<p>A ladder 15 m long, leans against a wall 7 m high, and a portion of the ladder protrudes over the wall such that its projection along the vertical is 3 m. How fast does the bottom start to slip away from the wall if the ladder slides down along the top edge of the wall at 2 m/s?</p>
Step-by-Step Solution
Key Concept: The ladder maintains constant length while sliding; use the constraint that the ladder length is fixed and applies to both segments (against wall and over it), then differentiate the Pythagorean relation with respect to time to find the rate of horizontal slip.
<p><strong>Step 1: Set up the geometry</strong></p><p>Let x = horizontal distance from wall to bottom of ladder, y = height on wall where ladder touches.</p><p>The ladder has two parts: against wall (length along ladder = √(x² + y²)) and protruding (3 m vertical).</p><p>Total ladder length: √(x² + y²) + √(0² + 3²) = 15 m is incorrect interpretation.</p><p><strong>Correct Setup:</strong> The ladder of length 15 m touches the wall at height y = 7 m and extends 3 m above vertically. If the ladder makes angle θ with ground: the portion against wall has length L₁, and the geometry gives us: L₁² = x² + 7².</p><p><strong>Step 2: Use ladder length constraint</strong></p><p>Since the ladder touches at y = 7 m and protrudes 3 m vertically: the total ladder length is 15 m.</p><p>If the top of ladder is at height 7 m on wall, and bottom is at distance x from wall: x² + 7² = (15 - 3)² = 144</p><p>Therefore: x² + 49 = 144</p><p>x² = 95, so x = √95 m</p><p><strong>Step 3: Differentiate the constraint</strong></p><p>From x² + y² = (ladder segment)², where y = 7 (constant at wall top):</p><p>Actually: x² + 49 = 144 (constant)</p><p>For the moving problem: 2x(dx/dt) + 2y(dy/dt) = 0</p><p>Given dy/dt = -2 m/s (sliding down), at y = 7:</p><p>2x(dx/dt) + 2(7)(-2) = 0</p><p>2√95(dx/dt) = 28</p><p>dx/dt = 28/(2√95) = 14/√95 = 14√95/95 ≈ 1.43 m/s</p><p><strong>∴ Answer: 14/√95 m/s or approximately 1.43 m/s</strong></p>
Correct Answer: 14