Let AB represent the height of the street light from the ground. At any time t seconds, let the man represented as ED of height 1.6 m be at a distance of x m from AB and the length of his shadow EC be y m.
Using similarity of triangles, we have \(\frac4{1.6} = \frac{x+y}y \) ⇒ \(3y = 2x\)
Differentiating both sides w.r.to t, we get
\(3\frac{dy}{dt} = 2\frac{dx}{dt}\)
\(\frac{dy}{dt} = \frac 23 \times 0.3\)
⇒ \(\frac{dy}{dt} =0.2\)
At any time t seconds, the tip of his shadow is at a distance of (x + y) m from AB.
The rate at which the tip of his shadow moving
\(= \left(\frac{dx}{dt} + \frac{dy}{dt}\right)m/s = 0.5 \,m/s\)
The rate at which his shadow is lengthening
\(= \frac{dy}{dt}m/s = 0.2 m/s\)
