To solve this problem, let’s analyze it with related rates:

1. Set Up the Problem with Similar Triangles:

   • The 6.2 ft tall man and his shadow form one triangle.
   • The 26 ft tall lamp post and the distance from the lamp post to the tip of the man’s shadow form another triangle.

   Since the triangles are similar, we can set up a proportion: $\frac{26}{x + y} = \frac{6.2}{y}$ where:

   • $x$ is the distance from the lamp post to the man,
   • $y$ is the length of the shadow.

2. Differentiate with Respect to Time:

   • Rewrite the proportion: $26y = 6.2(x + y)$ Simplifying, we get: $26y = 6.2x + 6.2y$ $19.8y = 6.2x$ or $y = \frac{6.2}{19.8}x = \frac{31}{99}x$

3. Differentiating Both Sides with Respect to Time (t): $\frac{dy}{dt} = \frac{31}{99} \frac{dx}{dt}$ Given $\frac{dx}{dt} = 3$ ft/s (since the man is moving away from the lamp post at 3 ft/s), substitute to find $\frac{dy}{dt}$: $\frac{dy}{dt} = \frac{31}{99} \cdot 3 = \frac{93}{99} = \frac{31}{33} \, \text{ft/s}$

4. Interpretation: Since the man is moving away from the light, the length of his shadow is increasing at a rate of $\frac{31}{33}$ ft/s.

Therefore, the correct answer should be:

• The length of his shadow is increasing at a rate of $\frac{31}{33}$ ft/s.

Would you like further details on this solution, or do you have any other questions?

Here are some related questions for further practice:

1. How would the rate change if the man’s height were different?
2. What happens to the shadow’s rate if the man moves faster?
3. How is the rate affected by the height of the lamp post?
4. Could this approach work with a lamp post on a slope?
5. What if the man walked towards the lamp instead of away?

Tip: Always keep track of what’s increasing or decreasing in related rate problems to avoid sign errors!