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To solve this problem, we can apply related rates with the Pythagorean theorem, as the ladder forms a right triangle with the wall and the ground.

1. **Define Variables:**
   - Let \( x \) be the distance from the bottom of the ladder to the base of the wall (horizontal distance).
   - Let \( y \) be the height of the ladder's top from the ground (vertical distance).
   - Let \( L = 18 \) feet be the length of the ladder (constant).
   - Given: \( \frac{dy}{dt} = -5 \) feet per second (the negative sign indicates \( y \) is decreasing as the ladder slips down).

2. **Use the Pythagorean theorem:**
   \[
   x^2 + y^2 = L^2
   \]
   Since \( L = 18 \), we have:
   \[
   x^2 + y^2 = 18^2 = 324
   \]

3. **Differentiate with respect to time \( t \):**
   \[
   2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0
   \]
   Simplify by dividing by 2:
   \[
   x \frac{dx}{dt} + y \frac{dy}{dt} = 0
   \]

4. **Substitute Known Values:**
   - At the moment we are interested in, \( x = 12 \) feet.
   - Using the Pythagorean theorem, solve for \( y \) when \( x = 12 \):
     \[
     12^2 + y^2 = 324
     \]
     \[
     144 + y^2 = 324
     \]
     \[
     y^2 = 180
     \]
     \[
     y = \sqrt{180} = 6\sqrt{5} \approx 13.42 \text{ feet}
     \]

   - Substitute \( y = 6\sqrt{5} \) and \( \frac{dy}{dt} = -5 \) ft/s.

5. **Solve for \( \frac{dx}{dt} \):**
   \[
   12 \frac{dx}{dt} + (6\sqrt{5})(-5) = 0
   \]
   \[
   12 \frac{dx}{dt} - 30\sqrt{5} = 0
   \]
   \[
   12 \frac{dx}{dt} = 30\sqrt{5}
   \]
   \[
   \frac{dx}{dt} = \frac{30\sqrt{5}}{12} = \frac{5\sqrt{5}}{2} \approx 5.59 \text{ feet per second}
   \]

So, the rate at which the bottom of the ladder slides along the ground is approximately **5.59 feet per second**.

Would you like more details on any step or have any questions?

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Here are five related questions to deepen understanding:

1. How would the solution change if the ladder were 20 feet long instead of 18 feet?
2. Why do we use a negative sign for the rate \( \frac{dy}{dt} \)?
3. How does the rate at which the bottom slides change if the ladder becomes almost vertical?
4. What happens to \( \frac{dx}{dt} \) if the ladder slips down the wall faster?
5. How can we confirm the calculations without re-solving the problem?

**Tip:** In related rate problems, carefully label each variable and rate of change to avoid confusion.

The top of a 18-foot ladder, leaning against a vertical wall, is slipping down the wall at the rate of 5 feet per second. How fast is the bottom of the ladder sliding along the ground when the bottom of the ladder is 12 feet away from the base of the wall?

Learn how to solve a related rates problem involving an 18-foot ladder slipping down a wall. This problem applies the Pythagorean theorem and differentiation to find the rate at which the ladder's bottom moves along the ground. Suitable for high school calculus students.

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