Question

A man, 2m tall, walks at the rate of `1 2/3` m/s towards a street light which is `5 1/3`m above the ground. At what rate is the tip of his shadow moving? At what rate is the length of the shadow changing when he is `3 1/3`m from the base of the light?

Answer 1

Let AB is the height of street light post and CD is the height of the man such that

AB = `5 1/3 = 16/3 "m"` and CD = 2 m

Let BC = x length (the distance of the man from the lamp post) and CE = y is the length of the shadow of the man at any instant.

From the figure, we see that 

ΔABE ~ Δ DCE   ......[By AAA Similarity]

∴ Taking ratio of their corresponding sides, we get

`"AB"/"CD" = "BE"/"CE"`

⇒ `"AB"/"CD" = ("BC" + "CE")/"CE"`

⇒ `(16/3)/2 = (x + y)/y`

⇒ `8/3 = (x + y)/y`

⇒ 8y = 3x + 3y

⇒ 8y – 3y = 3x

⇒ 5y = 3x

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

`"dy"/"dt" = 3 * "dx"/dt"`

⇒ `"dy"/"dt" = 3/5 * "dx"/"dt"`

⇒ `"dy"/"dt" = 3/5 * ((-5)/3)`   ......[∵ man is moving in opposite direction]

= – 1 m/s

Hence, the length of shadow is decreasing at the rate of 1 m/s.

Now let u = x + y   .....(u = Distance of the tip of shadow from the light post)

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

`"du"/"dt" = "dx"/"dt" + "dy"/dt"`

= `(- 1 2/3 - 1)`

= `-(5/3 + 1)`

= `- 8/3`

= `-2 2/3` m/s

Hence, the tip of the shadow is moving at the rate of `2 2/3` m/s towards the light post and the length of shadow decreasing at the rate of 1 m/s.