A man 2m tall, walks at the rate of `1 2/3m//s e c` towards a street light which is `5 1/3` m above the ground. At what rate is tip of his shadow moving? At what rate is the length of the shadow changing when he is `3 1/(13)m` from the base of the light?

Explanation

Let AB be the street light post and CD be the height of man i.e., CD = 2m.

Let BC= xm, CE=ym and `(dx)/(dt) =-5/3 m//s.`
From `DeltaABE` and `DeltaDCE`, we see that
`DeltaABE ~ DeltaDCE`
`therefore (AB)/(DC) = (BE)/(CE) rArr (16/3)/2 = (x+y)/(y)`
`rArr 16/6 = (x+y)/y`
`rArr 16y = 6x + 6y rArr 10y = 6x`
`rArr y=3/5x`
On differentiating both sides w.r.t. t, we get
`(dy)/(dt) = 3/5. (dx)/(dt) = 3/5. (-12/3)`
Since, man is moving towards the light post
`=3/5.(-5/3) = -1m//s`.
Let z=x+y
Now, differentiating both sides w.r.t. t, we get
`(dz)/(dt) = (dx)/(dt) + (dy)/(dt) = -(5/3+1)`
`=-8/3 = -2 2/3 m//s`.
Hence, the tip of shadow is moving at the rate of `22/3 m//s`
Hence, the tip of shadow is moving at the rate of `22/3 m//s` towards the light source and length of the shadow is decreasing at the rate of `1 m//s`.