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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

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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