1. **State the problem:** A street light is on top of a 20 ft pole. A 6 ft tall girl walks away from the pole at 6 ft/sec. We want to find: - The rate at which the tip of her shadow moves away from the pole when she is 49 ft away. - The rate at which her shadow lengthens at that moment. 2. **Set variables:** Let $x$ = distance of the girl from the pole (ft), $s$ = length of the shadow (ft). Given: $\frac{dx}{dt} = 6$ ft/sec, height of pole = 20 ft, height of girl = 6 ft. 3. **Use similar triangles:** The ratio of heights to shadow lengths is equal: $$\frac{20}{x + s} = \frac{6}{s}$$ Cross-multiplied: $$20s = 6(x + s)$$ $$20s = 6x + 6s$$ 4. **Simplify and solve for $s$:** $$20s - 6s = 6x$$ $$14s = 6x$$ $$s = \frac{6}{14}x = \frac{3}{7}x$$ 5. **Differentiate with respect to time $t$:** $$\frac{ds}{dt} = \frac{3}{7} \frac{dx}{dt}$$ Given $\frac{dx}{dt} = 6$ ft/sec, $$\frac{ds}{dt} = \frac{3}{7} \times 6 = \frac{18}{7} \approx 2.57 \text{ ft/sec}$$ 6. **Find the rate at which the tip of the shadow moves away from the pole:** The tip of the shadow is at distance $x + s$ from the pole. $$\frac{d}{dt}(x + s) = \frac{dx}{dt} + \frac{ds}{dt} = 6 + \frac{18}{7} = \frac{42}{7} + \frac{18}{7} = \frac{60}{7} \approx 8.57 \text{ ft/sec}$$ **Final answers:** - The tip of the shadow moves away from the pole at approximately $8.57$ ft/sec when the girl is 49 ft away. - The shadow lengthens at approximately $2.57$ ft/sec at that moment.