A ladder 8 m long rests against a vertical wall. If the bottom of the ladder slides away from the wall at the rate (speed) of \( \frac{2}{3} \mathrm{~m} / \mathrm{sec} \), how fast is the angle between the op of the ladder and the wall changing when the angle is \( \frac{\pi}{4} \mathrm{rad} \) ? \( \begin{array}{llll}\text { (A) } \frac{1}{2} \mathrm{rad} / \mathrm{sec} & \text { (B) } 2 \sqrt{2} \mathrm{rad} / \mathrm{s} & \text { (C) } \frac{\sqrt{2}}{4} \mathrm{rad} / & \text { (D) } \frac{\sqrt{2}}{2} \mathrm{rad} / \mathrm{sec}\end{array} \)

To find how fast the angle between the top of the ladder and the wall is changing, we'll use some trigonometry and rates of change. First, let’s denote the length of the ladder as \( L = 8 \, \text{m} \), the angle with the wall as \( \theta \), the height where the ladder contacts the wall as \( h \), and the distance from the wall where the ladder makes contact on the ground as \( x \). From the right triangle formed, we have: \[ \sin(\theta) = \frac{h}{L} \quad \text{and} \quad \cos(\theta) = \frac{x}{L} \] Differentiating with respect to time (t), we use the chain rule: \[ \frac{d\theta}{dt} = \frac{d}{dt}(\arctan(\frac{h}{x})) \] Given that the bottom of the ladder slides away from the wall at \( \frac{dx}{dt} = \frac{2}{3} \, \text{m/sec} \), we'll need to calculate \( h \) and \( x \) when \( \theta = \frac{\pi}{4} \). At \( \theta = \frac{\pi}{4} \), both \( h \) and \( x \) equal \( 4 \, \text{m} \) because of the properties of isosceles right triangles created by the ladder. Using the relationship: \[ \frac{d\theta}{dt} = \frac{1}{1 + \left(\frac{h}{x}\right)^2} \cdot \left(\frac{h \frac{dx}{dt} - x \frac{dh}{dt}}{x^2}\right) \] At \( \theta = \frac{\pi}{4} \) (where \( \tan(\theta) = 1 \)), we have: \[ \frac{d\theta}{dt} = \frac{1}{2} \cdot \frac{8 \cdot \frac{2}{3} - 8 \cdot 0}{16} = \frac{1}{2} \quad \text{rad/sec} \] So the answer is \( \frac{1}{2} \mathrm{ rad/sec} \). The selection based on this is (A) \( \frac{1}{2} \mathrm{ rad/sec} \).