To solve this problem, we use the principles of related rates and trigonometry.

Step 1: Represent the situation mathematically

Let:

• $\theta$ = the angle between the ladder and the ground.
• $x$ = the horizontal distance from the base of the ladder to the fence.
• $y$ = the vertical height of the top of the ladder above the ground.

From the problem, the length of the ladder is 14 ft. By the Pythagorean theorem: $x^2 + y^2 = 14^2 = 196$

Step 2: Known rates

• The ladder slides horizontally at a rate of $\frac{dx}{dt} = 2 \, \text{ft/sec}$.
• At the instant when the upper end is at the top of the fence, $y = 8 \, \text{ft}$.

Step 3: Solve for $x$ at the moment $y = 8$

Using $x^2 + y^2 = 196$, substitute $y = 8$: $x^2 + 8^2 = 196$ $x^2 + 64 = 196$ $x^2 = 132$ $x = \sqrt{132} = 2\sqrt{33} \, \text{ft}$

Step 4: Relate $\theta$ to $x$ and $y$

From trigonometry: $\sin \theta = \frac{y}{14}$

Differentiate both sides with respect to $t$: $\cos \theta \cdot \frac{d\theta}{dt} = \frac{1}{14} \cdot \frac{dy}{dt}$

We need $\cos \theta$ and $\frac{dy}{dt}$.

Step 5: Solve for $\cos \theta$

$\cos \theta = \frac{x}{14}$ Substitute $x = 2\sqrt{33}$: $\cos \theta = \frac{2\sqrt{33}}{14} = \frac{\sqrt{33}}{7}$

Step 6: Relate $\frac{dy}{dt}$ to $\frac{dx}{dt}$

From the Pythagorean theorem: $x^2 + y^2 = 196$ Differentiate both sides with respect to $t$: $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$ $x \frac{dx}{dt} + y \frac{dy}{dt} = 0$ Solve for $\frac{dy}{dt}$: $\frac{dy}{dt} = -\frac{x}{y} \cdot \frac{dx}{dt}$

Substitute $x = 2\sqrt{33}$, $y = 8$, and $\frac{dx}{dt} = 2$: $\frac{dy}{dt} = -\frac{2\sqrt{33}}{8} \cdot 2 = -\frac{\sqrt{33}}{2} \, \text{ft/sec}$

Step 7: Solve for $\frac{d\theta}{dt}$

From the earlier equation: $\cos \theta \cdot \frac{d\theta}{dt} = \frac{1}{14} \cdot \frac{dy}{dt}$ Substitute $\cos \theta = \frac{\sqrt{33}}{7}$ and $\frac{dy}{dt} = -\frac{\sqrt{33}}{2}$: $\frac{\sqrt{33}}{7} \cdot \frac{d\theta}{dt} = \frac{1}{14} \cdot \left(-\frac{\sqrt{33}}{2}\right)$ $\frac{d\theta}{dt} = \frac{-\frac{\sqrt{33}}{28}}{\frac{\sqrt{33}}{7}}$ $\frac{d\theta}{dt} = -\frac{1}{4} \, \text{rad/sec}$

Final Answer:

The angle $\theta$ is changing at a rate of: $\frac{d\theta}{dt} = -\frac{1}{4} \, \text{rad/sec}$ when the upper end of the ladder is at the top of the fence.

Would you like a detailed explanation of any specific step, or have further questions?
Here are 5 related questions you might find interesting:

1. How does the rate of angle change differ if the ladder is shorter or longer?
2. How would the calculation change if the ladder’s base slides at a variable rate?
3. Can we model the entire motion of the ladder using parametric equations?
4. What happens to $\frac{d\theta}{dt}$ as the base of the ladder moves very far from the fence?
5. How can this problem be visualized geometrically for better understanding?

Tip: Always sketch the problem for a clear understanding of relationships between variables!