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### How Fast is the Lighthouse Beam Moving?

```Date: 10/13/2002 at 20:29:57
From: Leigh
Subject: Rates and derivatives

A lighthouse is located on a small island 3 km 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 1 km away from P?
```

```
Date: 10/15/2002 at 13:45:29
From: Doctor Roy
Subject: Re: Rates and derivatives

Hi,

Thanks for writing to Dr. Math.

Let's draw a quick picture to make the situation clearer:

3km
L ---------------P
\             |
\           |
\         |
\       | 1km
x     \     |
\   |
\ |
A

Please note that the picture is not to scale. Point A is where
the beam of light hits the shore when it is 1 km away from P, and
x represents the length of the beam of light from the lighthouse.

We can find the value of x by using the Pythagorean theorem:

x^2 = 1^2 + 3^2

x^2 = 1 + 9

x^2 = 10

x = sqrt(10)

So, x is sqrt(10) km long, or roughly 3.16 km long.

Another fact we need to know is the rotational speed of the
lighthouse. The lighthouse beam makes 4 full revolutions in a minute.
So, we can express this as:

w = 4 rev/min

Do you remember the formula for the circumference of a circle? It is
given by:

C = 2*pi*r

The circumference is the same as the distance traveled around the
circle, or 1 revolution, so 1 rev = 2*pi*r

So the next step is to find the speed along the circumference of
a circle of an object moving at 4 rev/min. The radius will be
sqrt(10). It is sqrt(10), because that is the "current" distance from
the lighthouse to the point it lights on the shoreline. Remember that
this distance will change as the light moves. So, the light "at this
current moment" travels:

4 rev/min = 4*1 rev/min = 4 * (2*pi*r km) / min

= 4 * (2 * pi * sqrt(10) km) / min

= 8 pi sqrt(10) km/min

Note again that the distance of A from P makes a difference in the
speed. The lighthouse turns at the same rate, but the distance from the
lighthouse to the shoreline changes. This means the light will appear
to move faster the farther from P we get.

Note also that we're not done yet!  We have the speed of the light
along the circumference of the circle.  But we _want_ to find the
speed along the shorline.

3km
L ---------------P
\             |
\           |
\         |
\       | 1km
x     \     |
\   |
\ |  /
A/
/|
/  |
/
/  <---- The speed in _this_ direction
is 8 pi sqrt(10) km

Trigonometry works with speeds as well as with distances,

3km
L ---------------P
\             |
\           |
\         |
\       | 1km
x     \     |
\   |
\ |  /
A/
/|
v  / a| ?
/    |
/______|

so once you've figured out the measure of the angle marked 'a', you
can find the speed along the shore using the equation

?
cos(a) = -
v

Does this help?

- Doctor Roy, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 10/15/2002 at 22:14:33
From: leigh
Subject: Thank you (Rates and derivatives)

Thank you so much! I understand it now.

Leigh
```

```
Date: 06/13/2013 at 11:48:18
From: Josh
Subject: Related Rates Lighthouse Problem

This is regarding the previous answer by "Leigh" -- about the lighthouse
on an island.

The problem involves finding the speed of the light beam on a circle, then
breaking up that speed into a horizontal component for the answer, which
is given as approximately 75.4m/min.

I'm not sure if that solution given is correct -- but I also don't know
what is wrong with it.

Using calculus, the rate of change of the angle L with respect to time is

Let y = distance from beam to point P (segment AP in the diagram). The
goal is to find dy/dt.

y = 3 tan L
dy/dt = 3 sec^2 L x 8pi  (using chain rule)

When y = 1,

sec^2L = 10/9
dy/dt = 3(10/9) x 8 pi
= 80/3 pi m/min.  (or approximately 83.7m/min)

I'm pretty sure this solution is correct. It is also found in various
places on the Internet.

Assuming I am correct, I would really like to know what the error is in
the solution posted: it also makes sense to me!
```

```Date: 06/13/2013 at 16:35:16
From: Doctor Peterson
Subject: Re: Related Rates Lighthouse Problem

Hi, Josh.

This conversation between Leigh and Doctor Roy doesn't give an answer, so
evidently you've worked out the end result by continuing Doctor Roy's
work. He certainly used a different method than you did. I myself get your
answer (by your method) but with different units, 83.7 km/min (not m/min).

Let's continue Doctor Roy's work.

He says that the speed perpendicular to the radius, v, is 8 pi sqrt(10)
km/min, and that

s
cos(a) = -
v

Here, s is the speed we are looking for. Also, angle a is the same as the
angle L, so

tan(a) = 1/3

Putting this all together leads to

s = v cos(a)
= v/sec(a)
= v/sqrt(tan^2(a) + 1)
= v/sqrt(1/9 + 1)
= 8 pi sqrt(10) / sqrt(10/9)
= 8 pi sqrt(10) * 3/ sqrt(10)
= 24 pi
= 75.4 km/min

So, yes, that's where you got that number (though, again, you have the
wrong units).

Now, which is right? He has the right rotational velocity, 8 pi rad/min.
Let's try both methods without using specific numbers, so we can compare
them more closely. I'll use this picture:

L
|\
|a\
d |  \x
|   \
|    \
P-----A--------
y

Our method gives

dy/dt = d/dt(d tan(a))
= d sec^2(a) da/dt

This amounts to

L
|\
|a\
d |  \x         .
|   \    v .  .
|    \  .  a  .
P-----A--------
y      s

dy/dt = x da/dt * cos(a)
= sqrt(d^2 + y^2) cos(a) da/dt
= sqrt(d^2 + d^2tan^2(a)) cos(a) da/dt
= d sqrt(1 + tan^2(a)) cos(a) da/dt
= d sec(a) cos(a) da/dt

Hmmm ... if only that cos(a) were another sec(a), we'd agree. So I'm led
to think that the (implied) argument about vector components was wrong
(which is just where I was most unsure of it as I read it).

The picture for that part of the process should have been like this:

L
|\
|a\
d |  \x         .
|   \    v .   .
|    \  .  a    .
P-----A-----------
y        s

In other words, cos(a) = v/s rather than s/v. The speed s is not the
component of v along the coast, but a speed along the coast the component
perpendicular to the radius vector of which is v.

This is why I like to check this kind of reasoning against an alternative
method.

Also, because he left his work unfinished, it's quite possible Doctor Roy
missed this error, but would have caught it if he had continued. I've done
that sort of thing a few times!

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus

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