour to 2/3 miles per minute.

So, the light beam must turn 1/3 radians per minute to follow the boat as it passes directly opposite the station.

However, the book has written that the correct answer is $\frac{4}{9}.$ The book has given incorrect answers before. Is my answer incorrect, and, if so, where did I go wrong?

Thanks !

  • $\begingroup$ I redid the problem and got the same answer as you. $\endgroup$ – Matthew Leingang Aug 28 '19 at 19:18
  • $\begingroup$ Ok! Thanks Matthew! I’ll assume the book is wrong then. $\endgroup$ – Rafael Vergnaud Aug 28 '19 at 19:23

Apparently, the problem in the book was solved forgetting that the searchlight is stationed half a mile back of the shore. If the light were right on the shore, the answer $\frac49$ would be correct.

| cite | improve this answer | |
  • $\begingroup$ Good sleuthing. $\frac{4}{9}$ is $\frac{1}{1.5^2}$. $\endgroup$ – Matthew Leingang Aug 28 '19 at 19:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.