Question

A man, $2m$ tall, walks ar 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 rates is the tip of his shadow moving ? At what is the length of the shadow changing when he is $3\frac{1}{3}m$ from the base of the light?

Answer 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

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

From figure it is clear $△ABE$ and $△CDE$ are similar

$\therefore$ $\frac{AB}{CD}=\frac{BE}{DE}=\frac{AE}{CE}$ ----- ( 1 )

$AB=5\frac{1}{3}m$ and $CD=2m$ [ Given ]

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

$\Rightarrow$ $AE=x+y$

Now, substituting the values we get,

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

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

$\Rightarrow$ $16y=6x+6y$

$\Rightarrow$ $10y=6x$ ------ ( 2 )

Differentiating w.r.t we get,

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

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

$\Rightarrow$ $\frac{dy}{dx}=1$

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

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

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 ( 1 ),

$\frac{AB}{CD}=\frac{BE}{DE}$

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

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

Differentiating w.r.t $t$ bothe sides we get,

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

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

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

The rate at which the tip of the shadow moves is $\frac{8}{3}m/s$