Related Rates Part IV

Hckyplayer8

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Jun 9, 2019
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This whole section is kicking my butt.

The beacon on a lighthouse makes a revolution every 20 secs. The beacon is 300ft away from nearest point P on the shoreline. Find the rate at which the ray of light moves along the shoreline at a point 200ft from point P.

Let the lighthouse be the apex of the right triangle. The origin can be point P with the 200ft difference lining up on the x axis and the 300ft difference being the y axis.

The problem asks for the instantaneous velocity of the light at point 200ft, correct? So where the previous problems gave instantaneous velocity and line segment(s) to figure out angles and a line segment, this one gives the line segments and angle in order to derive velocity?
 
Here's a lesson I posted on another site:

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:

[MATH]\tan(\theta)=\frac{x}{y}[/MATH]
Now, let's differentiate with respect to time \(t[\), bearing in mind that while \(x\) is a function of \(t\), \(y\) is a constant.

[MATH]\sec^2(\theta)\cdot\d{\theta}{t}=\frac{1}{y} \cdot\d{x}{t}[/MATH]
Since we are being asked to find the speed of the spot, whose position is \(x\), we want to solve for [MATH]\d{x}{t}[/MATH]:

[MATH]\d{x}{t}=y\sec^2(\theta) \cdot\d{\theta}{t}[/MATH]
Let's let the angular velocity be given by:

[MATH]\omega=\d{\theta}{t}[/MATH]
and from the diagram and the Pythagorean theorem, we find:

[MATH]\sec^2(\theta)=\frac{x^2+y^2}{y^2}[/MATH]
Hence, we have:

[MATH]\d{x}{t}=\frac{\omega}{y}(y^2+x^2)[/MATH]
Now, the angular velocity is in radians per unit of time, and since there are \(2\pi\) radians per revolution, we may write:

[MATH]\omega=2\pi R[/MATH]
Hence:

[MATH]\d{x}{t}=\frac{2\pi R}{y}(y^2+x^2)[/MATH]
 
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