1. **Problem statement:** A 2-meter tall student walks away from a 5-meter tall light at 1.3 m/s. We want to find (a) the rate at which the tip of the shadow moves and (b) the rate at which the shadow length changes when the student is 3 meters from the light. 2. **Set variables:** Let $x$ = distance of student from light base (m), $s$ = length of shadow (m). 3. **Known rates:** $\frac{dx}{dt} = 1.3$ m/s (student walking away). 4. **Relation from similar triangles:** The lamp height to total shadow length equals student height to shadow length: $$\frac{5}{x+s} = \frac{2}{s}$$ 5. **Solve for $s$:** Cross multiply: $$5s = 2(x+s)$$ $$5s = 2x + 2s$$ $$5s - 2s = 2x$$ $$3s = 2x$$ $$s = \frac{2}{3}x$$ 6. **Differentiate both sides w.r.t. time $t$:** $$\frac{ds}{dt} = \frac{2}{3} \frac{dx}{dt}$$ 7. **Calculate $\frac{ds}{dt}$:** $$\frac{ds}{dt} = \frac{2}{3} \times 1.3 = \frac{2.6}{3} \approx 0.867 \text{ m/s}$$ 8. **Find rate of tip of shadow moving:** The tip of the shadow is at distance $x + s$ from the light, so $$\frac{d}{dt}(x+s) = \frac{dx}{dt} + \frac{ds}{dt} = 1.3 + 0.867 = 2.167 \text{ m/s}$$ **Final answers:** (a) The tip of the shadow moves at approximately $2.167$ m/s. (b) The shadow length increases at approximately $0.867$ m/s.