Mathematics

Problem

Water drains from a cylindrical tank of fixed radius

2\text{ m}

at a constant rate of

0.50\text{ m}^3\!\big/\!\text{min}

. If

h(t)

is the water depth in meters, what is the instantaneous rate of change

\dfrac{dh}{dt}

(in m/min) when

h=3\text{ m}

? a)

-\dfrac{1}{4\pi}

\dfrac{1}{8\pi}

-\dfrac{1}{8\pi}

-\dfrac{1}{32\pi}

Intuitive explanation

The key is to connect the volume leaving the tank to how fast the water level drops using the cylinder's geometry. Since volume $V = \pi r^2 h$ for a cylinder with fixed radius, the rate of volume change $\frac{dV}{dt}$ and the rate of height change $\frac{dh}{dt}$ are proportional—you just need to differentiate the volume formula with respect to time and solve for $\frac{dh}{dt}$ .

Intuitive explanation

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