Question #337946

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?


1
Expert's answer
2022-05-11T17:37:03-0400

The picture below corresponds to the problem described.


CD|CD| is the height of the man. CD=6ft.|CD|=6\,{\text{ft.}} EB|EB| is the height of the lamp post. EB=18ft.|EB|=18\,{\text{ft}}. Denote DE=x|DE|=x. In order to find AD|AD|, we write the definition of tan(α)\tan(\alpha), where α=<BAE\alpha=<BAE. We get: tan(α)=CDAD=BEAE\tan(\alpha)=\frac{|CD|}{|AD|}=\frac{|BE|}{|AE|}. AE=AD+DE=AD+x|AE|=|AD|+|DE|=|AD|+x. We receive: 6(AD+x)=18AD6(|AD|+x)=18|AD|. We get: AD=x2|AD|=\frac{x}2. Suppose that a man walked a distance ΔS=vΔt\Delta S=v\Delta t, where Δt\Delta t is a small period of time and v=4ft/sec.v=4\,{\text{ft}}/\text{sec}. We receive new segment CDC'D' of the length 6ft.6\,\text{ft.} and a new point AA', which is on the line BCBC'. DE=DE+vΔt=DE+4Δt=x+4Δt|D'E|=|DE|+v\Delta t=|DE|+4\Delta t=x +4\Delta t. Denote: α=<BAE\alpha'=<BA'E. We get: tan(α)=CDAD=BEAD+DEtan(\alpha')=\frac{|C'D'|}{|A'D'|}=\frac{|BE|}{|A'D'|+|D'E|}. From the latter we find: 6(AD+DE)=18AD6(|A'D'|+|D'E|)=18|A'D'|. Thus, AD=12DE=12(x+4Δt)|A'D'|=\frac{1}{2}|D'E|=\frac12(x+4\Delta t). ADAD=2Δt|AD|-|A'D'|=2\Delta t. From the latter we get that the shadow lengthen with a speed 2ft/sec2\,\text{ft}/\text{sec}.

Answer: the shadow lengthen with the speed 2ft/sec2\text{ft}/{\text{sec}}.


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