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

To solve the problem step by step, we will use the concept of similar triangles and related rates.

### Given:
- Height of the man (AB) = 2 m
- Height of the street light (CD) = \(5 \frac{1}{3} = \frac{16}{3}\) m
- Rate at which the man walks towards the street light = \(1 \frac{2}{3} = \frac{5}{3}\) m/s (towards the light, hence negative rate)
- Distance of the man from the base of the light (when we need to find the rate of shadow) = \(3 \frac{1}{13} = \frac{40}{13}\) m

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