A conical tank with height 12 meters and base radius 4 meters is being drained at a rate of 3 m³/min. How fast is the water level decreasing when the water is 6 meters deep?
["Title: Calculating the Rate at Which Water Level Drops in a Conical Tank", "When water drains from a conical tank at a steady rate, determining how quickly the water level drops is essential for engineering, reservoir management, and fluid dynamics. In this article, we analyze a specific scenario: a conical tank with a height of 12 meters and a base radius of 4 meters, currently filled with water at a depth of 6 meters, draining at 3 cubic meters per minute. We’ll calculate the rate at which the water level decreases at that moment.", "---", "### Understanding the Geometry of the Conical Tank", "The tank is a right circular cone with:", "- Height ( H = 12~\ ext{m} )\n- Base radius ( R = 4~\ ext{m} )\n- Current water depth ( h = 6~\ ext{m} )\n- Drain rate ( \frac{dV}{dt} = -3~\ ext{m}^3/\ ext{min} ) (negative because volume is decreasing)", "Because the tank is conical, the radius and height of the water surface are proportional. Using similar triangles:", "[\n\frac{r}{R} = \frac{h}{H} \implies r = \frac{R}{H} h = \frac{4}{12} h = \frac{1}{3} h\n]", "So at ( h = 6~\ ext{m} ), the radius of the water surface is:", "[\nr = \frac{1}{3} \ imes 6 = 2~\ ext{m}\n]", "---", "### Calculating Volume as a Function of Water Height", "The volume ( V ) of water in the cone is given by the cone volume formula:", "[\nV = \frac{1}{3} \pi r^2 h\n]", "Substitute ( r = \frac{1}{3}h ):", "[\nV = \frac{1}{3} \pi \left( \frac{1}{3}h \right)^2 h = \frac{1}{3} \pi \cdot \frac{1}{9} h^2 \cdot h = \frac{1}{27} \pi h^3\n]", "---", "### Differentiating Volume with Respect to Time", "To find the rate at which the water level decreases, we differentiate both sides with respect to time ( t ):", "[\n\frac{dV}{dt} = \frac{d}{dt} \left( \frac{1}{27} \pi h^3 \right) = \frac{1}{27} \pi \cdot 3h^2 \cdot \frac{dh}{dt} = \frac{\pi}{9} h^2 \frac{dh}{dt}\n]", "We are given ( \frac{dV}{dt} = -3~\ ext{m}^3/\ ext{min} ) and ( h = 6~\ ext{m} ). Plug in these values:", "[\n-3 = \frac{\pi}{9} (6)^2 \frac{dh}{dt}\n]", "[\n-3 = \frac{\pi}{9} \ imes 36 \ imes \frac{dh}{dt}\n]", "[\n-3 = 4\pi \frac{dh}{dt}\n]", "Solve for ( \frac{dh}{dt} ):", "[\n\frac{dh}{dt} = \frac{-3}{4\pi} ,\ ext{m/min}\n]", "---", "### Final Answer", "The water level is decreasing at a rate of:", "[\n\boxed{\frac{3}{4\pi}~\ ext{m/min}} \approx 0.238~\ ext{m/min}\n]", "---", "### Conclusion", "Even in a conical tank with non-linear geometry, fluid dynamics follows elegant mathematical principles. By expressing volume in terms of height and differentiating, we efficiently determined that when the water is 6 meters deep, the depth decreases at about 0.238 meters per minute, given a drainage rate of 3 m³/min. This calculation supports real-world applications in tank management and fluid flow analysis.", "---", "Keywords: conical tank drainage rate, water level rate of decrease, conical volume derivation, fluid dynamics calculation, fluid mechanics problems, draining cone tank geometry, rate of change in fluid volume."]
