**Toolbox:**

- If $ y = f(x)$, then $ \large\frac{dy}{dx}$ measures the rate of change of $y$ w.r.t $ x $.
- $ \bigg(\large \frac{dy}{dx} \bigg)_{x=x_0}$ represents the rate of change of $y$ w.r.t $x$ at $x=x_0$

Step 1

Let $AB$ be the lamp-post. Let at any time $t$, the man $CD$ be at $a$ distance $x$ meters from the lamp-post, and $y$ meters be the length of his shadow $DE$. It is given that

$ \large\frac{dx}{dt}=1 \large\frac{2}{3}m/s $

From the fig. it is clear $ \Delta ABE$ and $ \Delta CDE$ are similar

$ \therefore \large\frac{AB}{CD}= \large\frac{BE}{DE}=\frac{AE}{CE}$.......(i)

Given: $AB=5 \large\frac{1}{3}m$ and $CD=2m$

Let $BD=x\:\:and\:\:DE=y$

$\Rightarrow\: $ $AE=x+y$

Now substituting the values we get,

$ \large\frac{5 \large\frac{1}{3}}{2}= \large\frac{x+y}{y}$

$ \large\frac{16}{6}= \large\frac{x+y}{y}$

$ = 16y=6x+6y$

$ 10y=6x$.......(ii)

Differentiating w.r.t we get,

$ 10. \large\frac{dy}{dt}=6 \large\frac{dx}{dt}, but \large\frac{dx}{dt}=1 \large\frac{2}{3}= \large\frac{5}{3}$

$ \Rightarrow \large\frac{dy}{dx}= \large\frac{6}{10} \times \large\frac{5}{3}$

$ = 1$

Hence the shadow increases at the rate of $1m/s$

Step 2

$AB = 5 \large\frac{1}{3}m$

$CD=2m$

To find the rate at which the tip of the shadow $(E)$ moves,

we have to find the rate at which $BE=x+y$ moves.

Let $BE=x+y=p$

From (i) $ \large\frac{AB}{CD}= \large\frac{BE}{DE}$

$\Rightarrow\: \large\frac{8}{3}=\frac{x+y}{y}=\frac{p}{y}$

$\Rightarrow\:\large\frac{8}{3}$$y=p$

Differentiating w.r.t t both the sides we get

$\large\frac{8}{3}\frac{dy}{dt}=\frac{dp}{dt}$

But $\large\frac{dy}{dt}$$=1$.

$\therefore\:\large\frac{dp}{dt}=\frac{8}{3}$

$i.e.,$ The rate at which the tip of the shadow moves is $\large\frac{8}{3}$ $m/sec$