Applications of Derivatives
Rate of change of distance
Grade 12
Question:
<p>Two ships \(A\) and \(B\) start from points \(A\) and \(B\) respectively. Ship \(A\) travels \(20t\) and ship \(B\) travels \(30t\) in time \(t\). The angle between their directions is \(120°\). The rate of change of the distance between \(A'\) and \(B'\) at \(t = 0\) is:</p>
<p>\(\dfrac{260}{\sqrt{37}}\)</p>
<p>\(\dfrac{130}{\sqrt{37}}\)</p>
<p>\(\dfrac{520}{\sqrt{37}}\)</p>
<p>\(\dfrac{260}{37}\)</p>
Step-by-Step Solution
Key Concept: Use the law of cosines to express distance between ships as a function of time, then differentiate with respect to t and evaluate at t=0. The rate of change of distance is found by d(AB)/dt, which requires careful application of the chain rule to the distance formula.
<p><strong>Step 1:</strong> Let position of ship A at time t: distance from A = 20t</p><p>Let position of ship B at time t: distance from B = 30t</p><p>Angle between their paths = 120°</p><p><strong>Step 2:</strong> Using the law of cosines, the distance d between the two ships is:</p><p>d² = (20t)² + (30t)² - 2(20t)(30t)cos(120°)</p><p>d² = 400t² + 900t² - 2(20t)(30t)(-1/2)</p><p>d² = 400t² + 900t² + 600t²</p><p>d² = 1900t²</p><p>d = t√1900 = 10t√19</p><p><strong>Step 3:</strong> Differentiate with respect to t:</p><p>dd/dt = 10√19</p><p><strong>Step 4:</strong> At t = 0:</p><p>dd/dt|_{t=0} = 10√19 units per unit time</p><p>∴ Answer: A</p>
Correct Answer: A