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```Date: 10/13/2003 at 13:49:13
From: Chris
Subject: Mathematical Methods

A man 6 feet tall walks at 4 miles an hour directly away from a
lampost 18 feet tall. Why does his shadow lengthen at a constant rate?
What is this rate and how can I work it out?
```

```
Date: 10/15/2003 at 11:54:52
From: Doctor Luis
Subject: Re: Mathematical Methods

Hi Chris,

If you let h be the height of the man, L be the height of the
lamppost, x be the distance of the man from the lamppost, and y be the
length of the shadow, then you can see from the following diagram

that the proportion L/(x+y) = h/y holds, from similar triangles.

So we have Ly = h(x + y), or (L - h)y = hx.

Since x and y are related linearly, then so are dx/dt and dy/dt.  This
means that if dx/dt is constant, then dy/dt is also constant.  You can
use the equation between x and y to find the relationship between
dx/dt and dy/dt.

I hope this helped!  Let us know if you have any more questions.

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

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