Given: a 2m tall man walks at the rate of m/s towards a m tall street light.

To find: the rate at which the tip of the shadow is moving and also to find the rate at which the length of the shadow changing when he is m from the base of the light.

Explanation:

Here the street light is AB =

And man is DC = 2m

Let BC = x m and CE = y m

And given man walks towards the streetlight at a rate of , and it will negative because the man is moving towards the street light

Hence ………(i)

Now consider ΔABE and ΔDCE

∠DEC = ∠AEB (same angle)

∠DCE = ∠ABD = 90°

Hence by AA similarity,

ΔABE≅ΔDCE

Hence by CPCT,

Now substituting the values from the figure, we get

⇒ 16y = 6(x+y)

⇒ 16y = 6x+6y

⇒ 16y-6y = 6x

⇒ 10y = 6x

Now applying first derivative with respect to t, we get

Now substituting the value from equation (i) in above equation, we get

Hence the rate at which the tip of the shadow is moving is -1m/s, i.e., the length of the shadow is decreasing at the rate of 1m/s.

Let BE = z

So from fig,

z = x+y

Applying the first derivative on above equation with respect to t, we get

Now substituting the corresponding values, we get

Hence the tip of the shadow is moving at the rate of towards the light source.