A man 2 text{ m} tall walks at the rate of 1 | vedclass

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(A) Let $AB$ be the street light of height $H = 5 \frac{1}{3} = \frac{16}{3} 	ext{ m}$. Let $CD$ be the man of height $h = 2 	ext{ m}$. Let $AC = x$

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$-1$

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(A) Let $AB$ be the street light of height $H = 5 \frac{1}{3} = \frac{16}{3} 	ext{ m}$. Let $CD$ be the man of height $h = 2 	ext{ m}$. Let $AC = x$ be the distance of the man from the light and $CE = y$ be the length of his shadow. From similar triangles $	riangle ABE$ and $	riangle DCE$,we have: $\frac{AB}{CD} = \frac{AE}{CE} \Rightarrow \frac{16/3}{2} = \frac{x+y}{y}$ $\frac{8}{3} = \frac{x}{y} + 1$ $\Rightarrow \frac{x}{y} = \frac{5}{3}$ $\Rightarrow y = \frac{3}{5}x$ Differentiating with respect to time $t$: $\frac{dy}{dt} = \frac{3}{5} \frac{dx}{dt}$ Given that the man walks towards the light at $1 \frac{2}{3} 	ext{ m/s}$,so $\frac{dx}{dt} = -\frac{5}{3} 	ext{ m/s}$ (negative because $x$ is decreasing). Therefore,$\frac{dy}{dt} = \frac{3}{5} 	imes (-\frac{5}{3}) = -1 	ext{ m/s}$. The length of the shadow is changing at the rate of $-1 	ext{ m/s}$.

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