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Q.
A man $1.5\, m$ tall walks away from a lamp post $4.5 \, m$ high at a rate of $4\, km / hr$. How fast is the farther end of shadow moving on the pavement?
Application of Derivatives
Solution:
Let $AC$ be pole, $DE$ be man and $B$ be farther end of shadow as shown in figure From triangles $ABC$ and $DBE$
$ \frac{4.5}{x+y}=\frac{1.5}{y} $
$ 3 y=1.5 x$
$\frac{d y}{d t}=2,(x+y)=\frac{d x}{d t}+\frac{d y}{d t}$
