A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 5 ft/s along a straight path. How fast is the tip of his shadow moving when he is 35 ft from the pole?

To solve this problem, let's set up a relationship using similar triangles and related rates.

### Step-by-Step Solution

1. **Define the Variables:**
   - Let \( x \) be the distance between the man and the base of the pole, changing over time. The man is moving away from the pole at a speed of 5 ft/s, so \(\frac{dx}{dt} = 5 \text{ ft/s}\).
   - Let \( y \) be the length of the shadow of the man.

2. **Determine the Rate We Need to Find:**
   - We are asked to find how fast the tip of the shadow is moving. The tip of the shadow is at a distance \( x + y \) from the base of the pole, so we need to determine \(\frac{d(x + y)}{dt}\).

3. **Set Up a Relationship Using Similar Triangles:**
   - The height of the pole (15 ft) and the height of the man (6 ft) form similar triangles with the distances from the pole to the tip of the shadow.
   - Therefore, we can write:
     \[
     \frac{15}{x + y} = \frac{6}{y}
     \]

4. **Solve for \( y \):**
   - Cross-multiply to clear the fractions:
     \[
     15y = 6(x + y)
     \]
   - Distribute \( 6 \):
     \[
     15y = 6x + 6y
     \]
   - Rearrange to solve for \( y \):
     \[
     9y = 6x
     \]
     \[
     y = \frac{2}{3}x
     \]

5. **Differentiate Both Sides with Respect to \( t \):**
   - Since \( y = \frac{2}{3}x \), differentiate with respect to \( t \):
     \[
     \frac{dy}{dt} = \frac{2}{3} \frac{dx}{dt}
     \]
   - Substitute \(\frac{dx}{dt} = 5 \text{ ft/s}\):
     \[
     \frac{dy}{dt} = \frac{2}{3} \times 5 = \frac{10}{3} \text{ ft/s}
     \]

6. **Find the Rate at Which the Tip of the Shadow Moves:**
   - The speed of the tip of the shadow is \(\frac{d}{dt}(x + y)\).
   - Differentiate \( x + y \) with respect to \( t \):
     \[
     \frac{d}{dt}(x + y) = \frac{dx}{dt} + \frac{dy}{dt}
     \]
   - Substitute the values of \(\frac{dx}{dt} = 5\) and \(\frac{dy}{dt} = \frac{10}{3}\):
     \[
     \frac{d}{dt}(x + y) = 5 + \frac{10}{3} = \frac{15 + 10}{3} = \frac{25}{3} \text{ ft/s}
     \]

### Answer:
The tip of the shadow is moving at a rate of \(\frac{25}{3}\) ft/s, or approximately \(8.33\) ft/s, when the man is 35 ft from the pole.

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Would you like further details, or do you have any questions?

Here are five related questions for further practice:

1. How would the answer change if the pole were 20 ft tall instead of 15 ft?
2. What if the man was walking at a different speed, say 6 ft/s? How would that affect the rate of the shadow's tip movement?
3. If the man started closer to the pole, would the rate of the tip of the shadow change initially?
4. How would this problem change if the man’s height were different, say 5 ft?
5. What if the man was moving toward the pole instead of away? How would this affect the rate of the shadow's movement?

**Tip:** In related rates problems involving similar triangles, look for proportions to set up relationships, as they can simplify calculations significantly.

Learn how to determine the speed at which the tip of a shadow moves as a 6-ft man walks away from a 15-ft pole at 5 ft/s. This problem covers related rates and similar triangles, offering a step-by-step solution for advanced high school students.

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