A 53 ft long leans against the wall and slides towards the floor. The foot of the ladder is sliding way from the base of the wall at a rate of \( 4.5 \mathrm{ft} / \mathrm{sec} \). How fast is the top of the ladder sliding down the wall when the top of the ladder is \( 13 \mathrm{ft} \mathrm{from} \mathrm{the} \mathrm{ground?} \mathrm{(2dp)} 88] \)

To find out how fast the top of the ladder is sliding down the wall, we can use related rates in calculus. First, we denote the height of the ladder on the wall as \( y \), the distance from the wall to the foot of the ladder as \( x \), and the length of the ladder as \( L \). According to the Pythagorean theorem, we have: \[ x^2 + y^2 = L^2 \] Given that \( L = 53 \) ft, \( \frac{dx}{dt} = 4.5 \) ft/sec, and \( y = 13 \) ft, we need to first find \( x \) when \( y = 13 \). Using the Pythagorean theorem: \[ x^2 + 13^2 = 53^2 \] Calculating \( x^2 \): \[ x^2 + 169 = 2809 \implies x^2 = 2640 \implies x = \sqrt{2640} \approx 51.4 \, \text{ft} \] Now we differentiate the equation \( x^2 + y^2 = 53^2 \) with respect to time \( t \): \[ 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \] Solving for \( \frac{dy}{dt} \): \[ \frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt} \] Substituting the values: \[ \frac{dy}{dt} = -\frac{51.4}{13} \cdot 4.5 \] Calculating \( \frac{dy}{dt} \): \[ \frac{dy}{dt} \approx -\frac{51.4 \cdot 4.5}{13} \approx -17.5 \, \text{ft/sec} \] So the top of the ladder is sliding down the wall at approximately **17.5 ft/sec** when it is 13 ft from the ground. --- Fun Fact: Did you know that the ladder problem is often used in physics and math to demonstrate related rates — a concept that shows how different quantities change with respect to time? It's like a dance where each move affects the others! Real-World Twist: If you've ever watched a firefighter using a ladder, you've likely seen a live demonstration of this concept! When they're on the move and helping save the day, understanding how their equipment shifts is vital for ensuring safety and stability.