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To solve this problem, we need to apply related rates using similar triangles.

### Problem Breakdown:
1. We have a 26 ft tall street light and a 4 ft tall girl walking away from the pole at a speed of 3 ft/sec.
2. We need to find the rate at which the tip of her shadow is moving away from the top of the pole when she is 22 ft away from the pole.
3. The variables are:
   - Distance of the girl from the pole: \( x \)
   - Length of her shadow: \( s \)
   - Total distance from the pole to the tip of her shadow: \( L = x + s \)

### Step-by-Step Solution:
1. **Set up similar triangles:**
   Since the girl and the street light cast proportional shadows, we can set up a proportion based on their heights and shadow lengths.
   \[
   \frac{26}{L} = \frac{4}{s}
   \]
   where \( L = x + s \) is the distance from the pole to the tip of the shadow.

2. **Solve for \( s \) in terms of \( x \):**
   Rewrite the proportion:
   \[
   26s = 4L
   \]
   Substitute \( L = x + s \):
   \[
   26s = 4(x + s)
   \]
   Expanding and solving for \( s \):
   \[
   26s = 4x + 4s
   \]
   \[
   22s = 4x
   \]
   \[
   s = \frac{2x}{11}
   \]

3. **Differentiate with respect to time \( t \):**
   To find the rate at which the tip of the shadow is moving, we differentiate \( L = x + s \) with respect to time \( t \):
   \[
   \frac{dL}{dt} = \frac{dx}{dt} + \frac{ds}{dt}
   \]
   We know \( \frac{dx}{dt} = 3 \) ft/sec.

4. **Differentiate \( s = \frac{2x}{11} \):**
   Using \( s = \frac{2x}{11} \), we differentiate with respect to \( t \):
   \[
   \frac{ds}{dt} = \frac{2}{11} \cdot \frac{dx}{dt}
   \]
   \[
   \frac{ds}{dt} = \frac{2}{11} \cdot 3 = \frac{6}{11} \text{ ft/sec}
   \]

5. **Calculate \( \frac{dL}{dt} \):**
   Substitute \( \frac{dx}{dt} = 3 \) and \( \frac{ds}{dt} = \frac{6}{11} \):
   \[
   \frac{dL}{dt} = 3 + \frac{6}{11} = \frac{33 + 6}{11} = \frac{39}{11} \approx 3.545 \text{ ft/sec}
   \]

### Final Answer:
The rate at which the tip of her shadow is moving away from the top of the pole is approximately **3.545 ft/sec**.

Would you like more details on any part of this explanation or have any questions?

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### Related Questions:
1. How would the rate change if the height of the pole or the girl were different?
2. What if the girl’s speed varied as she walked away from the pole? How would that affect the rate?
3. How do similar triangles help in setting up related rates problems?
4. What if we were asked for the rate at which the shadow lengthens, rather than the tip of the shadow’s rate?
5. How does differentiating with respect to time help us find instantaneous rates in physics problems?

### Tip:
In related rates problems, always look for a relationship between variables, often using geometry or proportions, before differentiating.

A street light is at the top of a 26 ft pole. A 4 ft tall girl walks along a straight path away from the pole with a speed of 3 ft/sec. At what rate is the tip of her shadow moving away from the light (i.e., away from the top of the pole) when the girl is 22 ft away from the pole?

This problem explores related rates involving a street light and a girl walking away from it. Using similar triangles and differentiation, learn to find the rate at which the tip of her shadow moves from the light when she is 22 ft from the pole. Suitable for advanced high school students in Grades 11-12.

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