Question

A man of height 2 metres walks at a uniform speed of 6 km/hr away from a lamp post of 6 metres high. Find the rate at which the length of the shadow is increasing.

Answer 1

Let OA be the lamp post, MN the man, MB = x his shadow and OM = y the distance of the man from lamp post at time t.

Then, `(dy)/(dt)`= 6 km/hr is the rate at which the man is moving away from the lamp post. MN = 2 m, OA = 6 m               ...[Given]

`(dx)/(dt)` is the rate at which his shadow is increasing.

From the figure,

∆NMB ∼ ∆AOB

∴ `"MB"/"MN" = "OB"/"OA"`

∴ `x/2 = (x + y)/6`

∴ 6x = 2x + 2y

∴ 4x = 2y

∴ `x = (2y)/4`

∴ `x = y/2`

Differentiating w.r.t.t, we get,

`∴ (dx)/(dt) = 1/2. (dy)/(dt)`

= `1/2 × 6`

= 3 km/hr

Thus, the length of shadow is increasing at the rate of 3 km/hr.