Question

A man $160$ cm tall, walks away from a source of light situated at the top of a pole $6$ m high, at the rate of $1.1$ m/sec. How fast is the length of his shadow increasing when he is $1$ m away from the pole?

Answer 1

Let x$=$ distance of man from pole

Given $\frac{dx}{dt}=1.1$ m/s

$\Delta ABC\sim \Delta PRC$ (By AA rule)

$\frac{AB}{PR}=\frac{BC}{RC}$

$\Rightarrow \frac{6}{1.6}=\frac{x+y}{y}$

$\Rightarrow 6y=1.6x+1.6y$

$\Rightarrow 4.4y=1.6x$

$\Rightarrow y=\frac{4}{11}x$ we have to find ${\frac{dy}{dx}|}_{x=1m}$

$\therefore \frac{dy}{dt}=\frac{4}{11}\times \frac{dx}{dt}$

$\Rightarrow \frac{dy}{dt}=\frac{4}{11}\times 1.1$ m/s

$=0.4$ m/s

at $x=1$m, $\frac{dy}{dt}=0.4$ m/s.