Applications of Derivatives
Rate of change / Differential equation
Grade 12
Question:
<p>We have \(\dfrac{d(p(t))}{dt} = \dfrac{1}{2}p(t) - 450\). At \(t = 0\), \(p(0) = 50\). Find the time \(t\) when \(P(t) = 0\).</p>
<p>\(2\ln 18\)</p>
<p>\(\ln 18\)</p>
<p>\(2\ln 9\)</p>
<p>\(\ln 9\)</p>
Step-by-Step Solution
Key Concept: This is a first-order linear differential equation. Solve it by separation of variables or recognizing it as dp/dt = (1/2)(p - 900), which has equilibrium at p = 900. The general solution involves an exponential term that decays to this equilibrium.
<p><strong>Step 1:</strong> Rewrite the differential equation: dp/dt = (1/2)p - 450 = (1/2)(p - 900)</p><p><strong>Step 2:</strong> Separate variables: dp/(p - 900) = (1/2)dt</p><p><strong>Step 3:</strong> Integrate both sides: ln|p - 900| = (1/2)t + C</p><p><strong>Step 4:</strong> General solution: p(t) = 900 + Ae^(t/2), where A is the constant of integration</p><p><strong>Step 5:</strong> Apply initial condition p(0) = 50: 50 = 900 + A, so A = -850</p><p><strong>Step 6:</strong> Particular solution: p(t) = 900 - 850e^(t/2)</p><p><strong>Step 7:</strong> Set p(t) = 0: 0 = 900 - 850e^(t/2) ⟹ e^(t/2) = 900/850 = 18/17</p><p><strong>Step 8:</strong> Solve for t: t/2 = ln(18/17) ⟹ t = 2ln(18/17)</p><p>∴ Answer: A</p>
Correct Answer: A