Question

A cylindrical tank with radius 3 m is being filled with water at a rate of 4 m3/min. How fast is the height of the water increasing (in m/min)

Answer 1

Let R be the radius of the tank, H(t) the height of the water at time t, and V(t) the volume of the water. The quantities V(t), R and H(t) are related by the equation

V(t) = πR²H(t) ___(1)

The rate of increase of the volume is the derivative with respect to time,

dV/dt

and the rate of increase of the height is

dH/dt

We can therefore restate the given and the unknown as follows

Given:

dV/dt = 4m³/min

Unknown:

dH/dt

Now we take derivative of each side of (1) with respect to t:

dV/dt = πR²dH/dt

So

dH/dt = 1/πR²dV/dt

Substituting R = 3 m and dV/dt = 4m³/min we have

dH/dt = 1/π(3)² ⋅ 4 = 4/9π

Answer: the height of the water increasing at a rate of

dH/dt = 4/9π²

≈ 0.14 m/min