Science:Math Exam Resources/Courses/MATH102/December 2019/Question 10(b)

MATH102 December 2019

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Question 10(b)

A lighthouse is 1 km from the shore. Its beam rotates, taking 20 seconds for one full revolution. In the diagram at right, the dotted line is the beam of light; the solid line is the shortest line from the lighthouse to the shore; and $\theta$ is the angle between those two lines. (b) As the light beam turns, the point where it hits the shore moves. How fast is that point moving along the shoreline at the instant when it is closest to the lighthouse (when $\theta =0$)? Give your answer in km/sec. UBC Math 102 2019 Q10

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
$x$ is the distance between the point where the shoreline is closest to the light house and where the light hits the shoreline. What is the relationship between $x$ and $\theta$?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Observe that $x=\tan(\theta )$. Then we can get the rate in which the distance changes. ${\begin{aligned}{\frac {dx}{dt}}=\sec ^{2}(\theta ){\frac {d\theta }{dt}}=\sec ^{2}(\theta ){\frac {\pi }{10}}\end{aligned}}$. At $\theta =0$, ${\frac {dx}{dt}}=\sec ^{2}(0){\frac {\pi }{10}}={\frac {\pi }{10}}{\text{km/sec}}$.