Answer:
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