Applications of Derivatives
Rate of Change of Quantities
Grade 12
Question:
<p>Two ships <em>A</em> and <em>B</em> are sailing straight away from a fixed point <em>O</em> along routes such that \(\angle AOB\) is always 120°. At a certain instance, \(OA = 8\) km, \(OB = 6\) km and the ship <em>A</em> is sailing at the rate of 20 km/h while the ship <em>B</em> is sailing at the rate of 30 km/h. Then the distance between <em>A</em> and <em>B</em> is changing at the rate (in km/h)</p>
<p>\(\dfrac{260}{37}\)</p>
<p>\(\dfrac{80}{\sqrt{37}}\)</p>
<p>\(\dfrac{80}{37}\)</p>
<p>\(\dfrac{80}{37}\)</p>
Step-by-Step Solution
Key Concept: Use the cosine rule to express distance AB in terms of OA and OB, then differentiate with respect to time to find the rate of change. The constant angle of 120° is crucial—only the distances OA and OB change with time.
<p><strong>Step 1:</strong> Apply the cosine rule to find AB in terms of OA and OB:</p><p>AB² = OA² + OB² − 2(OA)(OB)cos(120°)</p><p>AB² = OA² + OB² − 2(OA)(OB)(−1/2)</p><p>AB² = OA² + OB² + (OA)(OB)</p><p><strong>Step 2:</strong> Substitute the given values at the instant:</p><p>AB² = 8² + 6² + (8)(6) = 64 + 36 + 48 = 148</p><p>AB = √148 = 2√37 km</p><p><strong>Step 3:</strong> Differentiate both sides with respect to time t:</p><p>2(AB)(dAB/dt) = 2(OA)(dOA/dt) + 2(OB)(dOB/dt) + (OB)(dOA/dt) + (OA)(dOB/dt)</p><p><strong>Step 4:</strong> Substitute dOA/dt = 20 km/h, dOB/dt = 30 km/h, OA = 8, OB = 6, AB = 2√37:</p><p>2(2√37)(dAB/dt) = 2(8)(20) + 2(6)(30) + (6)(20) + (8)(30)</p><p>4√37(dAB/dt) = 320 + 360 + 120 + 240 = 1040</p><p>dAB/dt = 1040/(4√37) = 260/√37 = 260√37/37 ≈ 42.8 km/h</p><p>∴ Answer: B</p>
Correct Answer: B