Lesson Related Rates: Lighthouse Beam

MarkFL

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A lighthouse is fixed \(y\) units from a straight shoreline. A spotlight in the lighthouse revolves at a rate of \(R\) revolutions per minute, shining a spot along the shoreline as it spins. At what rate is the spot moving when it is along the shoreline \(x\) units from the shoreline point closest to the lighthouse?

Consider the following diagram:

tml_lighthouse.png


As we can see, we may state:

\(\displaystyle \tan(\theta)=\frac{x}{y}\)

Now, let's differentiate with respect to time \(t[\), bearing in mind that while \(x\) is a function of \(t\), \(y\) is a constant.

\(\displaystyle \sec^2(\theta)\cdot\d{\theta}{t}=\frac{1}{y} \cdot\d{x}{t}\)

Since we are being asked to find the speed of the spot, whose position is \(x\), we want to solve for \(\displaystyle \d{x}{t}\):

\(\displaystyle \d{x}{t}=y\sec^2(\theta) \cdot\d{\theta}{t}\)

Let's let the angular velocity be given by:

\(\displaystyle \omega=\d{\theta}{t}\)

and from the diagram and the Pythagorean theorem, we find:

\(\displaystyle \sec^2(\theta)=\frac{x^2+y^2}{y^2}\)

Hence, we have:

\(\displaystyle \d{x}{t}=\frac{\omega}{y}(y^2+x^2)\)

Now, the angular velocity is in radians per unit of time, and since there are \(2\pi\) radians per revolution, we may write:

\(\displaystyle \omega=2\pi R\)

Hence:

\(\displaystyle \d{x}{t}=\frac{2\pi R}{y}(y^2+x^2)\)
 
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harpazo

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Amazing!