Question

A man of height $2$ m walks directly away from a lamp of height $5$m, on a level road at $3$ m/sec. The rate of which the length of his shadow is increasing is?

• A. $1$ m/sec
• B. $2$ m/sec
• C. $3$ m/sec
• D. $4$ m/sec

Answer 1

Answer: (B)

$2$ m/sec

Here, $AB$ is lamp and $CD$ is man

$dy=3$ m/sec

To calculate $dx$

$\tan \alpha =\frac{AB}{x+y}=\frac{CD}{x}$

$\frac{AB}{x+y}=\frac{CD}{x}\frac{5}{x+y}=\frac{2}{x}5x=2x+2y3x=2y3\frac{dx}{dt}=2\frac{dy}{dt}3\times \frac{dx}{dt}=2\times 3$

so , $\frac{dx}{dt}=2$ m/sec