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Shadow Moving Along a Wall


Date: 08/16/99 at 09:20:53
From: Kris 
Subject: Inverse trig functions


                                   B
                                -  |  -
                             -     |     - S
                           -       |      -
                          -        | C     -
                        L  ------------------
                          -        |       -
                           -       |      -
                             -     |    -
                                -  | -

The Problem:

At the front of a circular arena there is a light L. A boy starting 
from B runs at the rate of 3 m/s toward the center C. At what rate 
will his shadow be moving along the rear wall when he is halfway from 
B to C?

This problem was presented in my assignment and is the only one I 
can't seem to get right. I think the rate might be 8 m/s? (My friend 
used vector calculus, but we can't use this because our teacher said 
it wouldn't be relevant for this course.)

     I drew a Triangle LCS
     Angle CLS = Angle CSL

and from this I derived that

     Angle CLS + Angle CSL = Angle LCS

Then found out the height From C towards B within the Triangle LCS is 
equal to R (the radius) - 3t, since distance = vt.

then Angle BLC = Tan^-1(R-3t/R)

Letting Angle BLC = theta

     Angle LCS = 2theta

then

     2theta = 2Tan^-1(R-3t/R)

     d(theta)        -6
     -------- = ------------
        dt      R^2-(R-3t)^2


Then I found

     t = d/v
       = (R/2)(1/3)
       = R/6

I then substituted t into the above equation, and in the end the R's 
did not cancel out. Nobody in my class can get it either. Have I gone 
about this wrong?

Thanks heaps,
Kris


Date: 08/17/99 at 08:24:49
From: Doctor Jerry
Subject: Re: Inverse trig functions

Hi Kris,

Let's put in some coordinates. Let L and S have coordinates (-R,0) and 
(x,y). At time t the boy is at (0,R-3t). We can figure out equations 
for x and y. The line through L and the boy is:

         y-0 = ((R-3t)/R)(x+R)

The point (x,y) satisfies

     x^2+y^2 = R^2

Let's not try, however, to solve these equations simultaneously just 
yet. Let f(t) be the angle between the positive x-axis and the line 
from C to S. Letting the arc subtended by f(t) on the circle be s, we 
know that

     s = R*f(t)

We want to calculate ds/dt, right? We see that

     ds/dt = R*f'(t)

so that's what we'll work on.

Now, 

     tan(f(t)) = y/x

So, differentiating,

     sec^2(f(t))*f'(t) = [x*y' - x'*y]/x^2

When the boy is halfway to the center, we'll figure out x', y', x, 
and y. I'll outline this.

If you solve 

           y-0 = ((R-3t)/R)(x+R)
 and
       x^2+y^2 = R^2

simultaneously, when R-3t = R/2, you'll find

     x = 3R/5   and   y = 4R/5

Next, differentiate the equations

         y-0 = ((R-3t)/R)(x+R)
and
     x^2+y^2 = R^2

Replace x with 3R/5 and y with 4R/5, then solve for x' and y'. I think 
you'll get

     x' = 21/5   and   y' = -21/10

Put everything together, and I think you'll find that

     R*f'(t) = -231/50 m/s.

- Doctor Jerry, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Trigonometry

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