Differential Equations
Applications of differential equations
Grade 12
Question:
<p>Water is drained from a vertical cylindrical tank by opening a valve at the base of the tank. It is known that the rate at which the water level drops is proportional to the square root of water depth <em>y</em>, where the constant of proportionality \(k > 0\), depends on the acceleration due to gravity and the geometry of the hole. If <em>t</em> is measured in minutes and \(k = \dfrac{1}{15}\) then the time (in hour) to drain the tank if the water is 4m deep to start with is</p>
Step-by-Step Solution
Key Concept: The draining rate satisfies dy/dt = -k√y (negative because depth decreases). Separate variables and integrate to find time by solving ∫dy/√y = -k∫dt, then evaluate the definite integral from y=4 to y=0.
<p><strong>Step 1: Set up the differential equation</strong></p><p>Rate of water level drop is proportional to √y:</p><p>dy/dt = -k√y, where k = 1/15 (negative sign because water is draining)</p><p><strong>Step 2: Separate variables</strong></p><p>dy/√y = -k dt</p><p>y^(-1/2) dy = -(1/15) dt</p><p><strong>Step 3: Integrate both sides</strong></p><p>∫y^(-1/2) dy = -∫(1/15) dt</p><p>2y^(1/2) = -(1/15)t + C</p><p><strong>Step 4: Apply initial condition</strong></p><p>At t = 0, y = 4:</p><p>2√4 = C</p><p>C = 4</p><p>So: 2√y = 4 - (t/15)</p><p><strong>Step 5: Find time to drain (y = 0)</strong></p><p>2√0 = 4 - (t/15)</p><p>0 = 4 - (t/15)</p><p>t/15 = 4</p><p>t = 60 minutes</p><p><strong>Step 6: Convert to hours</strong></p><p>t = 60/60 = 1 hour</p><p>∴ <strong>Answer: 1 hour</strong></p>
Correct Answer: 1