Given as a man 180cm tall walks at a rate of 2 m/sec away; from a source of light that is 9 m above the ground
As to find the rate at which the length of his shadow increases when he is 3m away from the pole
Suppose AB be the lamp post and let MN be the man of height 180cm or 1.8m.
Suppose AL = l meter and MS be the shadow of the man
Suppose the of the shadow MS = s (shown in the below figure)
Given as the man walks at the speed of 2m/sec
So, dl/dt = 2m/sec ...(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 0.57 m/sec, and it is independent to the current distance of the man from the base of the light.
