Mathematics

Mathematics

Problem

A vertical cylindrical tank with radius 2 m is being filled with water at 120 L/min, but a leak allows water to escape at a rate of 2h2h L/min, where hh is the water height in meters. When h=3 mh = 3\text{ m}, what is dhdt\dfrac{dh}{dt}? (Use 1 m3=1000 L1\text{ m}^3 = 1000\text{ L}.) a) −0.1144π m/min-\dfrac{0.114}{4\pi}\text{ m/min} b) 0.1264π m/min\dfrac{0.126}{4\pi}\text{ m/min} c) 0.114π m/min\dfrac{0.114}{\pi}\text{ m/min} d) 0.1144π m/min\dfrac{0.114}{4\pi}\text{ m/min}
Intuitive explanation

This is a related rates problem where you need to connect the volume change of water to the height change. The key insight is to write the net rate of volume change (inflow minus outflow) and relate it to how fast the height is rising using the cylinder's cross-sectional area. Once you have dVdt\frac{dV}{dt} in terms of the leak rate, use V=πr2hV = \pi r^2 h to convert between volume and height rates.

Intuitive explanation

This is a related rates problem where you need to connect the volume change of water to the height change. The key insight is to write the net rate of volume change (inflow minus outflow) and relate it to how fast the height is rising using the cylinder's cross-sectional area. Once you have dVdt\frac{dV}{dt} in terms of the leak rate, use V=πr2hV = \pi r^2 h to convert between volume and height rates.

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