Given: man 2 metres high walks at a uniform speed of 5 km/hr away from a lamp - post 6 metres high
To find the rate at which the length of his shadow increases
Let AB be the lamp post and let MN be the man of height 2m.
Let AL = l meter and MS be the shadow of the man
Let length of the shadow MS = s
(as shown in the below figure)
Given man walks at the speed of 5 km/hr
So the rate at which the length of the man’s shadow increases will be \(\frac{ds}{dt}\)
Consider ΔASB
Now consider ΔMSN, we get
So from equation(ii) and (iii),
Applying derivative with respect to time on both sides we get,
Hence 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.
