Applications of Derivatives
Rate of Change
Grade 12
Question:
<p>A 2 m ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate 25 cm/s, then the rate (in cm/sec) at which the bottom of the ladder slides away from the wall on the horizontal ground when the top of the ladder is 1 m above the ground is ________ (up to three decimal places).</p>
Step-by-Step Solution
Key Concept: Use the Pythagorean theorem constraint (x² + y² = 4) and differentiate with respect to time to relate the rates of change of horizontal and vertical distances.
<p><strong>Step 1:</strong> Set up the constraint equation. Let y = height of ladder on wall, x = distance of base from wall. Since ladder length = 2 m: x² + y² = 4</p><p><strong>Step 2:</strong> Differentiate both sides with respect to time t: 2x(dx/dt) + 2y(dy/dt) = 0</p><p><strong>Step 3:</strong> Simplify: x(dx/dt) + y(dy/dt) = 0, so dx/dt = -y(dy/dt)/x</p><p><strong>Step 4:</strong> When y = 1 m, find x: x² + 1² = 4 ⟹ x² = 3 ⟹ x = √3 m</p><p><strong>Step 5:</strong> Given dy/dt = -25 cm/s (negative because height decreases). Substitute into the related rate equation:</p><p>dx/dt = -1 × (-25)/√3 = 25/√3 = 25√3/3 cm/s</p><p><strong>Step 6:</strong> Calculate: 25√3/3 = 25 × 1.732.../3 ≈ 14.434 cm/s</p><p>∴ Answer: <strong>14.434</strong></p>
Correct Answer: 14