This is a classic related rates problem involving a ladder leaning against a wall. Let's break it down and solve the problem step by step.

Problem Details:

• A 13-foot ladder is leaning against a vertical wall.
• The foot of the ladder is being pulled away from the wall at a rate of $\frac{dx}{dt} = 0.5 \text{ ft/s}$.
• You are asked to find the rate at which the top of the ladder is sliding down the wall when the foot of the ladder is 5 feet from the wall.

Step 1: Identify the variables

Let:

• $x$ be the distance from the foot of the ladder to the wall (horizontal distance).
• $y$ be the height of the top of the ladder from the ground (vertical distance).
• $L = 13$ feet is the constant length of the ladder.

We are given:

• $\frac{dx}{dt} = 0.5 \text{ ft/s}$, the rate at which the foot of the ladder is moving away from the wall.
• We need to find $\frac{dy}{dt}$, the rate at which the top of the ladder is sliding down when $x = 5$ feet.

Step 2: Relate $x$, $y$, and $L$ using the Pythagorean Theorem

Since the ladder, the wall, and the ground form a right triangle, we can use the Pythagorean Theorem: $x^2 + y^2 = L^2$ Substitute $L = 13$: $x^2 + y^2 = 13^2 = 169$

Step 3: Differentiate the equation with respect to time $t$

Differentiate both sides of $x^2 + y^2 = 169$ implicitly with respect to $t$: $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$ Simplify: $x \frac{dx}{dt} + y \frac{dy}{dt} = 0$ Now, solve for $\frac{dy}{dt}$: $y \frac{dy}{dt} = -x \frac{dx}{dt}$ $\frac{dy}{dt} = -\frac{x \frac{dx}{dt}}{y}$

Step 4: Find $y$ when $x = 5$ feet

From the Pythagorean Theorem: $x^2 + y^2 = 169$ Substitute $x = 5$: $5^2 + y^2 = 169$ $25 + y^2 = 169$ $y^2 = 144$ $y = 12 \text{ feet}$

Step 5: Calculate $\frac{dy}{dt}$

Now, substitute the values into the equation for $\frac{dy}{dt}$: $\frac{dy}{dt} = -\frac{5 \times 0.5}{12} = -\frac{2.5}{12} = -\frac{5}{24} \text{ ft/s}$

Final Answer:

The top of the ladder is sliding down the wall at a rate of $\frac{5}{24} \text{ ft/s}$ when the foot of the ladder is 5 feet away from the wall. This is approximately $0.2083 \text{ ft/s}$.

Would you like further details on any step or have any questions? Here are some related problems to consider:

1. How would the rate of sliding change if the ladder were longer, say 15 feet?
2. What happens to the rate of sliding when the foot of the ladder is pulled at a faster speed, such as 1 ft/s?
3. Can we find the time it takes for the ladder to completely fall to the ground?
4. What is the velocity of the top of the ladder when the foot is 8 feet away from the wall?
5. How would the problem change if the wall were inclined instead of vertical?

Tip: When solving related rates problems, always make sure to establish a relationship between variables before differentiating.