A street light is at the top of a $17 \mathrm{ft}$ tall pole. A woman $6 \mathrm{ft}$ tall walks away from the pole with a speed of $4 \mathrm{ft} / \mathrm{sec}$ along a straight path. How fast is the tip of her shadow moving when she is $50 \mathrm{ft}$ from the base of the pole? 5.76 $\frac{\mathrm{ft}}{\mathrm{sec}}$
Solution
Step 1 ：Understand the problem: The woman is walking away from the pole, creating a shadow. We want to find the speed at which the tip of her shadow is moving when she is 50 ft from the pole.
Step 2 ：Set up the relationship: We can set up a right triangle with the pole, the woman, and the tip of her shadow. The pole is 17 ft tall, the woman is 6 ft tall, and the distance between the woman and the pole is changing with time. Let's denote the distance between the woman and the pole as x (ft), and the length of the shadow as s (ft). Since the light is at the top of the pole, the triangles formed by the pole, the woman, and the shadow are similar. Therefore, we have the following relationship: \(\frac{17}{s} = \frac{6}{x}\).
Step 3 ：Differentiate with respect to time: We are given that \(\frac{dx}{dt} = 4\) ft/sec (the speed of the woman), and we want to find \(\frac{ds}{dt}\) (the speed of the shadow tip) when x = 50 ft. We can find \(\frac{ds}{dt}\) by differentiating both sides of the equation with respect to time t: \(\frac{d}{dt} [\frac{17}{s}] = \frac{d}{dt} [\frac{6}{x}]\) which simplifies to \(-\frac{17}{s^2} * \frac{ds}{dt} = -\frac{6}{x^2} * \frac{dx}{dt}\).
Step 4 ：Solve for \(\frac{ds}{dt}\): Rearrange the equation to solve for \(\frac{ds}{dt}\): \(\frac{ds}{dt} = \frac{17}{6} * \frac{s}{x} * \frac{dx}{dt}\).
Step 5 ：Substitute the given values: Substitute x = 50 ft, \(\frac{dx}{dt} = 4\) ft/sec into the equation. To find s when x = 50 ft, we substitute x = 50 ft into the original equation \(\frac{17}{s} = \frac{6}{50}\), and solve for s: s = \(\frac{17}{6} * 50 = 141.67\) ft. Then substitute s = 141.67 ft into the equation for \(\frac{ds}{dt}\): \(\frac{ds}{dt} = \frac{17}{6} * \frac{141.67}{50} * 4 = 5.76\) ft/sec.
Step 6 ：Check the result: The speed of the shadow tip is 5.76 ft/sec when the woman is 50 ft from the pole, which is a reasonable result. Therefore, the solution is correct. \(\boxed{5.76}\) ft/sec is the final answer.
