Conference rusure::math

  /  o
 / 45             Shoreline
/____________________________________________
\   o  !
 \45   !            Tangential velocity of beam = (2*pi)*(4)*(1/10) km/sec
  \    !4 km                                    = (4*pi/5)
   \   !            Vector component of beam parallel to shore = 
    \  !                                                         o
     \ !                                        = (4*pi/5)*cos(45 )
      \!                                        = (4*pi/5)*(sqrt(2)/2)
       Ship                                     = 1.777 km/sec

Hope I got my algebra right.  NCR (no calculus required)

The answer in my text book, is not what your answer is!!. However, I dont
known where you went wrong. If you want me to place the answer in a reply
to this note, tell me so.

		Your diagram does give a better idea of what is going on.


					Mike

What a sap I am!  I used the wrong value for the length of the beam.
What follows is the reply from the kind soul who pointed that out to me:

From:	GRUMAN::REILLY       "Matt Reilly LTN2-2/C05 (dtn) 229-6994" 20-NOV-1985 09:28
To:	GALLO::APPELLOF,REILLY      
Subj:	Related Rate Problem

Hmmm,

	I get something different....


	The distance from the light to the shoreline  when the beam
is at an angle to the shore is 4 * sqrt(2).

	The speed of the point of light is w * r where w is the 
angular velocity and r is the distance from the source of the light
to the point.  This gives us the tangential speed

	s = w * r = 2 * pi / 10   *  4 * sqrt(2) 

		  = 4 * sqrt(2) * pi / 5

	Taking the component along the shoreline

	s / cos(45 deg) = 4 * pi * sqrt(2) * 2 / 5 * sqrt(2)

	                = 8 * pi / 5

			= 5.0265 km/sec

	Is this right?

	(I think your problem is in the way you calculated the
tangential velocity of the beam.  When the beam is at 45 degrees
to the shoreline, r = 4 * sqrt(2) not 4.)


						matt

Boy are we confused!  The math muscles are creaky today.

How about:

Tangential velocity = (2*pi/10)*4*sqrt(2)  as in previous response.
                                                     o
Velocity along shoreline = (2*pi/10)*4*sqrt(2)*cos(45 )

Multiply by cos(45), don't divide.

Final answer:  4*pi/5

Are we there yet?

I can see it now....  Soon to appear in a journal near you....


	AN ITERATIVE METHOD FOR THE SOLUTION OF TRIGONOMETRY PROBLEMS


You are right...   Multiply by cos don't divide.

								matt
(GEEZ, I think I'll just strap myself into my chair and put away any
sharp objects...)

Matt's answer is correct, the answer is		8*PI
                                                ----  Km/sec.
						 5

Can someone show me how this answer can be derived by using derivatives?


						Thanks Mike

Wait a minute!  I thought it was 4*pi/5.  I'm so confused.

For a derivative point of view, how about this:
      
   <- x ->
__________
\        !                    tan(theta) = x/4  by definition, so
 \       !
  \      !                         x = 4 * tan(theta)
   \     ! 4 km
    \    !                        dx      d (tan(theta))
     \   !                        -- = 4* --------------
      \  !                        dt          dt
       \ !
        \! <- angle theta                        d (theta)
        ship                      You know that  --------- = (2*pi/10) rad/sec
                                                   dt

So a simple application of the chain rule at theta = (pi/4) should give
you the answer. (if you do the algebra right (ouch :-)))

This note is really from Tom Eggers.

Let Theta be angle of light beam off-normal to shore.
Let x be distance from where normal intersects shore to
	where beam hits shore.

Then:	tan (theta) = x/4km

	x = 4km * tan(theta)

	v = dx/dt = 4km * sec(theta)**2 * d(theta)/dt

When beam makes 45 degrees with shore, sec(theta)**2 = 2

Therefore:

	v = 4km * 2 * 2pi/10	=	8pi/5 km/sec

This method does use calculus (a slight disadvantage), but it
avoids any use of the length of the beam, the tangential velocity,
or converting the tangential velocity to shoreline velocity.

Now, what is the speed of the beam along the shore assuming that
light has a finite velocity. Perhaps change the light house into
a fog horn and consider the rate the noise propogates along the
shore.

having already typed in this file and then finding reply .8 I fiqured I
might as write it. Thanks again for everyones help.



Instead of using the chain rule, i found it easier to use implicit
differentiation.
      

tan(theta) = x/4  by definition, so

x = 4 * tan(theta)

using implicit differentiation,


dx      	     	  d(theta)
-- = 4* [ sec**2(theta)] --------
dt          		    dt

sec**2(theta) = ( 1/ cos(theta))**2

cos(theta) = .7071
sec**2(theta) = (1/.7071)**2  = 2.0

dx		d(theta)
-- = 4 * ( 2)	--------
dt		   dt


	      d(theta)	  2*PI
we known that -------- =  ---- sec
		 dt  	   10


dx		2*PI
-- = 4 * ( 2)	----  sec
dt		 10


dx	16*PI	
-- = 	---- km sec    or
dt	 10 	


the final answer ,

dx	8*PI	
-- = 	---- km sec    
dt	 5 	

				Thanks in setting up the equations
					correctly!!
						
						Mike

			

After some thought, I have decided that my first method with no calculus
is totally wrong.  The derivative method turns out correctly, and actually
is easier to understand.  Sorry about that.