Question

A man of height 1.5 meters walks towards a lamp post of height 4.5 meters, at the rate of `(3/4)` meter/sec. Find the rate at which (i) his shadow is shortening (ii) the tip of shadow is moving.

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 = (3)/(4)` is the rate at which the man is moving towards the lamp post.

`dx/dt` is the rate at which his shadow is shortening.

B is the tip of the shadow and it is at a distance of x + y from the post.

∴ `d/dt(x + y) = dx/dt + dy/dt` is the rate at which the tip of the shadow is moving.

From the figure,

`x/(1.5) = (x + y)/(4.5)`

∴ 45x = 15x + 15y
∴ 30x = 15y

∴ x = `(1)/(2)y`

∴ `dx/dt = (1)/(2).dy/dt = (1)/(2)(3/4) = (3/8)`meter/sec, and
`dx/dt + dy/dt = (3)/(8) + (3)/(4) = (9/8)`meter/sec

Hence,

(i) the shadow is shortening at the rate of `(3/8)`meter/sec, and

(ii) the tip of shadow is moving at the rate of `(9/8)`meter/sec.