A man of height $2 \mathrm{~m}$ walks at a uniform speed of $7 \mathrm{~m} / \mathrm{min}$ away from a lamp post of height $9 \mathrm{~m}$. The rate $(\mathrm{m} / \mathrm{min})$ at which the length of his shadow increases is

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Aicharya · Accepted Answer

Let $A B$ be the lamp-post and $P Q$ the man, $C P$ be his shadow at time $t$. Let $A P={ }^` x, P C=y$, Also $A B=9 \mathrm{~m}, P Q=2 \mathrm{~m}$ Now, $\triangle C A B$ and $\triangle C P Q$ are equiangular and hence similar. $\begin{aligned} & \therefore \quad \frac{P C}{A C}=\frac{P Q}{A B} \ & \Rightarrow \quad \frac{y}{x+y}=\frac{2}{9} \ & \Rightarrow \quad 9 y=2 x+2 y \ & \Rightarrow \quad 7 y=2 x \ & \Rightarrow \quad x=\frac{7}{2} y \ & \Rightarrow \quad \frac{d x}{d t}=\frac{7}{2} \frac{d y}{d t} \text { (differentiating w.r.t. } t \text {) } \ & \text { But } \quad \frac{d x}{d t}=7 \mathrm{~m} / \mathrm{min} \ & \therefore \quad 7=\frac{7}{2} \frac{d y}{d t} \ & \Rightarrow \quad \frac{d y}{d t}=2 \mathrm{~m} / \mathrm{min} \end{aligned}$ $\therefore$ Length of shadow is increases at $2 \mathrm{~m} / \mathrm{min}$.

A man of height \(2 \mathrm{~m}\) walks at a uniform speed of \(7 \mathrm{~m} / \mathrm{min}\) away from a…

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