Given as a man 2 metres high walks at a uniform speed of 5 km/hr. away from a lamp – post 6 metres high
As to find the rate at which the length of his shadow increases
Suppose AB be the lamp post and let MN be the man of height 2m.
Suppose AL = l meter and MS be the shadow of the man
Suppose length of the shadow MS = s (shown in the below figure)
Given as the man walks at the speed of 5km/hr
dl/dt = 5(km/h) ...(i)
Therefore, the rate at which the length of the man's shadow increases will be ds/dt
Considering ΔASB,
Then considering ΔMSN,
Therefore, from equation (ii) and (iii)
By applying derivative with respect to time on both sides
Thus, the rate at which the length of his shadow increases by 2.5 km/hr and it is independent to the current distance of the man from the base of the light.
