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Question

A lighthouse is located on a small island 3km away from the nearest point P on a straight shoreline and its light makes four revolutions per minute.

 

How fast is the beam of light moving along the shoreline when it is 1km from P?

Answer 1

It sounds like you're being asked to calculate the speed of the point of the light as it moves along the shore. So let's set up a triangle and some variables.

 

Call the spot of the beam along the shore S, and call the lighthouse L.  Then you have a right triangle SLP,

where the right angle is at P.

 

Now let x be the distance from P to the point of the beam on the shore.  The problem asking to find the speed with which S is moving, which is the same as asking for the speed at which x is changing, that is, dx/dt.

 

And finally, let θ be the angle L, at the lighthouse.

 

If you look at this triangle, you can see that x / 3 = tan θ.

 

So take the derivative of both sides with respect to t.

 

1 dx                                         dθ

-- -- = D_t tan θ = (1 + tan ²θ)  ---

3 dt                                          dt

 

So

 

dx                            dθ

--- = 3 (1 + tan² θ)  ---

dt                             dt

 

 

Now, tan θ is 1/3 at the moment the beam is 1 km from P (opposite over adjacent), so 1 + tan² θ = 1 + 1/9 = 10/9

 

And dθ/dt is 2π/4 per minute, since a complete revolution is 2π radians and this is done in 4 minutes.  So dθ/dt is equal to π/2.

 

So dx/dt = 3 (10/9) (π/2) = 5π/3 km per minute.

 

 

Answer 2

Let the zero of time ( t = 0) be such the the light is pointed directly at point P at t = 0.    As time advances, the distance along the shore to the point  hit by the beam increases.   Call this distance x.   When t = 0,  x =0 and more generally x is a function of time :   x(t).   Another variable is the angle, θ(t), between the normal to the shore (line between the lighthouse and point P) and the line of the beam at time t.

 

With the aid of a diagram, and a little trigonometry,   it can be worked out that   x(t) = 3 tan( θ(t) ).

 

The quantity being sought is  dx / dt.   The chain rule for derivatives, and the fact that the derivative of tan is sec² ,

gives    dx / dt =  3 sec²(θ) (dθ / dt) .  

 

Since the rate of rotation is 4 rpm,   it can be worked out that dθ / dt =  2 π 4 / 60  =  0.419 radian/sec.

 

When x = 1 km,    θ = arctan( 1/3) =  18.4 deg.     Plugging these values into the the expression for dx / dt

 

gives   dx / dt =  3 sec²(18.4) 0.419 =  1.40    (unit is km/sec).