A man $2 \mathrm{~m}$ tall, walks at the rate of $1 \frac{2}{3} \mathrm{~m} / \mathrm{s}$ towards a street light which is $5 \frac{1}{3} \mathrm{~m}$ above the ground. The rate at which the length of his shadow is changing when he is $3 \frac{1}{3} \mathrm{~m}$ away from the base of the…

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A man $2 \mathrm{~m}$ tall, walks at the rate of $1 \frac{2}{3} \mathrm{~m} / \mathrm{s}$ towards a street light which is $5 \frac{1}{3} \mathrm{~m}$ above the ground. The rate at which the length of his shadow is changing when he is $3 \frac{1}{3} \mathrm{~m}$ away from the base of the light is ____

Aicharya · Accepted Answer

$A B=$ street light, $C=$ Man. From $\triangle A B E$ and $\triangle D C E$, $\begin{aligned} \frac{A B}{C D} & =\frac{A E}{C E}=\frac{A C+C E}{C E} \ \Rightarrow \quad \frac{A B}{C D} & =\frac{A C}{C E}+1 \Rightarrow \frac{A C}{C E}=\frac{A B}{C D}-1 \ \Rightarrow \quad & \frac{A C}{C E}=\frac{5 \frac{1}{2}}{2}-1=\frac{5}{3} \Rightarrow C E=\frac{3}{5} A C \end{aligned}$ Differentiating with respect to time we get, $\frac{d}{d t} C E=\frac{3}{5}\left(\frac{d}{d t} A C\right) \quad \ldots (i)$ None $\frac{d}{d t} A C=-1 \frac{2}{3} \mathrm{~ms}^{-1}$ $\frac{d}{d t} A C=-\frac{5}{3} \mathrm{~ms}^{-1}$ Negative sign as $A C$ decreases with time. substituting in Eq. (i) we get. Rate of change of shadow length $=\frac{3}{5}\left(-\frac{5}{3}\right)=-1 \mathrm{~ms}^{-1}$

A man \(2 \mathrm{~m}\) tall, walks at the rate of \(1 \frac{2}{3} \mathrm{~m} / \mathrm{s}\) towards a…

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