009A Sample Final 2, Problem 5

A lighthouse is located on a small island 3km away from the nearest point  $P$  on a straight shoreline and its light makes 4 revolutions per minute. How fast is the beam of light moving along the shoreline on a point 1km away from  $P?$

Foundations:
When we see a problem talking about rates, it is usually a related rates problem.
Thus, we treat everything as a function of time, or  $t.$
We can usually find an equation relating one unknown to another, and then use implicit differentiation.
Since the problem usually gives us one rate, and from the given info we can usually find the values of variables at our particular moment in time, we can solve the equation for the remaining rate.

Solution:

Step 1:
We can begin this physical word problem by drawing a picture.
Insert picture
In the picture, we can consider the distance from the point  $P$  to the spot the light hits the shore to be the variable  $x.$
By drawing a right triangle with the beam as its hypotenuse, we can see that our variable  $x$  is related to the angle  $\theta$  by the equation
${\displaystyle \tan \theta \ =\ {\frac {\textrm {sideopp.}}{\textrm {sideadj.}}}\ =\ {\frac {x}{3}}.}$
This gives us a relation between the two variables.

Step 2:
Now, we use implicit differentiation to find
${\displaystyle \sec ^{2}\theta \cdot {\frac {d\theta }{dt}}\ =\ {\frac {1}{3}}\cdot {\frac {dx}{dt}}.}$
Rearranging, we have
${\displaystyle {\frac {dx}{dt}}\ =\ 3\sec ^{2}\theta \cdot {\frac {d\theta }{dt}}.}$
Again, everything is a function of time.

Step 3:
We want to know the rate that the beam is moving along the shore when we are one km away from the point  $P.$
This tells us that  $x=1.$
The problem also tells us that the lighthouse beam is revolving at 4 revolutions per minute.
However,  $\theta$  is measured in radians, and there are  $2\pi$  radians in a revolution.
Thus, we know
${\displaystyle {\frac {d\theta }{dt}}\ =\ 4\cdot 2\pi \ =\ 8\pi .}$
Finally, we require secant. Since we know  $x=1,$
we can solve the triangle to get that the length of the hypotenuse is
${\sqrt {1^{2}+3^{2}}}={\sqrt {10}}.$
This means that
${\displaystyle \sec \theta \ =\ {\frac {1}{\cos \theta }}\ =\ {\frac {\textrm {hyp.}}{\textrm {sideadj.}}}\ =\ {\frac {\sqrt {10}}{3}}.}$

Step 4:
Now, we can plug in all these values to find
${\begin{array}{rcl}\displaystyle {\frac {dx}{dt}}&=&\displaystyle {3\sec ^{2}\theta \cdot {\frac {d\theta }{dt}}}\\&&\\&=&\displaystyle {3\left({\frac {\sqrt {10}}{3}}\right)^{2}(8\pi )}\\&&\\&=&\displaystyle {3\left({\frac {10}{3}}\right)(8\pi )}\\&&\\&=&\displaystyle {80\pi {\text{ km/min.}}}\end{array}}$

Final Answer:
$80\pi {\text{ km/min}}$

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