Differential Equations
Applications of differential equations
Grade 12
Question:
<p>A and B are two separate reservoirs of water. The capacity of A is double that of B. Both the reservoirs are filled completely with water. Water is released simultaneously from both the reservoirs. For each of the reservoirs, the rate of flow out at any instant is proportional to the quantity of water left in the reservoir. After one hour, the quantity of water in A is 1.5 times the quantity of water in B. After how many hours from the time of release of water, do both A and B have the same quantity of water?</p>
<p>\(t = \frac{\log 2}{\log \frac{4}{3}}\) hours</p>
<p>\(t = \frac{\log 2}{\log \frac{3}{2}}\) hours</p>
<p>\(t = \frac{\log 4}{\log \frac{4}{3}}\) hours</p>
<p>\(t = \frac{2\log 2}{\log \frac{3}{2}}\) hours</p>
Step-by-Step Solution
Key Concept: Model each reservoir's water content as an exponential decay function Q(t) = Q₀e^(-kt), where k is the decay constant. Since decay rates differ between reservoirs, use the given condition after 1 hour to find the relationship between their constants, then solve for when Q_A(t) = Q_B(t).
<p><strong>Step 1: Set up differential equations</strong></p><p>For each reservoir: dQ/dt = -kQ, where k > 0 is the proportionality constant (which may differ for each reservoir).</p><p>Solution: Q(t) = Q₀e^(-kt)</p><p><strong>Step 2: Define initial conditions</strong></p><p>Let capacity of B = C, then capacity of A = 2C.</p><p>Q_A(0) = 2C, Q_B(0) = C</p><p>Q_A(t) = 2Ce^(-k_A·t), Q_B(t) = Ce^(-k_B·t)</p><p><strong>Step 3: Use the condition at t = 1 hour</strong></p><p>After 1 hour: Q_A(1) = 1.5·Q_B(1)</p><p>2Ce^(-k_A) = 1.5·Ce^(-k_B)</p><p>2e^(-k_A) = 1.5e^(-k_B)</p><p>e^(k_B - k_A) = 0.75</p><p>k_B - k_A = ln(0.75) = -ln(4/3)</p><p><strong>Step 4: Find when quantities are equal</strong></p><p>Q_A(t) = Q_B(t)</p><p>2Ce^(-k_A·t) = Ce^(-k_B·t)</p><p>2 = e^((k_A - k_B)·t)</p><p>2 = e^(ln(4/3)·t)</p><p>ln(2) = t·ln(4/3)</p><p>t = ln(2)/ln(4/3) = ln(2)/(ln(4) - ln(3)) = ln(2)/(2ln(2) - ln(3))</p><p><strong>Step 5: Simplify</strong></p><p>t = ln(2)/(2ln(2) - ln(3)) ≈ 0.693/(1.386 - 1.099) ≈ 0.693/0.287 ≈ <strong>2.4 hours</strong></p><p>Or expressed exactly: t = log₍₄/₃₎(2) hours</p><p>∴ Answer: A</p>
Correct Answer: A