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### The Moving Shadow

```
Date: 03/24/99 at 20:27:32
From: Isaac
Subject: Related Rates of a Shadow's Movement

I am not sure I did this problem right, because here the man walks
towards the light source, whereas in other problems of this nature he
usually walks away from the light. Please check to see whether I did
it right.

A man 6 feet tall walks at a rate of 5 feet per second toward a light
that is 20 feet above the ground.

a) At what rate is the tip of his shadow moving?
b) At what rate is the length of his shadow changing?

^ |`
20 |  `
v |    0
|   /|\`
|   / \   `
-------------
dx
---- = 5feet/sec
dt

20      y
--- =  -----
6     y - x

20y - 20x = 6y

14y = 20x

20x
y = ------
14

10x
y = -----
7
dy   10  dx
--- = ----------
dt    7  dt

dy   10        50
-- =  --- (5) = ---- ft/sec, since he is walking towards the light
dt    7           7
50
then it is negative, - ----  ft/sec
7

a) -(50/7) ft/sec
b) d(y-x)   dy   dx    -(50)
------ = ---- - ---- =  ----- - (5) = -(85/7) ft/sec
dt     dt    dt      7
```

```
Date: 03/26/99 at 13:59:44
From: Doctor Luis
Subject: Re: Related Rates of a Shadow's Movement

Your reasoning for the first part was correct; however, you could
have made dx/dt negative from the start. That is why the second part
(b) is not correct.

Here is a diagram:

P
|\
| \
|  \
|   \
|    \
|     \N
|     |\
|     | \
|     |  \
|_____|___\
O     M    S

According to the problem, OP is 20 ft, MN is 6 ft. The man, at M, is
walking TOWARD the light source at O. If x is the distance OM, then
dx/dt HAS to be negative because the x-values are decreasing.

Now, if y is the distance OS, (where the tip of the shadow is at S),
then, as you explained above, by similar triangles you have:

OP     OS
----- =  -----
MN     MS

20      y
---- = ------
6     y - x

and so you found

y = (10/7)x

Now if you just differentiate with respect to time, you get

dy/dt = (10/7)dx/dt
= (10/7)(-5)
= -50/7 ft/s

You can see that by making dx/dt negative from the start you avoid
the complication of changing the sign of dy/dt in the middle of the
problem. (That is why you did not get the last part.)

Given that, we find how fast the length of the shadow is changing:

d
--- (y - x) = dy/dt - dx/dt = (-50/7) - (-5) = -15/7 ft/sec
dt

Notice that the term -dx/dt makes a positive contribution to the rate
of change of the length of the shadow. This makes sense from a
physical standpoint, because if the tip of the shadow were to be held
still, the man, by walking away, would actually be increasing the
length of the shadow (y-x would be increasing if y were constant).
However, y is not constant, it is decreasing, and that's why it makes
a negative contribution to the rate of change of the length of the
shadow.

My recommendation to you is that you choose a consistent set of
directions from the start (for example in this problem we chose the
direction of increasing x-values to the right) and then, based on
that, give the time-derivatives the appropriate signs according to
whether the values whose rate of change they represent are increasing
or decreasing. If you do this from the beginning of the problem, you
will not have to be changing the signs every time.

What I am trying to say is you need to be consistent.

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

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