2022-04-08

A man 6 ft tall walks at a rate of 5 ft/sec away from a lamppost that is 12 ft high. At what rate is the length of his shadow changing when he is 25 ft away from the lamppost?

nick1337

Expert2022-06-08Added 699 answers

Let $x$ be the distance between the man and the street light and let $y$ be the length of his shadow. Note that we have two similar right triangles $OLM$ and $OAB$. Hence using the similar triangle property, we get

$\frac{5}{y}=\frac{12}{x+y}$

$\Rightarrow 5(x+y)=12y$

$\Rightarrow 5x=12-5y$

$\Rightarrow y=\frac{5}{7}x$

Differentiating with respect to time $t$ partially, we get

$\frac{dy}{dt}=\frac{5}{7}\frac{dx}{dt}$

Now Since the man is walking away's from the lamp at a rate ot $5$ $ft/s$.

$\Rightarrow \frac{dx}{dt}=5$

Hence at the moment when he is $25$ $ft$ aways from the lamppOst, the length of his shadow is changing at a rate of

$\frac{dy}{dt}=\frac{5}{7}*5=\frac{25}{7}ft/s\approx 3.6ft/s$

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