1. **State the problem:** A person 2 meters tall walks away from a streetlight 8 meters high. The shadow lengthens at a rate of $\frac{4}{9}$ meters per second. We need to find the rate at which the person is walking (rate of change of distance from the pole). 2. **Set variables:** Let $p$ = distance from the pole to the person (meters), $s$ = length of the shadow (meters). 3. **Given:** Height of pole = 8 m, Height of person = 2 m, $\frac{ds}{dt} = \frac{4}{9}$ m/s. 4. **Use similar triangles:** The triangles formed by the pole and shadow, and the person and shadow, are similar. $$\frac{8}{p + s} = \frac{2}{s}$$ Cross-multiplied: $$8s = 2(p + s)$$ Simplify: $$8s = 2p + 2s$$ $$8s - 2s = 2p$$ $$6s = 2p$$ $$p = 3s$$ 5. **Differentiate with respect to time $t$:** $$\frac{dp}{dt} = 3 \frac{ds}{dt}$$ 6. **Substitute known rate:** $$\frac{dp}{dt} = 3 \times \frac{4}{9} = \frac{12}{9} = \frac{4}{3}$$ 7. **Interpretation:** The person is walking away from the pole at a rate of $\frac{4}{3}$ meters per second. **Final answer:** $$\boxed{\frac{4}{3} \text{ meters per second}}$$