1. **Problem statement:** A street light is at the top of a 17 ft pole. A 4 ft tall girl walks away from the pole at 3 ft/sec. We want to find the rate at which the tip of her shadow moves away from the light when she is 34 ft away from the pole, and also how fast her shadow lengthens. 2. **Set variables:** Let $x$ = distance of the girl from the pole (ft), $s$ = length of her shadow (ft). Given $\frac{dx}{dt} = 3$ ft/sec. 3. **Similar triangles relation:** The pole and shadow form similar triangles: $$\frac{17}{s+x} = \frac{4}{s}$$ Cross multiply: $$17s = 4(s + x)$$ $$17s = 4s + 4x$$ Simplify: $$17s - 4s = 4x$$ $$13s = 4x$$ So, $$s = \frac{4x}{13}$$ 4. **Differentiate both sides with respect to $t$:** $$\frac{ds}{dt} = \frac{4}{13} \frac{dx}{dt}$$ Since $\frac{dx}{dt} = 3$, $$\frac{ds}{dt} = \frac{4}{13} \times 3 = \frac{12}{13}$$ ft/sec This is the rate at which the shadow lengthens. 5. **Rate of tip of shadow moving away from the light:** The tip of the shadow is $x + s$ from the pole: $$\frac{d}{dt}(x + s) = \frac{dx}{dt} + \frac{ds}{dt} = 3 + \frac{12}{13} = \frac{39}{13} + \frac{12}{13} = \frac{51}{13}$$ ft/sec **Final answers:** - Rate tip of shadow moves away from light: $\frac{51}{13}$ ft/sec - Rate shadow lengthens: $\frac{12}{13}$ ft/sec