To solve this problem, we need to use related rates in calculus. Let's define the variables and set up the relationships between them.

1. Let $L$ be the distance between the man and the lamp post.
2. Let $S$ be the length of the man's shadow.
3. Let $x$ be the total distance from the lamp post to the tip of the shadow, so $x = L + S$.

We know that the man is walking away from the lamp post at a speed of 7 ft/s, so $\frac{dL}{dt} = 7$ ft/s.

Given:

• Height of the lamp post = 13 ft
• Height of the man = 6 ft

Using similar triangles, we can write the following relationship between the heights and lengths: $\frac{13}{x} = \frac{6}{S}$ Solving for $x$: $13S = 6x \implies x = \frac{13}{6} S$ Since $x = L + S$: $\frac{13}{6} S = L + S$ Solving for $L$: $\frac{13}{6} S - S = L \implies \frac{13}{6} S - \frac{6}{6} S = L \implies \frac{7}{6} S = L \implies S = \frac{6}{7} L$

Now we differentiate both sides with respect to time $t$: $\frac{dS}{dt} = \frac{6}{7} \frac{dL}{dt}$ We know $\frac{dL}{dt} = 7$ ft/s: $\frac{dS}{dt} = \frac{6}{7} \times 7 = 6 \text{ ft/s}$

Thus, the length of the man's shadow is changing at a rate of 6 ft/s when he is 4 ft from the lamp post.

Would you like a more detailed explanation or have any other questions?

Here are 5 numbered questions you might ask next:

1. How do related rates problems work in general?
2. Can you solve another related rates problem?
3. How do you set up and solve similar triangles in word problems?
4. Can you explain the concept of differentiation in calculus?
5. How do you handle more complex related rates problems involving multiple variables?

Tip: When solving related rates problems, always start by identifying and relating all the variables involved before differentiating.