Question

A man 2 m tall walks away at a rate of 4 m/min from source of light 6 m high from the ground How fast is the length of his shadow changing?

Answer 1

see diagram, A man of 2m walks away at the rate of 4m/min from the source of light 6m high from the ground.

here, dl/dt = 4m/min

we have to find out , ds/dt

see other diagram,

here ∆SBA ~ ∆SNM

so, AB/MN = AS/MS

⇒6m/2m = (l + s)/s

⇒3 = (l + s)/s

⇒3s = l + s

⇒2s = l

differentiating both sides with respect to time,

i.e., 2ds/dt = dl/dt

ds/dt =(1/2) dl/dt

= 1/2 × 4m/min = 2m/min

hence rate of change in length of shadow is 2m/min.