To solve this problem, we use related rates and similar triangles.

Step 1: Understand the situation

We have:

1. A person who is 2 meters tall.
2. A streetlight that is 8 meters tall.
3. The person is walking away from the base of the streetlight at a speed of $\frac{dx}{dt} = 1.5 \, \text{m/s}$, where $x$ is the horizontal distance of the person from the streetlight's base.
4. Let $s$ be the length of the shadow.

We aim to find $\frac{ds}{dt}$, the rate at which the shadow's length is increasing, when the person is 5 meters away from the streetlight ($x = 5$).

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Step 2: Establish a geometric relationship

From similar triangles: $\frac{\text{Height of person}}{\text{Shadow length}} = \frac{\text{Height of streetlight}}{\text{Distance from streetlight to shadow tip}}.$ Substitute values: $\frac{2}{s} = \frac{8}{x + s}.$ Cross-multiply: $2(x + s) = 8s.$ Simplify: $2x + 2s = 8s.$ $2x = 6s \quad \Rightarrow \quad s = \frac{x}{3}.$

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Step 3: Differentiate the relationship

Differentiate $s = \frac{x}{3}$ with respect to $t$: $\frac{ds}{dt} = \frac{1}{3} \frac{dx}{dt}.$

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Step 4: Substitute known values

At $x = 5$ and $\frac{dx}{dt} = 1.5 \, \text{m/s}$: $\frac{ds}{dt} = \frac{1}{3}(1.5) = 0.5 \, \text{m/s}.$

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Final Answer:

The length of the shadow is increasing at a rate of 0.5 m/s when the person is 5 meters away from the base of the streetlight.

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Do you want a detailed step-by-step explanation or further clarification?

Related Questions:

1. What if the height of the person changes? How does it affect $\frac{ds}{dt}$?
2. How would the result change if the streetlight’s height were different?
3. What happens to the rate of the shadow’s increase if the walking speed is doubled?
4. Can you derive the formula for $\frac{ds}{dt}$ in general terms (not plugging in numbers)?
5. What is the shadow’s length when the person is 5 meters away?

Tip: Always identify relationships between variables geometrically before differentiating in related rate problems!