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[erix335]

Hi all, I'm working on some test corrections and can't seem to create an equation for this problem, or really even figure out where to start minus a diagram:

A man 6 feet tall walks away from a streetlight that is 18 feet tall. If the length of his shadow is changing at a rate of 3 feet per second when he is 25 feet away from the base of the light, how fast is he walking away from the light at this moment?

 

[tkhunny]

You have similar triangles. The heights are constant. The lengths are related. Let's see your efforts.

 

[erix335]

2s(ds/dt)= (25-3t)^2 + ?(18-6t)2
2(30.8 hyp of triangle) = 2(25-3t)3+2*18-6t)6
61.4=(50-6t)3+(36-12t)6
=366-90t
-304.6/-90=t
t=304.6/90 or 3.38....which seems to be a right answer but the second part of my equation had no real thought behind it.

 

[soroban]

Hello, erix335!

I have no idea where your equation came from . . .

A man 6 feet tall walks away from a streetlight that is 18 feet tall.
If the length of his shadow is changing at a rate of 3 ft/sec when he is 25 feet away from the base of the light,
how fast is he walking away from the light at this moment?

      A
      * 
      |   *   C
      |       * 
   18 |       |   * 
      |      6|       *
      |       |           *
      *-------*---------------* 
      B   x   D       s       E

\(\displaystyle AB\text{ is the streetlight: }AB = 18.\)
\(\displaystyle CD\text{ is the man: }CD = 6.\)

\(\displaystyle \text{Let }x = BD\text{, his distance from the streetlight.}\)
\(\displaystyle \text{Let }s = DE\text{, the length of his shadow.}\)
\(\displaystyle \text{We are given: }\,\frac{ds}{dt} = 3\text{ ft/sec}\)

\(\displaystyle \text{Since }\Delta ABE \sim \Dalta CDE: \;\frac{x+s}{18} \,=\,\frac{s}{6}\)

\(\displaystyle \text{Then: }\:6x+6s \:=\:18s \quad\Rightarrow\quad 6x\:=\:12s \quad\Rightarrow\quad x \:=\:2s\)

\(\displaystyle \text{Differentiate with respect to time: }\:\frac{dx}{dt} \:=\:2\frac{ds}{dt}\)

\(\displaystyle \text{Therefore: }\:\frac{dx}{dt} \:=\:2(3) \:=\:6\text{ ft/sec}\)