Question

A man 160 cm tall walks away from a source of light situated at the top of the pole 6m high at the rate of 1.1 m/sec. How fast is the length of the shadow increasing when he is 1m away from the pole.

Answer 1

Let the distance of the man from source of light be x

the distance of length of shadow from the man be y

Seeing figure attached, it is clear that triangles ABC and PQC are similar, thus, we can write the ratio of sides as

AB/BC = PQ/QC

600/(x + y) = 160/y

600y = 160x + 160y

160x = 440y

y = 4/11 x...(1)

Now, to get the rate of change of shadow, we want dy/dt

Differentiating (1) wrt to t we get

dy/dt = 4/11 dx/dt

As the man walks at 1.1 m/s or 110 cm/s i.e dx/dt, we have

dy/dt = 4/11 (110) = 40 cm/s [constant]

So, the length of the shadow is moving at 0.4 m/s and iis independent of the man's position from the light.