Answer to Question #337946 in Calculus for dme

Question

A man 6 ft. tall walks away from a lamp post 18 ft. high at the rate of 4 ft. per second. How fast does

the shadow of the man lengthen?

Answer 1

The picture below corresponds to the problem described.

$|CD|$ is the height of the man. $|CD|=6\,{\text{ft.}}$ $|EB|$ is the height of the lamp post. $|EB|=18\,{\text{ft}}.$ Denote $|DE|=x$. In order to find $|AD|$, we write the definition of $\tan(\alpha)$, where $\alpha=<BAE$. We get: $\tan(\alpha)=\frac{|CD|}{|AD|}=\frac{|BE|}{|AE|}$. $|AE|=|AD|+|DE|=|AD|+x$. We receive: $6(|AD|+x)=18|AD|$. We get: $|AD|=\frac{x}2$. Suppose that a man walked a distance $\Delta S=v\Delta t$, where $\Delta t$ is a small period of time and $v=4\,{\text{ft}}/\text{sec}.$ We receive new segment $C'D'$ of the length $6\,\text{ft.}$ and a new point $A'$, which is on the line $BC'$. $|D'E|=|DE|+v\Delta t=|DE|+4\Delta t=x +4\Delta t$. Denote: $\alpha'=<BA'E$. We get: $tan(\alpha')=\frac{|C'D'|}{|A'D'|}=\frac{|BE|}{|A'D'|+|D'E|}$. From the latter we find: $6(|A'D'|+|D'E|)=18|A'D'|$. Thus, $|A'D'|=\frac{1}{2}|D'E|=\frac12(x+4\Delta t)$. $|AD|-|A'D'|=2\Delta t$. From the latter we get that the shadow lengthen with a speed $2\,\text{ft}/\text{sec}$.

Answer: the shadow lengthen with the speed $2\text{ft}/{\text{sec}}$.