Because of a slippery surface, a 13-foot ladder leaning against a wall loses its footing and begins to slide away from the wall. When the foot of the ladder is 5 feet from the base of the wall, the top of the ladder is falling at a rate of 2 feet per second. How fast is the foot of the ladder moving from the wall at this instant? 

We can solve this problem using related rates from calculus. The situation involves a right triangle formed by the ladder, the wall, and the ground. As the top of the ladder slides down the wall, the foot of the ladder slides away from the base of the wall. We'll need to use the Pythagorean theorem and differentiate it with respect to time.

### Given:
- The length of the ladder is constant at \( 13 \) feet.
- The distance between the foot of the ladder and the base of the wall at a particular moment is \( x = 5 \) feet.
- The rate at which the top of the ladder is falling is \( \frac{dy}{dt} = -2 \) feet per second (negative because the height is decreasing).
- We need to find the rate at which the foot of the ladder is moving away from the wall at this instant, \( \frac{dx}{dt} \).

### Step 1: Use the Pythagorean theorem
The Pythagorean theorem gives the relationship between the height of the ladder on the wall \( y \), the distance of the foot of the ladder from the wall \( x \), and the length of the ladder (which is constant):
\[
x^2 + y^2 = 13^2
\]
\[
x^2 + y^2 = 169
\]

### Step 2: Differentiate with respect to time \( t \)
Implicitly differentiate both sides of the equation with respect to time \( t \):
\[
2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0
\]

### Step 3: Plug in known values
At the given moment, \( x = 5 \) feet, and the rate at which the top of the ladder is falling is \( \frac{dy}{dt} = -2 \) feet per second. To find \( y \) at this moment, use the Pythagorean theorem:
\[
5^2 + y^2 = 169
\]
\[
25 + y^2 = 169
\]
\[
y^2 = 144
\]
\[
y = 12 \text{ feet}
\]

Now, substitute \( x = 5 \), \( y = 12 \), and \( \frac{dy}{dt} = -2 \) into the differentiated equation:
\[
2(5) \frac{dx}{dt} + 2(12)(-2) = 0
\]
\[
10 \frac{dx}{dt} - 48 = 0
\]
\[
10 \frac{dx}{dt} = 48
\]
\[
\frac{dx}{dt} = \frac{48}{10} = 4.8 \text{ feet per second}
\]

### Conclusion:
The foot of the ladder is moving away from the wall at a rate of \( 4.8 \) feet per second when the foot of the ladder is 5 feet from the wall.

Would you like more details or have any questions?

Here are 5 related questions:
1. How would the speed change if the ladder were shorter?
2. What happens if the ladder is sliding down faster, say at 3 feet per second?
3. Can the same approach be used for a curved surface instead of a straight wall?
4. How would the problem change if the ladder was 20 feet long?
5. How can this be modeled in real-time for tracking the sliding speed?

**Tip:** In related rates problems, always identify how each quantity changes over time and express relationships using known geometric or physical principles.

Because of a slippery surface, a 13-foot ladder leaning against a wall loses its footing and begins to slide away from the wall. When the foot of the ladder is 5 feet from the base of the wall, the top of the ladder is falling at a rate of 2 feet per second. How fast is the foot of the ladder moving from the wall at this instant?

Solve the related rates problem of a 13-foot ladder sliding down a wall. Learn how to apply the Pythagorean theorem and differentiate to find how fast the foot of the ladder moves when the top falls at 2 feet per second. Suitable for high school students in Grades 11-12.

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