Question

A man of $2\,m$ height walks at a uniform speed of $6 \,km/h$ away from a lamp post of $6 \,m$ height. The rate at which the length of his shadow increases is

• A. $2\, km/h$
• B. $1\, km/h$
• C. $3\, km/h$
• D. $6\, km/h$

Answer 1

Answer: (C)

$3\, km/h$

Answer 2

In $\Delta ADC,$ $\tan \theta =\frac{6}{x+y}$ and in $\Delta BCE,$
$\therefore$ $\frac{2}{x}=\frac{6}{x+y}\Rightarrow x+y=3x$
$\Rightarrow$ $y=2x$
On differentiating w.r.t. t, we get
$\frac{dy}{dt}=2\frac{dx}{dt}$
$\Rightarrow$ $6=2\frac{dx}{dt}$ $\left( \because \frac{dy}{dt}=6given \right)$
$\Rightarrow$ $\frac{dx}{dt}=3\,km/h$