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# A man ,2m tall,walks at the rate of $1\frac{2}{3}m/s$ towards a street light which is $5\frac{1}{3}m$ above the ground.At what rate is the tip of his shadow moving?At what rate is the length of the shadow changing when he is $3\frac{1}{3}m$ from the base of the light?

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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 $\large\frac{8}{3}$ $m/sec$

edited Jan 21, 2014