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This problem involves related rates and similar triangles.

Let's define the variables:

- Let \( x \) be the distance between the woman and the streetlight.
- Let \( y \) be the length of her shadow.
- The height of the streetlight is \( 30 \) feet.
- The woman’s height is \( 5 \) feet.

The relationship between \( x \) and \( y \) can be set up by using similar triangles.

### Step 1: Setting up the Similar Triangles
The two triangles are:
1. The large triangle formed by the streetlight and the tip of the shadow.
2. The smaller triangle formed by the woman and her shadow.

Using similar triangles:
\[
\frac{30}{x + y} = \frac{5}{y}
\]

### Step 2: Solving for \( y \) in Terms of \( x \)
Cross-multiplying gives:
\[
30y = 5(x + y)
\]

Expanding and simplifying:
\[
30y = 5x + 5y
\]
\[
25y = 5x
\]
\[
y = \frac{1}{5}x
\]

### Step 3: Differentiating with Respect to Time
Since we want to find the rate at which the tip of her shadow is moving, we need the rate of change of \( x + y \) with respect to time, \( \frac{d}{dt}(x + y) \).

Differentiating both sides of \( y = \frac{1}{5}x \) with respect to \( t \):
\[
\frac{dy}{dt} = \frac{1}{5} \frac{dx}{dt}
\]

### Step 4: Substituting Known Values
- The woman walks toward the light at \( \frac{dx}{dt} = -7 \) ft/s (negative because \( x \) is decreasing).
- When \( x = 11 \) feet, substitute to find \( y \):
  \[
  y = \frac{1}{5}(11) = \frac{11}{5} = 2.2 \text{ feet}
  \]

Now, calculate \( \frac{dy}{dt} \):
\[
\frac{dy}{dt} = \frac{1}{5}(-7) = -\frac{7}{5} = -1.4 \text{ ft/s}
\]

### Step 5: Finding \( \frac{d}{dt}(x + y) \)
Since \( x + y \) represents the position of the tip of her shadow,
\[
\frac{d}{dt}(x + y) = \frac{dx}{dt} + \frac{dy}{dt}
\]
Substitute \( \frac{dx}{dt} = -7 \) and \( \frac{dy}{dt} = -1.4 \):
\[
\frac{d}{dt}(x + y) = -7 - 1.4 = -8.4 \text{ ft/s}
\]

### Conclusion
The tip of her shadow is moving at a rate of \( 8.4 \) ft/s away from the light source.

---

Would you like further details on any of the steps or have questions? 

Here are 5 related questions to consider:

1. How would the rate change if the woman were walking away from the streetlight?
2. What if the streetlight height or her walking speed changed? How would that affect the result?
3. How can similar triangle principles be applied in other related rate problems?
4. What if the woman’s height were different? How would the calculations change?
5. How would the shadow rate change if the woman were closer to the streetlight?

**Tip:** In related rates, identifying similar triangles and relationships is key to setting up the problem correctly.

A 5-foot-tall woman walks at 7 ft/s toward a street light that is 30 ft above the ground. What is the rate of change of the length of her shadow when she is 11 ft from the street light? At what rate is the tip of her shadow moving?

This problem involves related rates and similar triangles to determine the rate of change of a shadow’s length. A 5-foot-tall woman walks at 7 ft/s toward a streetlight that is 30 feet above the ground. The solution calculates the rate at which the shadow tip moves when the woman is 11 feet from the streetlight.

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