A man, 2m tall, walks at the rate of m/s towards

Solution:

Let AB is the height of street light post and CD is the height of the man such that

AB = 5(1/3) = 16/3 m and CD = 2 m

Let BC = x length (the distance of the man from the lamp post)

And CE = y is the length of the shadow of the man at any instant.

It’s seen from the figure that,

∆ ABE ~ ∆ DCE [by AAA similarity criterion]

Now, taking ratio of their corresponding sides, we have

$\frac{A B}{C D}=\frac{B E}{C E} \Rightarrow \frac{A B}{C D}=\frac{B C+C E}{C E}$

$\frac{16 / 3}{2}=\frac{x+y}{y} \Rightarrow \frac{8}{3}=\frac{x+y}{y}$

$8 y=3 x+3 y \quad \Rightarrow 8 y-3 y=3 x \quad \Rightarrow 5 y=3 x$

Differentiating both sides w.r.t, t, we have

$5 \cdot \frac{d y}{d t}=3 \cdot \frac{d x}{d t}$

$\Rightarrow \frac{d y}{d t}=\frac{3}{5} \cdot \frac{d x}{d t} \Rightarrow \frac{d y}{d t}=\frac{3}{5} \cdot\left(-1 \frac{2}{3}\right)=\frac{3}{5} \cdot\left(\frac{-5}{3}\right)$

$[\because$ man is moving in opposite direction $]$

$=-1 \mathrm{~m} / \mathrm{s}$

So, the length of shadow is decreasing at the rate of 1 m/s.

Now, let u = x + y

(where, u = distance of the tip of shadow from the light post)

On differentiating both sides w.r.t. t, we get

$2 \frac{2}{3}$

$\frac{d u}{d t}=\frac{d x}{d t}+\frac{d y}{d t}$

$=\left(-1 \frac{2}{3}-1\right)=-\left(\frac{5}{3}+1\right)=-\frac{8}{3}=-2 \frac{2}{3} \mathrm{~m} / \mathrm{s}$

Therefore, the tip of the shadow is moving at the rate of m/s towards the light post and the length of shadow decreasing at the rate of 1 m/s.